appt itude

APTITUDE

Numbers

Introduction:

Natural Numbers:

All positive integers are natural numbers.Ex 1,2,3,4,8,......

There are infinite natural numbers and number 1 is the least natural number.Based on divisibility there would be two types of natural numbers. They are

Prime and composite.

Prime Numbers:

A natural number larger than unity is a prime number if it does not have other divisors except for itself and unity.Note:-Unity i e,1 is not a prime number.

Properties Of Prime Numbers:

->The lowest prime number is 2.->2 is also the only even prime number.->The lowest odd prime number is 3.->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However, if a number on being divided by 6 gives a remainder 1 or 5 need not be prime. ->The remainder of division of the square of a prime number p>=5 divide by 24 is 1.->For prime numbers p>3, p²-1 is divided by 24.->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b² is composite.->The remainder of the division of the square of a prime number p>=5divided by 12 is 1.

Process to Check A Number s Prime or not:

Take the square root of the number.Round of the square root to the next highest integer call this number as Z.Check for divisibility of the number N by all prime numbers below Z. If there is no numbers below the value of Z which divides N then the numberwill be prime.

Example 239 is prime or not?√239 lies between 15 or 16.Hence take the value of Z=16.Prime numbers less than 16 are 2,3,5,7,11 and 13.

239 is not divisible by any of these. Hence we can conclude that 239 is a prime number.

Composite Numbers:

The numbers which are not prime are known as composite numbers.

Co-Primes:

Two numbers a an b are said to be co-primes,if their H.C.F is 1.Example (2,3),(4,5),(7,9),(8,11).....Place value or Local value of a digit in a Number:

place value:

Example 689745132Place value of 2 is (2*1)=2Place value of 3 is (3*10)=30 and so on.Face value:-It is the value of the digit itself at whatever place it may be.

Example 689745132Face value of 2 is 2.Face value of 3 is 3 and so on. Tests of Divisibility:

Divisibility by 2:-A number is divisible by 2,if its unit's digit is any of 0,2,4,6,8.

Example 84932 is divisible by 2,while 65935 is not.Divisibility by 3:-A number is divisible by 3,if the sum of its digits isdivisible by 3.

Example 1.592482 is divisible by 3,since sum of its digits5+9+2+4+8+2=30 which is divisible by 3.

Example 2.864329 is not divisible by 3,since sum of its digits 8+6+4+3+2+9=32 which is not divisible by 3.

Divisibility by 4:-A number is divisible by 4,if the number formed by last two digits is divisible by 4.

Example 1.892648 is divisible by 4,since the number formed by the last two digits is 48 divisible by 4.

Example 2.But 749282 is not divisible by 4,since the number formed by the last two digits is 82 is not divisible by 4.

Divisibility by 5:-A number divisible by 5,if its unit's digit is either 0 or 5.

Example 20820,50345Divisibility by 6:-If the number is divisible by both 2 and 3.

example 35256 is clearly divisible by 2sum of digits =3+5+2+5+21,which is divisible by 3Thus the given number is divisible by 6.

Divisibility by 8:-A number is divisible by 8 if the last 3 digits of the number are divisible by 8.

Divisibility by 11:-If the difference of the sum of the digits in theodd places and the sum of the digitsin the even places is zero or divisible by 11.

Example 4832718(8+7+3+4) - (1+2+8)=11 which is divisible by 11.

Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.

Divisibility by 7,11,13:-The difference of the number of its thousandsand the remainder of its division by 1000 is divisible by 7,11,13.

BASIC FORMULAE:

->(a+b)²=a²+b²+2ab->(a-b)²=a²+b²-2ab->(a+b)²-(a-b)²=4ab->(a+b)²+(a-b)²=2(a²+b²)->a²-b²=(a+b)(a-b)->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)->a³+b³=(a+b)(a²+b²-ab)->a³-b³=(a-b)(a²+b²+ab)->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)->If a+b+c=0 then a³+b³+c³=3a b c

DIVISION ALGORITHM

If we divide a number by another number ,then Dividend = (Divisor * quotient) + Remainder

MULTIPLICATION BY SHORT CUT METHODS

1.Multiplication by distributive law:

a)a*(b+c)=a*b+a*cb)a*(b-c)=a*b-a*c

Examplea)567958*99999=567958*(100000-1)567958*100000-567958*156795800000-56795856795232042

b)978*184+978*816=978*(184+816)978*1000=978000

2.Multiplication of a number by 5n:-Put n zeros to the right of the

multiplicand and divide the number so formed by 2n

Example 975436*625=975436*54=9754360000/16=609647500.

PROGRESSION:

A succession of numbers formed and arranged in a definite order accordingto certain definite rule is called a progression.

1.Arithmetic Progression:-If each term of a progression differs from itspreceding term by a constant.This constant difference is called the common difference of the A.P.The n th term of this A.P is Tn=a(n-1)+d.The sum of n terms of A.P Sn=n/2[2a+(n-1)d].

xImportant Results:

a.1+2+3+4+5......................=n(n+1)/2.b.12+22+32+42+52......................=n(n+1)(2n+1)/6.c.13+23+33+43+53......................=n2(n+1)2/4

2.Geometric Progression:-A progression of numbers in which every term bears a constant ratio with ts preceding term.i.e a,a r,a r2,a r3...............In G.P Tn=a r n-1Sum of n terms Sn=a(1-r n)/1-r

Problems

1.Simplify a.8888+888+88+8 b.11992-7823-456

Solution: a.88888888889872b.11992-7823-456=11992-(7823+456)=11992-8279=3713

2.What could be the maximum value of Q in the following equation?5PQ+3R7+2Q8=1114

Solution: 5 P Q 3 R 7 2 Q 8 11 1 42+P+Q+R=11 Maximum value of Q =11-2=9 (P=0,R=0)

3.Simplify: a.5793405*9999 b.839478*625

Solution: a. 5793405*9999=5793405*(10000-1)

57934050000-5793405=57928256595b. 839478*625=839478*54=8394780000/16=524673750.

4.Evaluate 313*313+287*287

Solution: a²+b²=1/2((a+b)²+(a-b)²)1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )½(360000+676)=180338 5.Which of the following is a prime number? a.241 b.337 c.391

Solution:a.24116>√241.Hence take the value of Z=16.Prime numbers less than 16 are 2,3,5,7,11 and 13.241 is not divisible by any of these. Hence we can conclude that 241 is a prime number. b. 33719>√337.Hence take the value of Z=19.Prime numbers less than 16 are 2,3,5,7,11,13 and 17.337 is not divisible by any of these. Hence we can conclude that 337 is a prime number. c. 391 20>√391.Hence take the value of Z=20.Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.391 is divisible by 17. Hence we can conclude that 391 is not a prime number.

6.Find the unit's digit n the product 2467 153 * 34172?

Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172Now 7 4 gives unit digit 17 152 gives unit digit 17 153 gives 1*7=7.Also 172 gives 1Hence unit's digit in the product =7*1=7.

7.Find the total number of prime factors in 411 *7 5 *112 ? Solution: 411 7 5 112= (2*2) 11 *7 5 *112= 222 *7 5 *112Total number of prime factors=22+5+2=29

8.Which of the following numbers s divisible by 3?a.541326b.5967013 Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.

9.What least value must be assigned to * so that th number 197*5462 is divisible by 9? Solution: Let the missing digit be xSum of digits = (1+9+7+x+5+4+6+2)=34+xFor 34+x to be divisible by 9 , x must be replaced by 2

The digit in place of x must be 2.

10.What least number must be added to 3000 to obtain a number exactlydivisible by 19?

Solution:On dividing 3000 by 19 we get 17 as remainderTherefore number to be added = 19-17=2.

11.Find the smallest number of 6 digits which is exactly divisible by 111?

Solution:Smallest number of 6 digits is 100000On dividing 10000 by 111 we get 100 as remainderNumber to be added =111-100=11.Hence,required number =10011.

12.On dividing 15968 by a certain number the quotient is 89 and the remainderis 37.Find the divisor?

Solution:Divisor = (Dividend-Remainder)/Quotient=(15968-37) / 89 =179.

13.A number when divided by 342 gives a remainder 47.When the same numberis divided by 19 what would be the remainder?

Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.The given number when divided by 19 gives 18 K + 2 as quotient and 9 as remainder. 14.A number being successively divided by 3,5,8 leaves remainders 1,4,7respectively. Find the respective remainders if the order ofdivisors are reversed?

Solution:Let the number be x.

3 x 5 y - 1 8 z - 4 1 - 7 z=8*1+7=15y=5z+4 = 5*15+4 = 79x=3y+1 = 3*79+1=238Now 8 238 5 29 - 6 3 5 - 4 1 - 2Respective remainders are 6,4,2.

15.Find the remainder when 231 is divided by 5?

Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as 4*4*4 gives unit digit 4unit digit of 231 is 8.Now 8 when divided by 5 gives 3 as remainder.231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7?

Solution:The required numbers are 14,21,28,...........,84This is an A.P with a=14,d=7.Let it contain n terms then T =84=a+(n-1)d=14+(n-1)7=7+7n7n=77 =>n=11.

17.Find the sum of all odd numbers up to 100?

Solution:The given numbers are 1,3,5.........99.This is an A.P with a=1,d=2.Let it contain n terms 1+(n-1)2=99=>n=50Then required sum =n/2(first term +last term)=50/2(1+99)=2500.

18.How many terms are there in 2,4,6,8..........,1024?

Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2Let the number of terms be nthen 2*2 n-1=10242n-1 =512=29n-1=9 n=10.

19.2+22+23+24+25..........+28=?

Solution:Given series is a G.P with a=2,r=2 and n=8.Sum Sn=a(1-r n)/1-r=Sn=2(1-28)/1-2.=2*255=510.

20.A positive number which when added to 1000 gives a sum ,which is greater than when it is multiplied by 1000.The positive integer is?a.1 b.3 c.5 d.7

Solution:1000+N>1000Nclearly N=1.

21.The sum of all possible two digit numbers formed from three different one digit natural numbers when divided by the sum of the original three numbers is equal to?a.18 b.22 c.36 d. none

Solution:Let the one digit numbers x,y,zSum of all possible two digit numbers==(10x+y)+(10x+z)+(10y+x)+(10y+z)+(10z+x)+(10z+y)= 22(x+y+z)Therefore sum of all possible two digit numbers when divided by sum of one digit numbers gives 22.

22.The sum of three prime numbers is 100.If one of them exceeds another by36 then one of the numbers is?

a.7 b.29 c.41 d67.

Solution:x+(x+36)+y=1002x+y=64Therefore y must be even prime which is 22x+2=64=>x=31.Third prime number =x+36=31+36=67.

23.A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as remainder .The number is?a.1220 b.1250 c.22030 d.220030.

Solution:Number=(555+445)*(555-445)*2+30=(555+445)*2*110+30=220000+30=220030.

24.The difference between two numbers s 1365.When the larger number is divided by the smaller one the quotient is 6 and the remainder is 15.The smaller number is?a.240 b.270 c.295 d.360

Solution:Let the smaller number be x, then larger number =1365+xTherefore 1365+x=6x+155x=1350 => x=270Required number is 270.

25.In doing a division of a question with zero remainder,a candidate took 12 as divisor instead of 21.The quotient obtained by him was 35.The correct quotient is?a.0 b.12 c.13 d.20

Solution:Dividend=12*35=420.Now dividend =420 and divisor =21.Therefore correct quotient =420/21=20.

H.C.F AND L.C.M

Facts And Formulae:

1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :

The H.C.F of two or more than two numbers

is the greatest

number that divides each of them exactly.

There are two methods :

1.Factorization method: Express each one

of the given numbers as the product of prime factors.

The product of least powers of common prime factors gives HCF.

Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72

Sol: The prime numbers given common numbers are 2,5,7

Therefore HCF is 22 * 5 *72 .

2.Division Method : Divide the larger number by

smaller one. Now divide the divisor by remainder. Repeat the process

of dividing preceding number last obtained till zero is obtained as

number. The last divisor is HCF.

Example: Find HCF of 513, 1134, 1215

Sol:

1134) 1215(1 1134

---------- 81)1134(14 81

----------- 324 324

----------- 0

----------- HCF of this two numbers is 81.

81)513(6 486

-------- 27)81(3 81 ----- 0 ---

HCF of 81 and 513 is 27.

3.Least common multiple[LCM] : The least number which is

divisible by each one of given numbers is LCM.

There are two methods for this:

1.Factorization method :

Resolve each one into product of prime factors.

Then LCM is product of highest powers

of all factors.

2.Common division method.

Problems:

1.The HCF of 2 numbers is 11 and LCM is 693.If one

of numbers is 77.find other.

Sol: Other number = 11 * 693/77=99.

2.Find largest number of 4 digits divisible by 12,15,18,27

Sol: The largest number is 9999.

LCM of 12,15,18,27 is 540.

on dividing 9999 by 540 we get 279 as remainder.

Therefore

number =9999 â€“ 279 =9720.

3.Find least number which when divided by 20,25,35,40

leaves remainders 14,19,29,34.

Sol:

20â€“14=6 25-19=6 35-29=6 40-34=6

Therefore number =LCM of (20,25,35,40) - 6=1394

4.252 can be expressed as prime as :

2 252 2 126

3 63 3 21

7

prime factor is 2 *2 * 3 * 3 *7

5.1095/1168 when expressed in simple form is

1095)1168(1 1095 ------

73)1095(15 73

--------- 365 365

--------- 0

----------So, HCF is 73

Therefore

1095/1168 = 1095/73/1168/73= 15/16

6.GCD of 1.08,0.36,0.9 is

Sol:

HCF of 108,36,90 36)90(2 72 ----

18)36(2 36 ---- 0 ----

HCF is 18.

HCF of 18 and 108 is 18 18)108(6 108

------- 0

--------Therefore HCF =0.18

7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.

Sol:

Let the numbers be x.

Three numbers are x,2x,3x

Therefore

HCF is 2x)3x(1 2x

----- x)2x(2 2x

-------- 0

-------------

HCF is x so, x is 12

Therefore numbers are 12,24,36.

8.The sum of two numbers is 216 and HCF is 27.

Sol:

Let numbers are 27a + 27 b =216

a + b =216/27=8

Co-primes of 8 are (1,7) and (3,5)

numbers=(27 * 1 ), (27 * 7) =27,89

9.LCM of two numbers is 48..The numbers are in ratio 2:3.

The sum of numbers is

Sol:

Let the number be x.

Numbers are 2x,3x

LCM of 2x,3x is 6x

Therefore

6x=48

x=8.

Numbers are 16 and 24

Sum=16 +24=40.

10.HCF and LCM of two numbers are 84 and 21.If ratio of

two numbers is 1:4.Then largest of two numbers is

Sol:

Let the numbers be x,4x

Then x * 4x = 84 * 21

x2 =84 * 21 /4

x = 21

Largest number is 4 * 21.

11.HCF of two numbers is 23,and other factors of LCM are

13,14.Largest number is

Sol:

23 * 14 is Largest number.

12.The maximum number of students among them 1001 pens

and 910 pencils can be distributed in such a way that

each student gets same number of pens and pencils is ?

Sol:

HCF of 1001 and 910

910)1001(1 910

------------ 91)910(10 910

-------- 0

--------- Therefore HCF=91

13.The least number which should be added to 2497 so that sum is

divisible by 5,6,4,3 ?

Sol: LCM of 5,6,4,3 is 60.

On dividing 2497 by 60 we get 37 as remainder.

Therefore number to added is 60 â€“ 37 =23.

Answer is 23.

14.The least number which is a perfect square and is divisible by

each of numbers 16,20,24 is ?

Sol: LCM of 16,20,24 is 240.

2 * 2*2*2*3*5=240

To make it a perfect square multiply by 3 * 5

Therefore 240 * 3 * 5=3600

Answer is 3600.

DECIMAL FRACTIONS

1.Decimal fractions: Fractionin which denominations are powers of 10 are decimalfractions.1 /10 = 0.1, 1 / 100 = 0.01

2.Convertion of Decimal into fraction:-eg: 0.25 = 25/100 = 1/43.i) If numerator and denominator contain same number of decimal places, then we remove decimal sign. Thus, 1.84/2.99 =184/299

PROBLEMS:

1.0.75 =75/100 =3/4

2.Find porducts= 6.3204*100= 632.04

3.2.61*1.3=261*13=3393some of decimal places 2 +1 =3= 3.393

4.If 1/3.718 =0.2689,then find value of 1/0.0003718 ?Sol: 10000/3.718 =10000*1/3.718=10000*0.2689= 2689

5.Find fractions :i) 0.37 = 37/99ii)3.142857 =3+0.142857=3 +142857/999999= 3 142857/ 999999 iii) 0.17=17-1/90 =16/90=8/45iv)0.1254 =1254 -12/9900 =1242/9900=69/550

6.Fraction 101 27/100000Sol: 101+27/100000=101+0.00027=101.00027

7.If 47.2506 =4A + 7/B +2C + 5/D + 6E then 40+7+0.2+0.05+0.0006Sol: compairing terms 4A= 40 => A=107/B = 7 => B=12C= 0.2=> C=0.15/D= 0.05=>D=5/0.05 =>5*100/5 =1006E= 0.0006=> E= 0.00015A + 3B+6C+ D+ 3E = 5*10+ 3*1+ 6*0.1 + 100+ 3*0.0001=50+3+0.6+100+0.0003=153.6003

8.4.036 divided by 0.04Sol: 4.036/0.04 =4036/4 =100.99.[ 0.05/0.25 + 0.25/ 0.05]3Sol: =>[5/25 + 25/5] = [1/5+ 5]3=26/53=5.23= 140.603

10.The least among the following :-a. 0.2 b.1/0.2 c. 0.2 d. 0.2210/2 =5 0.2222 0.04 0.04 < 0.2 < 0.22 --------<5Since 0.04 is least (0.2)2 is least.

11.Let F= 0.84181Sol: when F is written as a fraction in lowest terms, denominator exceeds numerator by 84181 -841 /99000 = 83340/99000 =463/550Required distence = (550 â€“ 463) = 87

12.2 .75 + 3.78Sol: [-2+0.75]+[-3+0.78]=-5+[0.75+0.78]= -5+1.53=-5+1+0.53= -4+0.53

= 4.53

13.the sum of first 20 terms of series is 1/5*6 +1/6*7+1/7*8-------------Sol: [1/5 -1/6]+[1/6-1/7]+[1/7-1/8]+------------------------= [1/5-1/25]=4/25=0.16

14.13 +23+ ------------+93 =2025Sol: value of (0.11) 3+ (0.22) 3+---------(0.99)3 =>(0.11) [1+2+--------+9]=0.001331*2025=2.695275

15.(0.96)3 â€“ (0.1)3/ (0.96)2 +0.096 +(0.1)2Sol: formula => a3 -b3/a2 +ab +b2 =a -b(0.96-0.1)=0.86

16.3.6*0.48*2.50 / 0.12*0.09*0.5Sol: 36*48*250/12*9*5=800

17.find x/y = 0.04/1.5= 4/150 =2/75find y-x/y+x (1- x/y) / (1+ x/y)1 - 2/75 /1 +2/75 =73/77

18.0.3467+0.1333Sol: 3467 -34/9900 + 1333-13/9900= 3433 +1320/9900= 4753/9900= 4801 -48/9900 =0.4301

Simplifications

Introduction:

'BODMAS' rule: This rule depicts the correct sequence in whichthe operations are to be executed, so as to find out the value ofa given expression.

Here B stands for Bracket, O for Of, D for Division, M for Multiplication, A for Addition and S for Subtraction.

First of all the brackets must be removed, strictly in the order () , {} , [].

After removing the brackets, we want use the following operations:1.Of 2. Division 3. Multiplication 4. Addition 5. Subtraction

Modulus of a real number:Modulus of a real number is a defined as |a| = a, if a>0 or -a, if a < 0;

Problems:

1.(5004 /139) – 6= ?

Sol: Expression = 5004/ 139 – 6 = 36 – 6 = 30;

2.What mathematical operations should come at the place of ? in theequation : (2 ? 6 – 12 / 4 + 2 = 11) ?

Sol: 2 ? 6 = 11 + 12 / 4 – 2= 11 + 3 – 2= 122 * 6 = 12

3.( 8 / 88) * 8888088 = ?

Sol : (1/11) * 8888088 = 808008

4.How many 1/8's are there in 371/2 ?

Sol: (371/2) /(1/8)= (75/2) /(1/8) = 300

5.Find the values of 1/2*3 +1/3*4 +1/4*5+ .................+1/9*10 ?

Sol: 1/2*3 +1/3*4+1/4*5+ ..................+1/9*10= [½ -1/3] +[ 1/3 – ¼] + [¼- 1/5] +...............+[1/9-1/10]= [ ½ – 1/10] = 4/15 = 2/5

6.The value of 999 of 995/999* 999 is:

Sol: [1000- 4/1000]*999 = 999000-4

= 998996

7.Along a yard 225m long, 26 trees are planted at equal distance, onetree being at each end of the yard. what is the distance between two consecutive trees ?

Sol: 26 trees have 25 gaps between them.Hence , required distance = 225/ 25 m= 9m

8.In a garden , there are 10 rows and 12 columns of mango trees. thedistance between the two trees is 2 m and a distance of one meter is left from all sides of the boundary of the length of the garden is :

Sol: Each row contains 12 plants.leaving 2 corner plants, 10 plants in between have 10 * 2 meters and 1 meter on each side is left. length = (20 + 2) m = 22m

9.Eight people are planning to share equally the cost of a rental car,if one person with draws from the arrangement and the others share equally the entire cost of the car, then the share of each of theremaining persons increased by?

Sol: Original share of one person = 1/8 new share of one person = 1/7 increase = 1/7 – 1/8 = 1/56 required fractions = (1/56)/(1/8) = 1/7

10.A piece of cloth cost Rs 35. if the length of the piece would have been 4m longer and each meter cost Re 1 less , the cost would have remained unchanged. how long is the piece?

Sol: Left the length of the piece be x m.then, cost of 1m of piece = Rs [35 / x]35/ x – 35 /x+4 = 1x + 4 – x = x(x+ 4)/35x2 + 4x – 140 = 0 x= 10

11.A man divides Rs 8600 among 5sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five as much as each nephew. how much does eachdaughter receive ?

Sol:Let the share of each nephew be Rs x.then, share of each daughter Rs 4x.share of each son = 5x Rsso, 5 *5x+ 4 * 4x + 2x =86002x + 16x + 25x= 860043x = 8600x = 200share of each daughter = 4 * 200 = Rs 800

12.A man spends 2/5 of his salary on house rent, 3/10 of his salary on food, and 1/8 of his salary on conveyance. if he has Rs 1400 left with him, find his expenditure on food and conveyance?

Sol: Part of the salary left = 1-[2/5 +3/10+1/9]= 1- 33/40=7/40Let the monthly salary be rs xthen, 7/40 of x = 1400x= [1400*40]/7x= 8000Expenditure on food = 3/10*8000 =Rs 2400Expenditure on conveyance= 1/8*8000 =Rs 1000

Averages

Formula:

1.Average=Sum of quantities/Number of quantities.

2.Suppose a man covers a certain distance at x kmph and an equal distance at y kmph ,then the average speed during the whole journey is (2xy/x+y) kmph.

Examples: 1.Find the average of all these numbers.142,147,153,165,157.

Solution:142 147 153 165 157Here consider the least number i.e, 142 comparing with others,142 147 153 165 157+5 +11 +23 +15Now add 5+11+23+15 = 52/5 = 10.8Now add 10.8 to 142 we get 152.8(Average of all these numbers).Answer is 152.8

2.Find the average of all these numbers.4,10,16,22,28

Solution:4,10,16,22,28As the difference of number is 6Then the average of these numbers is central one i.e, 16.Answer is 16.

3.Find the average of all these numbers.4,10,16,22,28,34.

Solution:Here also difference is 6.Then middle numbers 16,22 take average of these

two numbers 16+22/2=19Therefore the average of these numbers is 19.Answer is 19.

4.The average marks of a marks of a student in 4 Examination is 40.If he got 80 marks in 5th Exam then what is his new average.

Solution:4*40+80=240Then average means 240/5=48.Answer is 48.

5.In a group the average income of 6 men is 500 and thatof 5 women is 280, then what is average income of the group.

Solution:6*500+5*280=4400then average is 4400/11=400.Another Method: here consider for 6 men6 men â€“ each 500.so 5th women is 280.then 500-280=220.then 220*6/11=120.therefore 120+280=400.Answer is 400.

6.The average weight of a class of 30 students is 40 kgs if the teacher weight is included then average increases by 2 kgs thenfind the weight of the teacher?

Solution:30 students average weight is 40 kgs.So,when teacher weight is added it increases by 2 kgsso total 31 persons ,therefore 31*2=62.Now add the average weight of all student to it we get teachers weight i.e, 62+40=102 kgs.Answer is 102 kgs.

7.The average age of Mr and Mrs Sharma 4 years ago is 28 years .If the present average age of Mr and Mrs Sharma and their son is 22 years. What is the age of their son.

Solution:4 years ago their average age is 28 years.So their present average age is 32 years.32 years for Mr and Mrs Sharma then 32*2=64 years.Then present age including their son is 22 years.So 22*3 =66 years.Therefore son age will be 66-64 = 2 years.Answer is 2 years.

8.The average price of 10 books is increased by 17 Rupees when one of them whose value is Rs.400 is replaced by a new book. What is the price of new book?

Solution:

10 books Average increases by 17 Rupeesso 10*17= 170.so the new book cost is more and by adding its cost averageincrease,therefore the cost of new book is 400+170=570Rs.Answer is 570 Rs.

9.The average marks of girls in a class is 62.5. The average marksof 4 girls among them is 60.The average marks of remaining girls is 63,then what is the number of girls in the class?

Solution:Total number of girls be x+4.Average marks of 4 girls is 60.therefore 62.5-60=2.5then 4*2.5 =10.the average of remaining girls is 63here 0.5 difference therefore 0.5*x=10(since we got from 4 girls)(this is taken becoz both should be equal)x=10/0.5x=20.This clear says that remaining are 20 girlstherefore total is x+4=20+4=24 girlsAnswer is 24 girls. 10.Find the average of first 50 natural numbers.

Solution:Sum of the Natural Numbers is n(n+1)/2therefore for 50 Natural numbers 50*51/2=775.the average is 775/50=15.5Answer is 15.5 .

11.The average of the first nine prime number is?

Solution:Prime numbers are 2,3,5,7,11,13,17,19,23therefore 2+3+5+7+11+13+17+19+23=100then the average 100/9= 11 1/9.Answer is 11 1/9.

12.The average of 2,7,6 and x is 5 and the average of and the average of 18,1,6,x and y is 10 .what is the value of y?

Solution:2+7+6+x/4=5=>15+x=20=>x=5.18+1+6+x+y/5=10=>25+5+y=50=>y=20.

13.The average of a non-zero number and its square is 5 times the number.The number is

Solution:The number be x

then x+x2/2=5x=>x2-9x=0=>x(x-9)=0therefore x=0 or x=9.The number is 9.

14.Nine persons went to a hotel for taking their meals . Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 then the average expenditure of all the nine. What was the total money spent by them?

Solution:The average expenditure be x.then 8*12+(x+8)=9x=>96+x+8=9x.=>8x=104=>x=13Total money spent =9x=>9*13=117Answer is Rs.117

15.The average weight of A.B.C is 45 Kgs.If the average weight ofA and B be 40 Kgs and that of Band C be 43 Kgs. Find the weight of B?

Solution:The weight of A,B,Care 45*3=135 Kgs.The weight of A,B are 40*2=80 Kgs.The weight of B,C are 43*2=86 Kgs.To get the Weight of B.(A+B)+(B+C)-(A+B+C)=80+86-135B=31 kgs.Answer is 31 Kgs.

16.The sum of three consecutive odd number is 48 more than the average of these number .What is the first of these numbers?

Solution:let the three consecutive odd numbers are x, x+2, x+4.By adding them we get x+x+2+x+4=3x+6.Then 3x+6-(3x+6)/3=38(given)=>2(3x+6)=38*3.=>6x+12=114=>6x=102=>x=17.Answer is 17.

17.A family consists of grandparents,parents and three grandchildren.The average age of the grandparents is 67 years,that of parents is 35years and that of the grand children is 6 years . What is the averageage of the family?

Solution:grandparents age is 67*2=134parents age is 35*2=70grandchildren age is 6*3=18therefore age of family is 134+70+18=222average is 222/7=31 5/7 years.

Answer is 31 5/7 years. 18.A library has an average of 510 visitors on Sundays and 240 onother days .The average number of visitors per day in a month 30 days beginning with a Sunday is?

Solution:Here specified that month starts with Sunday so, in a month there are 5 Sundays.Therefore remaining days will be 25 days.510*5+240*25=2550+6000=8550 visitors.The average visitors are 8550/30=285.Answer is 285.

19.The average age of a class of 39 students is 15 years .If the age of the teacher be included ,then average increases by 3 months. Find the age of the teacher.

Solution: Total age for 39 persons is 39*15=585 years.Now 40 persons is 40* 61/4=610 years(since 15 years 3 months=15 3/12=61/4)Age of the teacher =610-585 years=>25 years.Answer is 25 years.

20.The average weight of a 10 oarsmen in a boat is increasesby 1.8 Kgs .When one of the crew ,who weighs 53 Kgs is replaced by new man. Find the weight of the new man.

Solution: Weight of 10 oars men is increases by 1.8 Kgsso, 10*1.8=18 Kgstherefore 53+18=71 Kgs will be the weight of the man.Answer is 71 Kgs.

21.A bats man makes a score of 87 runs in the 17th inningand thus increases his average by 3. Find the average after 17th inning.

Solution: Average after 17 th inning =xthen for 16th inning is x-3.Therefore 16(x-3)+87 =17x=>x=87-48=>x=39.Answer is 39.

22.The average age of a class is 15.8 years .The average ageof boys in the class is 16.4 years while that of the girlsis 15.4 years .What is the ratio of boys to girls in the class.

Solution: Ratio be k:1 then k*16.4 + 1*15.4 = (k+1)*15.8=>(16.4-15.8)k=15.8-15.4=>k=0.4/0.6=>k=2/3therefore 2/3:1=>2:3

Answer is 2:3

23.In a cricket eleven ,the average of eleven players is28 years .Out of these ,the average ages of three groups of players each are 25 years,28 years, and 30 years respectively. If in these groups ,the captain and theyoungest player are not included and the captain iseleven years older than the youngest players , what is the age of the captain?

Solution: let the age of youngest player be xthen ,age of the captain =(x+11)therefore 3*25 + 3*28 + 3*30 + x + x+11=11*28=>75+84+90+2x+11=308=>2x=48=>x=24.Therefore age of the captain =(x+11)= 24+11= 35 years.Answer is 35 years. 24.The average age of the boys in the class is twicethe number of girls in the class .If the ratio ofboys and girls in the class of 36 be 5:1, what is the total of the age (in years) of the boys in the class?

Solution: Number of boys=36*5/6=30Number of girls =6Average age of boys =2*6=12 years Total age of the boys=30*12=360 yearsAnswer is 360 years.

25.Five years ago, the average age of P and Q was15 years ,average age of P,Q, and R today is20 years,how old will R be after 10 years?

Solution: Age of P and Q are 15*2=30 yearsPresent age of P and Q is 30+5*2=40 years.Age of P Q and R is 20*3= 60 years.R ,present age is 60-40=20 yearsAfter 10 years =20+10=30 years.Answer is 30 years.

26.The average weight of 3 men A,B and C is 84 Kgs. Another man D joins the group and the average nowbecomes 80 Kgs.If another man E whose weight is3 Kgs more than that of D ,replaces A then theaverage weight B,C,D and E becomes 79 Kgs. The weight of A is.

Solution:Total weight of A, B and C is 84 * 3 =252 Kgs.Total weight of A,B,C and Dis 80*4=320 KgsTherefore D=320-252=68 Kgs.E weight (68+3)=71 kgsTotal weight of B,C,D and E = 79*4=316 Kgs(A+B+C+D)-(B+C+D+E)=320-316 =4KgsA-E=4KgsA-71=4 kgs

A=75 KgsAnswer is 75 kgs

27.A team of 8 persons joins in a shooting competition.The best marksman scored 85 points.If he had scored92 points ,the average score for the team wouldhave been 84.The team scored was.

Solution: Here consider the total score be x.therefore x+92-85/8=84=>x+7=672=>x=665.Answer is 665

28.A man whose bowling average is 12.4,takes 5 wicketsfor 26 runs and there by decrease his average by 0.4.The number of wickets,taken by him before his last match is:

Solution: Number of wickets taken before last match be x.therefore 12.4x26/x+5=12(since average decrease by 0.4 therefore 12.4-0.4=12)=>12.4x+2612x+60=>0.4x=34=>x=340/4=>x=85.Answer is 85.

29.The mean temperature of Monday to Wednesday was 37 degreesand of Tuesday to Thursday was 34 degrees .If thetemperature on Thursday was 4/5th that of Monday. The temperature on Thursday was:

Solution:The total temperature recorded on Monday,Wednesday was 37*3=111.The total temperature recorded on Tuesday,Wednesday,Thursday was 34*3=102.and also given that Th=4/5M=>M=5/4Th(M+T+W)-(T+W+Th)=111-102=9M-Th=95/4Th-Th=9Th(1/4)=9=>Th=36 degrees.

30. 16 children are to be divided into two groups A and Bof 10 and 6 children. The average percent marks obtainedby the children of group A is 75 and the average percentmarks of all the 16 children is 76. What is the averagepercent marks of children of groups B?

Solution: Here given average of group A and whole groups .So,(76*16)-(75*10)/6=>1216-750/6=>466/6=233/3=77 2/3Answer is 77 2/3.

31.Of the three numbers the first is twice the second andthe second is twice the third .The average of the reciprocalof the numbers is 7/72,the number are.

Solution:Let the third number be xLet the second number be 2x.Let the first number be 4x.Therefore average of the reciprocal means 1/x+1/2x+1/4x=(7/72*3)7/4x=7/24=>4x=24x=6.Therefore First number is 4*6=24.Second number is 2*6=12Third number is 1*6=6Answer is 24,12,6.

32.The average of 5 numbers is 7.When 3 new numbersare added the average of the eight numbers is 8.5.The average of the three new number is:

Solution: Sum of three new numbers=(8*8.5-5*7)=33Their average =33/3=11.Answer is 11.

33.The average temperature of the town in the first four days of a month was 58 degrees. The average for the second ,third,fourth and fifth days was60 degree .If the temperature of the first andfifth days were in the ratio 7:8 then what isthe temperature on the fifth day?

Solution : Sum of temperature on 1st 2nd 3rd and 4th days =58*4=232 degrees.Sum of temperature on 2nd 3rd 4thand 5th days =60*4=240 degreesTherefore 5th day temperature is 240-232=8 degrees.The ratio given for 1st and 5th days be 7x and 8x degrees then 8x-7x=8=>x=8.therefore temperature on the 5th day =8x=8*8=64 degrees.

Problems on Ages

Simple problems:

1.The present age of a father is 3 years more than three times the age of his son.Three years hence,fatherâ€™s age will be 10 years more than twice the age of the son.Find the present age of the father.

Solution: Let the present age be 'x' years. Then father's present age is 3x+3 years. Three years hence (3x+3)+3=2(x+3)+10 x=10 Hence father's present age = 3x+3 = 33 years.

2. One year ago the ratio of Ramu & Somu age was 6:7respectively. Four years hence their ratio would become 7:8. How old is Somu.

Solution: Let us assume Ramu &Somu ages are x &y respectively. One year ago their ratio was 6:7 i.e x-1 / y-1 = 7x-6y=1 Four years hence their ratios,would become 7:8 i.e x-4 / y-4 = 7 / 8 8x-7y=-4 From the above two equations we get y= 36 years. i.e Somu present age is 36 years.

3. The total age of A &B is 12 years more than the total age of B&C. C is how many year younger than A.

Solution: From the given data A+B = 12+(B+C) A+B-(B+C) = 12 A-C=12 years. C is 12 years younger than A

4. The ratio of the present age of P & Q is 6:7. If Q is 4 years old than P. what will be the ratio of the ages of P & Q after 4 years.

Solution: The present age of P & Q is 6:7 i.e P / Q = 6 / 7 Q is 4 years old than P i.e Q = P+4. P/ P+4 = 6/7 7P-6P = 24, P = 24 , Q = P+4 =24+4 = 28 After 4 years the ratio of P &Q is P+4:Q+4 24+4 : 28+4 = 28:32 = 7:8

5. The ratio of the age of a man & his wife is 4:3.After 4 years this ratio will be 9:7. If the time of marriage the ratio was 5:3, then how many years ago were they married.

Solution: The age of a man is 4x . The age of his wife is 3x. After 4 years their ratio's will be 9:7 i.e 4x+4 / 3x+4 = 9 / 7 28x-27x=36-28 x = 8.

Age of a man is 4x = 4*8 = 32 years. Age of his wife is 3x = 3*8 = 24 years. Let us assume 'y' years ago they were married , the ratio was 5:3 ,i.e 32-y / 24-y = 5/ 3 y=12 years i.e 12 years ago they were married

6. Sneh's age is 1/6th of her father's age.Sneh's father's age will be twice the age of Vimal's age after 10 years. If Vimalâ€™s eight birthday was celebrated two years before,then what is Sneh's present age.

a) 6 2/3 years b) 24 years c) 30 years d) None of the above

Solution: Assume Snehâ€™s age is 'x' years. Assume her fathers age is 'y' years. Snehâ€™s age is 1/6 of her fathers age i.e x = y /6. Fatherâ€™s age will be twice of Vimal's age after 10 years. i.e y+10 = 2( V+10)( where 'V' is the Vimal's age) Vimal's eight birthday was celebrated two years before, Then the Vimal's present age is 10 years. Y+10 = 2(10+10) Y=30 years. Sneh's present age x = y/6 x = 30/6 = 5 years. Sneh's present age is 5 years.

7.The sum of the ages of the 5 children's born at the intervals of 3 years each is 50 years what is the age of the youngest child.

a) 4 years b) 8 years c) 10 years d)None of the above

Solution: Let the age of the children's be x ,x+3, x+6, x+9, x+12. x+(x+3)+(x+6)+(x+9)+(x+12) = 50 5x+30 = 50 5x = 20 x=4. Age of the youngest child is x = 4 years.

8. If 6 years are subtracted from the present age of Gagan and the remainder is divided by 18,then the present age of his grandson Anup is obtained. If Anup is 2 years younger to Madan whose age is 5 years,then what is Gagan's present age.

a) 48 years b)60 years c)84 years d)65 years

Solution: Let us assume Gagan present age is 'x' years.

Anup age = 5-2 = 3 years. (x-6) / 18 = 3 x-6 = 54 x=60 years

9.My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my father was 26 years of age when i was born. If my sister was 4 years of age when my brother was born,then what was the age my father and mother respectively when my brother was born.

a) 32 yrs, 23yrs b)32 yrs, 29yrs c)35 yrs,29yrs d)35yrs,33 yrs

Solution: My brother was born 3 years before I was born & 4 years after my sister was born. Father's age when brother was born = 28+4 = 32 years. Mother's age when brother was born = 26-3 = 23 years.

SURDS AND INDICES

Simple problems:

1. Laws of Indices: (i) am * an = a(m+n) (ii) am / an = a(m-n) (iii) (am)n = a(m*n) (iv) (ab)n = an * bn (v) (a/b)n = an / bn (vi) a0 = 1

2.Surds :Let 'a' be a rational number & 'n' be a positive integer such that a1/n = nth root a is irrational.Then nth root a is called 'a' surd of 'n'.

Problems:- (1) (i) (27)2/3 = (33)2/3 = 32 = 9. (ii) (1024)-4/5 = (45)-4/5 = (4)-4= 1/(4)4 = 1/256. (iii)(8/125)-4/3 =((2/5)3)-4/3 = (2/5)-4 = (5/2)4 = 625/16

(2) If 2(x-1)+ 2(x+1) = 1280 then find the value of x .

Solution: 2x/2+2x.2 = 1280 2x(1+22) = 2*1280 2x = 2560/5 2x = 512 => 2x = 29 x = 9

(3) Find the value of [5[81/3+271/3]3]1/4

Solution: [5[(23)1/3+(33)1/3]3]1/4 [5[2+3]3]1/4 [54]1/4 => 5.

(4) If (1/5)3y= 0.008 then find the value of (0.25)y

Solution: (1/5)3y = 0.008 (1/5)3y =[0.2]3

(1/5)3y =(1/5)3 3y= 3 => y=1. (0.25)y = (0.25)1 => 0.25 = 25/100 = 1/4

(5) Find the value of (243)n/5 * 32n+1 / 9n * 3 n-1 Solution: (35)n/5 * 32n +1 / (32)n * 3n-1 33n+1 / 33n-1 3 33n+1 * 3-3n+1 => 32 =>9.

(6) Find the value of (21/4-1)( 23/4 +21/2+21/4+1) Solution: Let us say 21/4 = x (x-1)(x3+x2+x+1) (x-1)(x2(x+1)+(x+1)) (x-1) (x2+1) (x+1) [(x-1)(x+1) = (x2-1)] (x2+1) (x2-1) => (x4-1) ((21/4))4 - 1) = > (2-1) = > 1.

(7) If x= ya , y = zb , z = xc then find the value of abc.

Solution: z= xc z= (ya)c [ x= ya ] z= (y)ac z= (zb)ac [y= zb] z= zabc abc = 1

(8)Simplify (xa/xb)a2+ab+b2*(xb/xc)b2+bc+c2*(xc/xa)c2+ca+a2

Solution:[xa-b]a2+ab+b2 * [xb-c]b2+bc+c2 * [xc-a]c2+ca+a2 [ (a-b)(a2+ab+b2) = a3-b3]

from the above formula => xa3-b3 xb3-c3 xc3-a3 => xa3-b3+b3-c3+c3-a3 => x0 = 1

(9) (1000)7 /1018 = ?

(a) 10 (b) 100 (c ) 1000 (d) 10000 Solution: (1000)7 / 1018 (103)7 / (10)18 = > (10)21 / (10)18 => (10)21-18 => (10)3 => 1000 Ans :( c )

(10) The value of (8-25-8-26) is (a) 7* 8-25 (b) 7*8-26 (c ) 8* 8-26 (d) None

Solution: ( 8-25 - 8-26 ) => 8-26 (8-1 ) => 7* 8-26 Ans: (b)

(11) 1 / (1+ an-m ) +1/ (1+am-n) = ? (a) 0 (b) 1/2 (c ) 1 (d) an+m Solution: 1/ (1+ an/am) + 1/ ( 1+ am/an) => am / (am+ an ) + an /(am +an ) => (am +an ) /(am + an) => 1 Ans: ( c)

(12) 1/(1+xb-a+xc-a)+1/(1+xa-b+xc-b)+1/(1+xb-c+xa-c)=? (a) 0 (b) 1 ( c ) xa-b-c (d) None of the above

Solution: 1/ (1+xb/xa+xc/xa) + 1/(1+xa/xb +xc/xb) + 1/(1+xb/xc +xa/xc) => xa /(xa +xb+xc) + xb/(xa +xb+xc) +xc/(xa +xb+xc) =>(xa +xb+xc) /(xa +xb+xc) =>1 Ans: (b)

(13) If x=3+2 âˆš2 then the value of (âˆšx â€“ 1/ âˆšx) is [ âˆš=root]

(a) 1 (b) 2 (c ) 2âˆš2 ( d) 3âˆš3

Solution: (âˆšx-1/âˆšx)2 = x+ 1/x-2 => 3+2âˆš2 + (1/3+2âˆš2 )-2 => 3+2âˆš2 + 3-2âˆš2 -2 => 6-2 = 4 (âˆšx-1/âˆšx)2 = 4 =>(âˆšx-1/âˆšx)2 = 22 (âˆšx-1/âˆšx) = 2.

Ans : (b)

(14) (xb/xc)b+c-a (xc/xa)c+a-b (xa/xb)b+a-c = ?

(a) xabc (b) 1 ( c) xab+bc+ca (d) xa+b+c

Solution: [xb-c]b+c-a [xc-a]c+a-b [xa-b]a+b-c =>x(b-c)(b+c-a) x(c-a)(c+a-b) x(a-b)(a+b-c) =>x(b2-c2-ab-ac) x(c2-a2-bc-ab) x(a2-b2-ac-bc) =>x(b2-c2-ab-ac+c2-a2-bc-ab+a2-b2-ac-bc) => x0 =>1 Ans: (b) (15) If 3x-y = 27 and 3x+y = 243 then x is equal to (a) 0 (b) 2 (c ) 4 (d) 6

Solution: 3x-y = 27 => 3x-y = 33 x-y= 3 3x+y = 243 => 3x+y = 35 x+y = 5 From above two equations x = 4 , y=1 Ans: (c )

(16) If ax = by = cz and b2 = ac then â€˜yâ€™equals (a)xz/x+z (b)xz/2(x-z) (c)xz/2(z-x) (d)2xz/x+z

Solution: Let us say ax = by = cz = k ax =k => [ax]1/x = k1/x => a = k1/x Simillarly b = k1/y c = k1/z b2 = ac [k1/y]2=k1/xk1/z =>k2/y = k1/x+1/z => 2/y = 1/x+1/y =>y= 2xz/x+z Ans: (d)

(17) ax = b,by = c ,cz = a then the value of xyz is is (a) 0 (b) 1 (c ) 1/abc (d) abc Solution: ax = b

(cz)x = b [cz = a] by)xz = b [by = c] =>xyz =1 Ans: (b)

(18) If 2x = 4y =8z and (1/2x +1/4y +1/6z) =24/7 then the value of 'z' is (a) 7/16 (b) 7 / 32 (c ) 7/48 (d) 7/64 Solution: 2x = 4y=8z 2x = 22y = 23z x= 2y = 3z Multiply above equation with â€˜ 2â€™ 2x = 4y= 6z (1/2x+1/4y+1/6z) = 24/7 =>(1/6z+1/6z+1/6z) = 24/7 => 3 / 6z = 24/7 => z= 7/48 Ans: ( c)

PERCENTAGES

EXAMPLE PROBLEMS:

EXAPLE PROBLEMS:

1 . Express the following as a fraction.a) 56%

SOLUTION: 56/100=14/25

b) 4% SOLUTION:

4/100=1/25c) 0.6%

SOLUTION: 0.6/100=6/1000=3/500

d) 0.08% SOLUTION:

0.08/100=8/10000=1/1250

2.Express the following as decimalsa) 6%

SOLUTION: 6% = 6/100=0.06

b) 0.04% SOLUTION:

0.04% = 0.04/100=0.0004

3 . Express the following as rate percent.i).23/36

SOLUTION: = (23/36*100) %

= 63 8/9%

ii).6 Â¾ SOLUTION:

6 Â¾ =27/4 (27/4 *100) % =675 %

4.Evaluate the following: 28% of 450 + 45% of 280 ? SOLUTION:

=(28/100) *450 + (45/100) *280 = 28 * 45 / 5

= 252

5.2 is what percent of 50? SOLUTION:

Formula : (IS / OF ) *100 % = 2/50 *100

= 4%

6.Â½ is what percent of 1/3? SOLUTION:

=( Â½) / (1/3) *100 % = 3/2 *100 % = 150 %

7.What percent of 2 Metric tonnes is 40 Quintals? SOLUTION:

1 metric tonne =10 Quintals So required percentage = (40/(2*10)) *100 % = 200%

8.Find the missing figure . i) ? % of 25 = 2.125 SOLUTION :

Let x% of 25 = 2.125. then (x/100) *25 =2.125

x = 2.125 * 4 = 8.5

ii) 9% of ? =6.3 SOLUTION:

Let 9 % of x = 6.3. Then 9/100 of x= 6.3

so x = 6.3 *100/7 = 70.

9.Which is the greatest in 16 2/3 %, 2/15,0.17? SOLUTION:

16 2/3 % = 50/3 % =50/3 * 1/100

=1/6 = 0.166

2 / 15 =0.133 So 0.17 is greatest number in the given

series.

10.If the sales tax be reduced from 3 Â½ % to 3 1/3 % ,then what difference

does it make to a person who purchases an article with marked price of RS 8400? SOLUTION:

Required difference = 3 Â½ % of 8400 â€“ 3 1/3 % of 8400 =(7/2-10/3)% of 8400

=1/6 % of 8400 = 1/6* 1/100* 8400

= Rs 14. 11. A rejects 0.08% of the meters as defective .How many will he examine

to reject 2? SOLUTION:

Let the number of meters to be examined be x. Then 0.08% of x=2. 0.08/100*x= 2 x= 2 * 100/0.08

=2 * 100 * 100/8 = 2500

12.65 % of a number is 21 less than 4/5 of that number. What is the number?

SOLUTION: Let the number be x. 4/5 x- (65% of x) = 21 4/5x â€“ 65/100 x=21

15x=2100 x=140

13. Difference of two numbers is 1660.If 7.5 % of one number is 12.5% of the other number. Find two numbers?

SOLUTION: Let the two numbers be x and y.

7.5% of x=12.5% of y So 75x=125 y 3x=5y x=5/3y. Now x-y=1660

5/3y-y=1660 2/3y=1660 y=2490

So x= 2490+1660 =4150.

So the numbers are 4150 , 1660.

14. In expressing a length 81.472 KM as nearly as possible with 3 significant digits ,Find the % error?

SOLUTION: Error= 81.5-81.472=0.028

So the required percentage = 0.028/81.472*100% = 0.034%

15. In an election between two persons ,75% of the voters cast their votes out of which 2% are invalid. A got 9261 which 75% of the total

valid votes. Find total number of votes? SOLUTION:

Let x be the total votes. valid votes are 98% of 75% of x. So 75%(98%(75% of x))) = 9261

==> 75/100 *98 /100 * 75 100 *x = 9261 x= 1029 * 4 *100 *4 / 9

= 16800

So total no of votes = 16800

16 . A's maths test had 75 problems i.e 10 arithmetic, 30 algebra and 35 geometry problems. Although he answered 70% of arithmetic , 40% of algebra and 60 % of geometry problems correctly he didn't pass

the test because he got less than 60% of the problems right. How many more questions he would have needed to answer correctly to get a

60% passing grade.

SOLUTION: 70% of 10 =70/100 * 10

=7 40% of 30 = 40 / 100 * 30

= 12 60 % of 35 = 60 / 100 *35 = 21

So correctly attempted questions = 7 + 12 + 21

=40.

Questions to be answered correctly for 60% grade

=60% of 75

= 60/100 *75

=45.

So required questions=45-40 = 5

17 . If 50% of (x â€“ y) = 30% of (x + y) then what percent of x is y ? SOLUTION:

50/100(x-y) =30/100(x+y) Â½ (x-y)= 3/10(x+y)

5x-5y=3x+3y x=4y

So Required percentage =y/x*100 %

=y/4y *100 %

= 25%.

18 . If the price of tea is increased by 20% ,find how much percent must a

householder reduce her consumption of tea so as not to increase the

expenditure? SOLUTION:

Reduction in consumption= R/(100+R) *100%

=20/120 *100

= 16 2/3 %

19.The population of a town is 176400 . If it increases at the rate of 5%

per annum ,what will be the population 2 years hence? What was it 2 years ago?

SOLUTION: Population After 2 years = 176400[1+5/100]2 =1

76400 * 21/20 *21/20 =194481

Population 2 years ago = 176400/(1+5/100)2 = 176400 * 20/21

*20/ 21 =160000

20.1 liter of water is add to 5 liters of a 20 % solution of alcohol in water .

Find the strength of alcohol in new solution? SOLUTION:

Alcohol in 5 liters = 20% of 5 =1 liter Alcohol in 6 liters of new mixture =

1liter So % of alcohol is =1/6 *100=16 2/3%

21. If A earns 33 1/3 more than B .Then B earns less than A by what percent?

SOLUTION: 33 1/3 =100 / 3

Required Percentage = (100/3)/(100 + (100/3)) *100 %

= 100/400 *100 = 25 %

22. A school has only three classes which contain 40,50,60 students respectively . The pass percent of these classes are 10, 20 and 10

respectively . Then find the pass percent in the school. SOLUTION:

Number of passed candidates = 10/100*40+20/100 *50+10/100 * 60

=4+10+6 =20

Total students in school = 40+50+60 =150 So required percentage = 20/150 *100

= 40 /3

=13 1/3 %

23. There are 600 boys in a hostel . Each plays either hockey or football

or both .If 75% play hockey and 45 % play football ,Find how many play both?

SOLUTION: n(A)=75/100 *600 =450

n(B) = 45/100 *600 = 270

n(A^B)=n(A) + n(B) â€“ n(AUB)

=450 + 270 -600 =120

So 120 boys play both the games.

24.A bag contains 600 coins of 25p denomination and 1200 coins of 50pdenomination. If 12% of 25p coins and 24 % of 50p coins are removed,

Find the percentage of money removed from the bag ? SOLUTION:

Total money = (600 * 25/100 +1200 *50/100) =Rs 750

25p coins removed = 12/100 *600 =72 50p coins removed = 24/100 *1200

=288 So money removed =72 *1/4 +288 *1/2

= Rs 162 So required percentage=162/750 *100

=21 .6%

25. P is six times as large as Q.Find the percent that Q is less than P? SOLUTION:

Given that P= 6Q So Q is less than P by 5Q.

Required percentage= 5Q/P*100 % =5/6 * 100

% =83 1/3%

26.For a sphere of radius 10 cm ,the numerical value of surface area is what percent of the numerical value of its volume?

SOLUTION: Surface area = 4 *22/7 *r2

= 3/r(4/3 * 22/7 * r3)

=3/r * VOLUME Where r = 10 cm

So we have S= 3/10 V =3/10 *100 % of V

= 30 % of V So surface area is 30 % of Volume.

27. A reduction of 21 % in the price of wheat enables a person to buy 10 .5

kg more for Rs 100.What is the reduced price per kg. SOLUTION:

Let the original price = Rs x/kg Reduced price =79/100x /kg

==> 100/(79x/100)-100/x =10.5 ==> 10000/79x-100/x=10.5 ==> 10000-7900=10.5 * 79 x

==> x= 2100/10.5 *79 So required price = Rs (79/100

*2100/10.5 *79) /kg = Rs 2 per kg.

28.The length of a rectangle is increased by 60 % .By what percent would the width have to be decreased to maintain the same area?

SOLUTION: Let the length =l,Breadth= b.

Let the required decrease in breadth be x %

then 160/100 l *(100-x)/100 b=lb 160(100-x)=100 *100

or 100-x =10000/160 =125/2 so x = 100-125/2

=75/2 =37.5

Profit and Loss

Important Facts:

Cost Price: The price at which an article is purchased,is called its cost price,abbreviated as C.P.

Selling Price: The price at which an article is sold,is called its selling price,abbreviated as S.P.

Profit or Gain: If S.P. Is greater than C.P. The selleris said to have a profit or gain.

Loss:if S.P. Is less than C.P., the seller is said to have incurred a loss.

Formulae 1.Gain=(S.P-C.P)2.Loss=(C.P-S.P)3.Loss or Gain is always reckoned on C.P.4.Gain%=(gain*100)/C.P5.Loss%=(loss*100)/C.P6.S.P=[(100+gain%)/100]*C.P7.S.P=[(100-loss%)/100]*C.P8.C.P=(100*S.P)/(100+gain%)9.C.P=(100*S.P)/(100-loss%)10.If an article is sold at a gain of say,35%,then S.P=135% of C.P.11.If an article is sold at a loss of say,35%,then S.P=65% of C.P.12.When a person sells two similar items, one at a gain of say,x%,and the other at a loss of x%,then the seller always incurs a loss given by Loss%=[common loss and gain %/10]2=(x/10)213.If a trader professes to sell his goods at cost price,but uses false weight,then Gain%=[(error/(true value-error))*100]%14.Net selling price=Marked price-Discount

Simple Problems

1.A man buys an article for Rs.27.50 and sells it for Rs.28.60

Find the gain percent.

Sol: C.P=Rs 27.50 S.P=Rs 28.60 then Gain=S.P-C.P=28.60-27.50=Rs 1.10 Gain%=(gain*100)/C.P% =(1.10*100)/27.50%=4%

2.If a radio is purchased for Rs 490 and sold for Rs 465.50 Find the loss%?

Sol: C.P=Rs 490 S.P=Rs 465.50 Loss=C.P-S.P=490-465.50=Rs 24.50 Loss%=(loss*100)/C.P% =(24.50*100)/490%=5%

3.Find S.P when C.P=Rs 56.25 and Gain=20%

Sol: S.P=[(100+gain%)/100]*C.P S.P=[(100+20)/100]56.25=Rs 67.50

4.Find S.P when C.P=Rs 80.40,loss=5%

Sol: S.P=[(100-loss%)/100]*C.P S.P=[(100-5)/100]*80.40=Rs 68.34

5.Find C.P when S.P=Rs 40.60,gain=16%?

Sol: C.P=(100*S.P)/(100+gain%) C.P=(100*40.60)/(100+16)=Rs 35

6.Find C.P when S.P=Rs 51.70 ,loss=12%?

Sol: C.P=(100*S.P)/(100-loss%) C.P=(100*51.70)/(100-12)=Rs 58.75

7.A person incurs 5% loss by selling a watch for Rs 1140 . At what price should the watch be sold to earn 5% profit?

Sol: Let the new S.P be Rs x then, (100-loss%):(1st S.P)=(100+gain%):(2nd S.P) (100-5)/1140=(100+5)/x x=(105*1140)/95=Rs 1260

8.If the cost price is 96% of the selling price,then what isthe profit percent?

Sol: let S.P=Rs 100 then C.P=Rs 96 profit=S.P-C.P=100-96=Rs 4 profit%=(profit*C.P)/100% =(4*96)/100=4.17%

9.A discount dealer professes to sell his goods at cost pricebut uses a weight of 960 gms for a Kg weight .Find his gain %?

Sol: Gain%=[(error*100)/(true value-error)]% =[(40*100)/1000-40)]%=25/6%

10.A man sold two flats for Rs 675,958 each .On one he gains 16% while on the other he losses 16%.How much does he gain orlose in the whole transaction?

Sol: loss%=[common loss or gain%/10]2=(16/10)2=2.56%

11.A man sold two cows at Rs 1995 each. On one he lost 10% andon the other he gained 10%.what his gain or loss percent?

Sol: If loss% and gain% is equal to 10 then there is no loss or no gain.

12.The price of an article is reduced by 25% in order to restore the must be increased by ?

Sol: [x/(100-x)]*100 =[25/(100-25)]*100 =(25/75)*100=100/3%

13.Two discounts of 40% and 20% equal to a single discount of?

Sol: {[(100-40)/100]*[(100-20)/100]}%=(60*80)/(100*100)% =48% single discount is equal to (100-48)%=52% Difficult Problems

1.The cost of an article including the sales tax is Rs 616.The rate of sales tax is 10%,if the shopkeeper has made a profitof 12%,then the cost price of the article is?

Sol: 110% of S.P=616 S.P=(616*100)/110=Rs 560 C.P=(100*S.P)/(100+gain%) C.P. =(100*560)/(100+12)=Rs 500

2.Sam purchased 20 dozens of toys at the rate of 375 Rs per dozen.He sold each one of then at the rate of Rs 33.What was his percentage profit?

Sol: C.P of one toy=Rs 375/12=Rs 31.25 S.P of one toy=Rs 33 profit=S.P-C.P=33-31.25=Rs 1.75 profit %=(profit/C.P)*100 =(1.75/31.25)*100 profit% =5.6%

3.Two third of consignment was sold at a profit of 5% and the remainder at a loss of 2%.If the total was Rs 400,the value of the consignment was?

Sol: let the total value be Rs x value of 2/3=2x/3, value of 1/3=x/3 total S.P value be Rs x total S.P=[(105% of 2x/3)+(98% of x/3)] =(105*2x)/(100*3)+(98/100)+x/3 =308x/300

(308x/300)-x=400 8x/300=400 x=(300*400)/8=Rs 15000

4.Kunal bought a suitcase with 15% discount on the labelled price. He sold the suitcase for Rs 2880 with 20% profit on the labelledprice .At what price did he buy the suitcase?

Sol: let the labelled price be Rs x then 120% of x=2880 x=(2880*100)/120=Rs 2400 C.P=85% of the 2400 (85*2400)/100=Rs 2040

5.A tradesman gives 4% discount on the marked price and gives article free for buying every 15 articles and thus gains 35%. Themarked price is above the cost price by

Sol: let the C.P of each article be Rs 100 then C.P of 16 articles=Rs (100*16)=Rs 1600 S.P of 15 articles =1600*(135/100)=Rs 2160 S.P of each article =2160/15=Rs 144 If S.P is Rs 96, marked price =Rs 100 If S.P is Rs 144,marked price =(100/96)*144=Rs 15000 therefore marked price=50% above C.P

6.By selling 33m of cloth ,one gains the selling price of 11m.Find the gain percent?

Sol: gain=S.P of 33m-C.P of 33m =11m of S.P S.P of 22m=C.P of 33m let C.P of each meter be Rs 1,then C.P of 22m=Rs 22 S.P of 22m=Rs 33 gain=S.P-C.P=33-22=Rs 11 gain%=(gain/C.P)*100 =(11/22)*100=50%

7.The price of a jewel, passing through three hands, rises on the whole by 65%.if the first and second sellers earned 20% and 25% profit respectively,find the percentage profit earned by the third seller?

Sol: let the original price of the jewel be Rs P and let the profit earned by the third seller be x% then (100+x)% of 125% of P=165% of P [(100+x)/100]*(125/100)*(120/100)*P=(165/100)*P 100+x=(165*100*100)/(125*120) 100+x=110 x=10%

8.When a producer allows 36% commission on the retail price of hisproduct ,he earns a profit of 8.8%.what would be his profitpercent if the commission is reduced by 24%

Sol: let retail price =Rs 100 commission=Rs 36 S.P=retail price-commission=100-36=Rs 64 But profit=8.8% C.P=(100*C.P)/(gain+100)=(100*64)/(100+8.8)=Rs 1000/17 new commission=Rs 12 new S.P=100-12=Rs 88 gain=88-(1000/17)=Rs 496/17 gain%=gain*100/C.P =(496*17*100)(17*1000) gain%=49.6%

9.Vikas bought paper sheets for Rs 7200 and spent Rs 200 on transport. Paying Rs 600,he had 330 boxes made,which he soldat Rs 28 each. His profit percentage is

Sol: total investments=7200+200+600=Rs 8000 total receipt=330*28=Rs 9240 gain=S.P-C.P =total receipt-total investments gain=9240-8000=Rs 1240 gain% =gain*100/C.P=1240*100/8000=15.5%

10.A person earns 15% on investment but loses 10% on another investment .If the ratio of the two investments be 3:5 ,what is thegain or loss on the two investments taken together?

Sol: let the investments be 3x and 5x then total investment=8x total receipt=115% of 3x+90% of 5x =115*3x/100+90*5x/100=7.95x loss=C.P-S.P=8x-7.95x=0.05x loss%=.05x*100/8x=0.625%

11.The profit earned by selling an article for Rs 900 is double theloss incurred when the same article is sold for Rs 490 .At whatprice should the article be sold to make 25% profit?

Sol: let C.P be Rs x 900-x=2(x-450) 3x=1800 x=Rs 600 C.P=Rs 600 , gain required=25% S.P=(100+gain%)*C.P/100 S.P=(100+25)*600/100=Rs 750

12.If an article is sold at 5% gain instead of 5% loss,the sellergets Rs 6.72 more. The C.P of the article is?

Sol: let C.P be Rs x 105% of x-95% of x=6.72 (105/100)*x-(95/100)*x=6.72 x/10=6.72 x=Rs 67.21.

RATIO AND PROPORTION

Important Facts:

1.Ratio : The ratio of two qualities a and b in the same units, is the fraction a/b and we write it as a:b. In the ratio, a:b, we call â€˜aâ€™ as the first term of antecedent and b, the second term consequent.

Ex: The ratio 5:9 represents 5/9 with antecedent=5 ,consequent=9

3Rule: The multiplication or division of each term of 9 ratio by the same non-zero number does not affect

the ratio.

4.Proportion: The equality of two ratios is called proportion. If a:b=c:d, we write a:b::c:d and we say that a,b,c and d are in proportion. Here a and b are called extremes, while b and c are called mean terms.

Product of means=product of extremes Thus, a:b::c:d => (b*c)=(a*d)

5.Fourth proportional: If a:b::c:d, then d is called

the fourth proportional to a,b and c.

6.Third proportional: If a:b::b:c, then c is called third proportional to a and b.

7.Mean proportional: Mean proportional between a and b is SQRT(a*b).

COMPARISION OF RATIOS:

We say that (a:b)>(c:d) => (a/b)>(c/d)

8.Compounded ratio: The compounded ratio of the ratios (a:b), (c:d),(e:f) is (ace:bdf).

9.Duplicate Ratio: If (a:b) is (a2: b2 )

10.Sub-duplicate ratio of (a:b) is (SQRT(a):SQRT(b))

11.Triplicate ratio of (a:b) is (a3: b3 )

12.Sub-triplicate ratio of (a:b) is (a1/3: b1/3 ).

13.If a/b=c/d, then (a+b)/(a-b)=(c+d)/(c-d) (componend o and dividend o)

VARIATION:

14.we say that x is directly proportional to y, if x=ky for some constant k and we write.

15.We say that x is inversely proportional to y, if xy=k for some constant and we write.

16. X is inversely proportional to y.

If a/b=c/d=e/f=g/h=k then k=(a+c+e+g)/(b+d+f+h) If a1/b1,a2/b2,

a3/b3..............an/bn are unequal fractions then the ratio.

a1+a2+a3+..........an/(b1+b2+b3+...............bn) lies between the lowest & the highest of the three

fractions.

SIMPLE PROBLEMS

1.If a:b =5:9 and b:c=4:7 Find a:b:c?Sol: a:b=5:9 and b:c=4:7=4*9/4:9*4/9=9:63/9

a:b:c=5:9:63/9=20:36:63

2.Find the fourth proportion to 4,9,12Sol: d is the fourth proportion to a,b,c

a:b=c:d 4:9=12:x

4x=9*12=>x=27

3.Find third proportion to 16,36

Sol: if a:b=b:c then c is the third proportion to a,b

16:36=36:x 16x=36*36 x=81

4.Find mean proportion between 0.08 and 0.18Sol: mean proportion between a and b=square root

of ab mean proportion =square root of

0.08*0.18=0.12

5.If a:b=2:3 b:c=4:5, c:d=6:7 then a:b:c:d isSol: a:b=2:3 and b:c=4:5=4*3/4:5*3/4=3:15/4 c:d=6:7=6*15/24:7*15/24=15/4:35/8 a:b:c:d=2:3:15/4:35/8=16:24:30:35

6.2A=3B=4C then A:B:C?

Sol: let 2A=3B=4C=k then A=k/2, B=k/3, C=k/4 A:B:C=k/2:k/3:k/4=6:4:3

7.15% of x=20% of y then x:y isSol: (15/100)*x=(20/100)*y 3x=4y x:y=4:3

8.a/3=b/4=c/7 then (a+b+c)/c=

Sol: let a/3=b/4=c/7=k (a+b+c)/c=(3k+4k+7k)/7k=2

9.Rs 3650 is divided among 4 engineers, 3 MBAâ€™s and 5 CAâ€™s such that 3 CAâ€™s get as much as 2 MBAâ€™s and 3 Engâ€™s as much as 2 CAâ€™s .Find the share of

an MBA.Sol: 4E+3M+5C=3650

3C=2M, that is M=1.5C 3E=2C that is E=.66 C

Then, (4*0.66C)+(3*1.5C)+5C=3650 C=3650/1

2.166 C=300

M=1.5 and C=450.

DIFFICULT PROBLEMS

1.Three containers A,B and C are having mixtures of milk and water in the ratio of 1:5 and 3:5 and 5:7

respectively. If the capacities of the containers are in the ratio of all the three containers are in the ratio 5:4:5, find the ratio of milk to water, if the

mixtures of all the three containers are mixed together.

Sol: Assume that there are 500,400 and 500 liters respectively in the 3 containers.

Then ,we have, 83.33, 150 and 208.33 liters of milk in each of the three containers.

Thus, the total milk is 441.66 liters. Hence, the amount of water in the mixture is 1400-

441.66=958.33liters. Hence, the ratio of milk to water is 441.66:958.33

=> 53:115(using division by .3333) The calculation thought process should be

(441*2+2):(958*3+1)=1325:2875 Dividing by 25 => 53:115.

2.A certain number of one rupee,fifty parse and twenty five paise coins are in the ratio of 2:5:3:4, add up to Rs 210. How many 50 paise coins were there?

Sol: the ratio of 2.5:3:4 can be written as 5:6:8 let us assume that there are 5 one rupee coins,6 fifty paise coins and 8 twenty-five paise

coins in all. their

value=(5*1)+(6*.50)+(8*.25)=5+3+2=Rs 10 If the total is Rs 10,number of 50 paise

coins are 6. if the total is Rs 210, number of 50 paise

coins would be 210*6/10=126.

3.The incomes of A and B are in the ratio of 4:3 and their expenditure are in the ratio of 2:1 . if each

one saves Rs 1000,what are their incomes?

Sol: Ratio of incomes of A and B=4:3 Ratio of expenditures of A and B=2:1

Amount of money saved by A=Amount of money saved by B=Rs 1000

let the incomes of A and B be 4x and 3x respectively

let the expense of A and B be 2y and 1yrespectively

Amount of money saved by A=(income-expenditure)=4x-2y=Rs 1000

Amount of money saved by B=3x-y=Rs 1000 this can be even written as 6x-2y=Rs 2000

now solve 1 and 3 to get x=Rs 500

therefore income of A=4x=4*500=Rs 2000 income of B=3x=3*500=Rs 1500

4.A sum of Rs 1162 is divided among A,B and C. Such that 4 times A's share share is equal to 5 times B's share and 7 times C's share . What is the share of C?Sol: 4 times of A's share =5 times of B's share=7

times of C's share=1 therefore , the ratio of their share

=1/4:1/5:1/7 LCM of 4,5,7=140

therefore, Â¼:1/5:1/7=35:28:20 the ratio now can be written as

35:28:20 therefore C's

share=(20/83)*1162=20*14=Rs 280.

5.The ratio of the present ages of saritha and her mother is 2:9, mother's age at the time of saritha's birth was 28 years , what is saritha's present age?

Sol: ratio of ages of saritha and her mother =2:9

let the present age of saritha be 2x years. then the mother's

present age would be 9x years Difference in their ages =28

years 9x-2x=28 years

7x=28=>x=4 therefore saritha's age=2*4=8

years

Partnership

Important Facts:

Partnership:When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership.

Ratio of Division of Gains:

1.When the investments of all the partners are do the same time, the gain or loss is distributed among the partners in the ratio of theirinvestments.

Suppose A and B invest Rs x and Rs y respectively for a year in a business, then at the end of the year: (A's share of profit):(B's share of profit)=x:y

2.When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital*numberof units of time). Now gain or loss is divided in the ratio of thesecapitals.

Suppose A invests Rs x for p months and B invests Rs y for q months,then (A's share of profit):(B's share of profit)=xp:yq

3.Working and sleeping partners:A partner who manages the business isknown as working partner and the one who simply invests the money is a sleeping partner.

Formulae

1.When investments of A and B are Rs x and Rs y for a year in a business ,then at the end of the year (A's share of profit):(B's share of profit)=x:y

2.When A invests Rs x for p months and B invests Rs y for q months,then A's share profit:B's share of profit=xp:yq

Short cuts:

1.In case of 3 A,B,C investments then individual share is to be found then A=16000 , B=32,000 , C=40,000

Sol: A:B:C=16:32:40 =2:4:5`then individual share can be easily known.

2.If business mans A contributes for 5 months and B contributes for 9 months then share of B in the total profit of Rs 26,8000 ,A = Rs 15000,

B =Rs 12000

Sol: 15000*5 : 12000*9 25 : 36 for 36 parts=268000*(36/61) =Rs 158.16 Back Difficult problems:

1.P and Q started a business investing Rs 85,000 and Rs 15,000 respectively. In what ratio the profit earned after 2 years be dividedbetween P and Q respectively?

Sol: 85,000*2 : 15,000*2 17*2 : 3*2=34:6

2.A,B and C started a business by investing Rs 1,20,000, Rs 1,35,000 and Rs 1,50,000.Find the share of each ,out of an annual profit ofRs 56,700?

Sol: Ratio of shares of A,B and C=Ratio of their investments 120,000:135,000:150,000

=8:9:10 A's share=Rs 56,700*(8/27)=Rs 16,800 B's share =Rs 56,700*(9/27)=Rs 18,900 C's share =Rs 56,700*(10/27)=Rs 21,000

3.3 milkman A,B,C rented a pasture A grazed his 45 cows for 12 days B grazed his 36 cows for 15 days and c 60 cows for 10 days.If b's share of rent was Rs 540 What is the total rent?

Sol: 45*12:36*15:60*10 =9:9:10 9 parts is equal to Rs 540 then one part is equal to Rs 60 total rent=60*28=Rs 1680

4.Ramu and Krishna entered into a partnership with Rs 50,000 and Rs 60,000, after 4 months Ramu invested Rs 25,000 more while Krishnawithdraw Rs 20,000 . Find the share of Ramu in the annual profit ofRs 289,000.

Sol: Ramu : Krishna=50,000*4+75,000*8:60,000*4+40,000*8 =10:7 Ramu's annual profit=289000*(10/17)=Rs 170,000

5.A,B,C enter into partnership .A invests 3 times as much as B invests and B invests two third of what C invests. At the end of the year ,the profit earned is Rs 6600. what is the share of B?

Sol: let C's capital =Rs x B's capital=Rs (2/3)*x

A's capital =3*(2/3)*x=Rs 2x ratio of their capitals=2x:(2/3)*x:x =6x:2x;3x

B's share =Rs 6600(2/11)=Rs 1200

6.A,B and C enter into a partnership by investing in the ratio of 3:2:4.After one year ,B invests another Rs 2,70,000 and C,at the end of 2 years, also invests Rs 2,70,000.At the end of 3 years ,profit are sharedin the ratio of 3:4:5.Find the initial investment of each?

Sol: Initial investments of A,B,c be Rs 3x, Rs 2x, Rs 4x then for 3 years (3x*36):[(2x*12)+(2x+270000)*24]:[(4x*24)+(4x+270000)*12]=3:4:5 108x:(72x+640,000):(144x+3240000)=3:4:5 108x:72x+6480000:144x+3240000=3:4:5 (108x)/(72x+6480000)=3/4 432x=216x+19440000 216x=19440000 x=Rs 90000 A's initial investment=3x=3*90,000=Rs 2,70,000 B's initial investment=2x=2*90,000=Rs 1,80,000 C's initial investment=4x=4*90,000=Rs 3,60,000

CHAIN RULE1.DIRECT PROPORTION: Two Quantities are said to be directly proportional,if on the increase (or decrease) of th one, the other increases(or decreases) to the same extent.

Ex:(i) Cost is directly proportional to the number of articles.(More articles, More cost).

(ii)Work done is directly proportional to the number of men working on it. (More men, more work).

2.INDIRECT PROPORTION: Two Quantities are said to be indirectly proportional,if on the increase of the one , the other decreases to the same extent and vice-versa.

Ex:(i) The time taken by a car covering a certain distance is inversely proportional to th speed of the car.(More speed, less is the time taken to cover the distance).

(ii)Time taken to finish a work is inversely proportional to the number of persons working at it. (More persons, less is the time taken to finish a job).

NOTE: In solving Questions by chain rule, we compare every item with the term to be found out.

SIMPLE PROBLEMS1)If 15 toys cost Rs.234, what do 35 toys cost ?

Sol: Let the required cost be Rs. x then more toys more cost(direct proportion)

15:35:: 234:x (15*x)=(234*35) x=(234*35) /(15)= 546 Rs

2)If 36 men can do a piece of work in 25hours, in how many hours will 15men do it?

Sol: Let the required number of hours be x. less men more hours(Indirect proportion). 15:36::25:x (15*x)=(36*25) x=(36*25) /15 x=60 For 15 men it takes 60 hours.

3)If 9 engines consume 24metric tonnes of coal, when each is working 8 hours a day, how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of former type consume as much as 4 engines of latter type?

Sol: Let 3 engines of former type consume 1 unit in 1 hour. 4 engines of latter type consume 1 unit in 1 hour. 1 engine of former type consumes 1/3 unit in 1 hour. 1 engine of latter type consumes Â¼ unit in 1 hour. Let required consumption of coal be x units. Less engines, less coal consumed.(direct) More working hours, more coal consumed(direct) Less rate of consumption, less coal consumed (direct) 9:8 8:13 :: 24:x 1/3:1/4

(9*8*(1/3)*x)=(8*13*(1/4)*24)

24x=624

x=26 metric tonnes.

COMPLEX PROBLEMS1)A contract is to be completed in 46 days and 117 men were set to work,each working 8 hours a day. After 33 days, 4/7 of the work is completed.How many additional men may be employed so that the work may be completedin time, each man now working 9 hours a day? Sol: 4/7 of work is completed . Remaining work=1- 4/7 =3/7 Remaining period= 46-33 =13 days Less work, less men(direct proportion) less days, more men(Indirect proportion) More hours/day, less men(Indirect proportion)

work 4/7:3/7 Days 13:33 :: 117:x

hrs/day 9:8 (4/7)*13*9*x=(3/7)*33*8*117 x=(3*33*8*117) / (4*13*9) x=198 men So, additional men to be employed=198 -117=81

2)A garrison had provisions for a certain number of days. After 10 days, 1/5 of the men desert and it is found that the provisions will now last just as long as before. How long was that?

Sol: Let initially there be x men having food for y days. After, 10 days x men had food for ( y-10)days Also, (x -x/5) men had food for y days. x(y-10)=(4x/5)*y => (x*y) -50x=(4(x*y)/5) 5(x*y)-4(x*y)=50x x*y=50x y=50

3)A contractor undertook to do a certain piece of work in 40 days. He engages 100 men at the beginning and 100 more after 35 days and completesthe work in stipulated time. If he had not engaged the additional men, how many days behind schedule would it be finished?

Sol: 40 days- 35 days=5 days =>(100*35)+(200*5) men can finish the work in 1 day. 4500 men can finish it in 4500/100= 45 days This s 5 days behind the schedule.

4)12 men and 18 boys,working 7 Â½ hors a day, can do a piece if work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working 9 hours a day?

Sol: 1man =2 boys 12men+18boys=>(12*2+18)boys=42 boys let the required number of boys=x 21 men+x boys =>((21*2)+x) boys =>(42+x) boys less days, more boys(Indirect proportion) more hours per day, less boys(Indirect proportion)

days 50:60 hrs/day 9:15/2 :: 42:(42+x) work 1:2 (50*9*1*(42+x))=60*(15/2)*2*42 (42+x)= (60*15*42)/(50*9)= 84 x=84-42= 42 =42 42 days behind the schedule it will be finished.

Time and Work

Important Facts:

1.If A can do a piece of work in n days, then A's 1 day work=1/n

2.If A's 1 day's work=1/n, then A can finish the work in n days.

Ex: If A can do a piece of work in 4 days,then A's 1 day's work=1/4. If A's 1 day’s work=1/5, then A can finish the work in 5 days

3.If A is thrice as good workman as B,then: Ratio of work done byA and B =3:1. Ratio of time taken by A and B to finish a work=1:3

4.Definition of Variation: The change in two different variables follow some definite rule. It said that the two variables vary directly or inversely.Its notation is X/Y=k, where k is called constant. This variation is called direct variation. XY=k. This variation is called inverse variation.

5.Some Pairs of Variables:

i)Number of workers and their wages. If the number of workers increases, their total wages increase. If the number of days reduced, there will be less work. If the number of days is increased, there will be more work. Therefore, here we have direct proportion or direct variation.

ii)Number workers and days required to do a certain work is an example of inverse variation. If more men are employed, they will require fewer days and if there are less number of workers, more days are required.

iii)There is an inverse proportion between the daily hours of a work and the days required. If the number of hours is increased, less number of days are required and if the number of hours is reduced, more days are required.

6.Some Important Tips:

More Men -Less Days and Conversely More Day-Less Men.More Men -More Work and Conversely More Work-More Men.More Days-More Work and Conversely More Work-More Days.Number of days required to complete the given work=Total work/One day’s work.

Since the total work is assumed to be one(unit), the number of daysrequired to complete the given work would be the reciprocal of oneday’s work.Sometimes, the problems on time and work can be solved using the proportional rule ((man*days*hours)/work) in another situation.

7.If men is fixed,work is proportional to time. If work is fixed, then time is inversely proportional to men therefore, (M1*T1/W1)=(M2*T2/W2)

Problems

1)If 9 men working 6 hours a day can do a work in 88 days. Then 6 menworking 8 hours a day can do it in how many days?

Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2) so (9*6*88/1)=(6*8*d/1) on solving, d=99 days.2)If 34 men completed 2/5th of a work in 8 days working 9 hours a day.How many more man should be engaged to finish the rest of the work in6 days working 9 hours a day?

Sol: From the above formula i.e (m1*t1/w1)=(m2*t2/w2) so, (34*8*9/(2/5))=(x*6*9/(3/5)) so x=136 men number of men to be added to finish the work=136-34=102 men

3)If 5 women or 8 girls can do a work in 84 days. In how many days can10 women and 5 girls can do the same work?

Sol: Given that 5 women is equal to 8 girls to complete a work so, 10 women=16 girls. Therefore 10women +5girls=16girls+5girls=21girls. 8 girls can do a work in 84 days then 21 girls ---------------? answer= (8*84/21)=32days. Therefore 10 women and 5 girls can a work in 32days

4)Worker A takes 8 hours to do a job. Worker B takes 10hours to do the same job. How long it take both A & B, working together but independently,to do the same job?

Sol: A's one hour work=1/8. B's one hour work=1/10 (A+B)'s one hour work=1/8+1/10 =9/40 Both A & B can finish the work in 40/9 days

5)A can finish a work in 18 days and B can do the same work in half the time taken by A. Then, working together, what part of the same work they can finish in a day?

Sol: Given that B alone can complete the same work in days=half the time taken by A=9days A's one day work=1/18 B's one day work=1/9 (A+B)'s one day work=1/18+1/9=1/6

6)A is twice as good a workman as B and together they finish a piece of work in 18 days.In how many days will A alone finish the work.

Sol: if A takes x days to do a work then B takes 2x days to do the same work =>1/x+1/2x=1/18 =>3/2x=1/18 =>x=27 days. Hence, A alone can finish the work in 27 days.

7)A can do a certain work in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?

Sol: Ratio of time taken by A&B=160:100 =8:5 Suppose B alone takes x days to do the job. Then, 8:5::12:x => 8x=5*12 => x=15/2 days.

8)A can do a piece of work n 7 days of 9 hours each and B alone can do it in 6 days of 7 hours each. How long will they take to do it working together 8 2/5 hours a day?

Sol: A can complete the work in (7*9)=63 days B can complete the work in (6*7)=42 days => A's one hour's work=1/63 and B's one hour work=1/42 (A+B)'s one hour work=1/63+1/42=5/126 Therefore, Both can finish the work in 126/5 hours. Number of days of 8 2/5 hours each=(126*5/(5*42))=3days

9)A takes twice as much time as B or thrice as much time to finish a piece of work. Working together they can finish the work in 2 days. B can do the work alone in ? Sol: Suppose A,B and C take x,x/2 and x/3 hours respectively finish the work then 1/x+2/x+3/x=1/2 => 6/x=1/2 =>x=12 So, B takes 6 hours to finish the work.

10)X can do ¼ of a work in 10 days, Y can do 40% of work in 40 days and Z can do 1/3 of work in 13 days. Who will complete the work first?

Sol: Whole work will be done by X in 10*4=40 days. Whole work will be done by Y in (40*100/40)=100 days. Whole work will be done by Z in (13*3)=39 days Therefore,Z will complete the work first. Complex Problems

1)A and B undertake to do a piece of workfor Rs 600.A alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they can finish it in 3 days, Find the share of each?

Sol: C's one day's work=(1/3)-(1/6+1/8)=1/24

Therefore, A:B:C= Ratio of their one day’s work=1/6:1/8:1/24=4:3:1 A's share=Rs (600*4/8)=300 B's share= Rs (600*3/8)=225 C's share=Rs[600-(300+225)]=Rs 75

2)A can do a piece of work in 80 days. He works at it for 10 days & then B alonefinishes the remaining work in 42 days. In how much time will A and B, workingtogether, finish the work?

Sol: Work done by A in 10 days=10/80=1/8 Remaining work=(1-(1/8))=7/8 Now, work will be done by B in 42 days. Whole work will be done by B in (42*8/7)=48 days Therefore, A's one day's work=1/80 B’s one day's work=1/48 (A+B)'s one day's work=1/80+1/48=8/240=1/30 Hence, both will finish the work in 30 days.

3)P,Q and R are three typists who working simultaneously can type 216 pages in 4 hours In one hour , R can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as many pages as P can during seven hours. How many pages does each of them type per hour?

Sol:Let the number of pages typed in one hour by P, Q and R be x,y and z respectively Then x+y+z=216/4=54 ---------------1 z-y=y-x => 2y=x+z -----------2 5z=7x => x=5x/7 ---------------3 Solving 1,2 and 3 we get x=15,y=18, and z=21

4)Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?

Sol: Number of pages typed by Ronald in one hour=32/6=16/3 Number of pages typed by Elan in one hour=40/5=8 Number of pages typed by both in one hour=((16/3)+8)=40/3 Time taken by both to type 110 pages=110*3/40=8 hours.

5)Two workers A and B are engaged to do a work. A working alone takes 8 hoursmore to complete the job than if both working together. If B worked alone, he would need 4 1/2 hours more to compete the job than they both working together. What time would they take to do the work together.

Sol: (1/(x+8))+(1/(x+(9/2)))=1/x =>(1/(x+8))+(2/(2x+9))=1/x => x(4x+25)=(x+8)(2x+9) => 2x2 =72

=> x2 = 36 => x=6 Therefore, A and B together can do the work in 6 days.

6)A and B can do a work in12 days, B and C in 15 days, C and A in 20 days.If A,B and C work together, they will complete the work in how many days?

Sol: (A+B)'s one day's work=1/12; (B+C)'s one day's work=1/15; (A+C)'s one day's work=1/20; Adding we get 2(A+B+C)'s one day's work=1/12+1/15+1/20=12/60=1/5 (A+B+C)'s one day work=1/10 So, A,B,and C together can complete the work in 10 days.

7)A and B can do a work in 8 days, B and C can do the same wor in 12 days. A,B and C together can finish it in 6 days. A and C together will do it in how many days?

Sol: (A+B+C)'s one day's work=1/6; (A+B)'s one day's work=1/8; (B+C)'s one day's work=1/12; (A+C)'s one day's work=2(A+B+C)'s one day's work-((A+B)'s one day work+(B+C)'s one day work) = (2/6)-(1/8+1/12) =(1/3)- (5/24) =3/24 =1/8 So, A and C together will do the work in 8 days.

8)A can do a certain work in the same time in which B and C together can do it.If A and B together could do it in 10 days and C alone in 50 days, then B alonecould do it in how many days?

Sol: (A+B)'s one day's work=1/10; C's one day's work=1/50 (A+B+C)'s one day's work=(1/10+1/50)=6/50=3/25 Also, A's one day's work=(B+C)’s one day's work From i and ii ,we get :2*(A's one day's work)=3/25 => A's one day's work=3/50 B's one day’s work=(1/10-3/50) =2/50 =1/25 B alone could complete the work in 25 days.

9) A is thrice as good a workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:

Sol: Ratio of times taken by A and B=1:3.

If difference of time is 2 days , B takes 3 days If difference of time is 60 days, B takes (3*60/2)=90 days So, A takes 30 days to do the work=1/90 A's one day's work=1/30; B's one day's work=1/90; (A+B)'s one day's work=1/30+1/90=4/90=2/45 Therefore, A&B together can do the work in 45/2days 10) A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A&B,working together, finish the work?

Sol: Work Done by A n 10 days =10/80=1/8 Remaining work =1-1/8=7/8 Now 7/8 work is done by B in 42 days Whole work will be done by B in 42*8/7= 48 days => A's one day's work =1/80 and B's one day's work =1/48 (A+B)'s one day's work = 1/80+1/48 = 8/240 = 1/30 Hence both will finish the work in 30 days.

11) 45 men can complete a work in 16 days. Six days after they started working,so more men joined them. How many days will they now take to complete the remaining work?

Sol: M1*D1/W1=M2*D2/W2 =>45*6/(6/16)=75*x/(1-(6/16)) => x=6 days

12)A is 50% as efficient as B. C does half the work done by A&B together. If C alone does the work n 40 days, then A,B and C together can do the work in:

Sol: A's one day's work:B's one days work=150:100 =3:2 Let A's &B's one day's work be 3x and 2x days respectively. Then C's one day's work=5x/2 => 5x/2=1/40 => x=((1/40)*(2/5))=1/100 A's one day's work=3/100 B's one day's work=1/50 C's one day's work=1/40 So, A,B and C can do the work in 13 1/3 days.

13)A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days A alone can finish the remaining work?

Sol: B's 10 day's work=10/15=2/3

Remaining work=(1-(2/3))=1/3 Now, 1/18 work is done by A in 1 day. Therefore 1/3 work is done by A in 18*(1/3)=6 days.

14)A can finish a work in 24 days, B n 9 days and C in 12 days. B&C start the work but are forced to leave after 3 days. The remaining work done by A in:

Sol: (B+C)'s one day's work=1/9+1/12=7/36 Work done by B & C in 3 days=3*7/36=7/12 Remaining work=1-(7/12)=5/12 Now , 1/24 work is done by A in 1 day. So, 5/12 work is done by A in 24*5/12=10 days

15)X and Y can do a piece of work n 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of work. How long did the work last?

Sol: work done by X in 4 days =4/20 =1/5 Remaining work= 1-1/5 =4/5 (X+Y)'s one day's work =1/20+1/12 =8/60=2/15 Now, 2/15 work is done by X and Y in one day. So, 4/5 work will be done by X and Y in 15/2*4/5=6 days Hence Total time taken =(6+4) days = 10 days

16)A does 4/5 of work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

Sol: Whole work is done by A in 20*5/4=25 days Now, (1-(4/5)) i.e 1/5 work is done by A& B in days. Whole work will be done by A& B in 3*5=15 days =>B's one day's work= 1/15-1/25=4/150=2/75 So, B alone would do the work in 75/2= 37 ½ days.

17) A and B can do a piece of work in 45 days and 40 days respectively. They began to do the work together but A leaves after some days and then B completed the remaining work n 23 days. The number of days after which A left the work was

Sol: (A+B)'s one day's work=1/45+1/40=17/360 Work done by B in 23 days=23/40 Remaining work=1-(23/40)=17/40 Now, 17/360 work was done by (A+B) in 1 day. 17/40 work was done by (A+B) in (1*(360/17)*(17/40))= 9 days So, A left after 9 days.

18)A can do a piece of work in 10 days, B in 15 days. They work for 5 days. The rest of work finished by C in 2 days. If they get Rs 1500 for the whole

work, the daily wages of B and C are

Sol: Part of work done by A= 5/10=1/2 Part of work done by B=1/3 Part of work done by C=(1-(1/2+1/3))=1/6 A's share: B's share: C's share=1/2:1/3:1/6= 3:2:1 A's share=(3/6)*1500=750 B's share=(2/6)*1500=500 C's share=(1/6)*1500=250 A's daily wages=750/5=150/- B's daily wages=500/5=100/- C's daily wages=250/2=125/- Daily wages of B&C = 100+125=225/-

19)A alone can complete a work in 16 days and B alone can complete the same in 12 days. Starting with A, they work on alternate days. The total work willbe completed in how many days?

(a) 12 days (b) 13 days (c) 13 5/7 days (d)13 ¾ days

Sol: (A+B)'s 2 days work = 1/16 + 1/12 =7/48 work done in 6 pairs of days =(7/48) * 6 = 7/8 remaining work = 1- 7/8 = 1/8 work done by A on 13th day = 1/16 remaining work = 1/8 – 1/16 = 1/16 on 14th day, it is B’s turn 1/12 work is done by B in 1 day. 1/16 work is done by B in ¾ day. Total time taken= 13 ¾ days. So, Answer is: D 20)A,B and C can do a piece of work in 20,30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?

Sol: A's two day's work=2/20=1/10 (A+B+C)'s one day's work=1/20+1/30+1/60=6/60=1/10 Work done in 3 days=(1/10+1/10)=1/5 Now, 1/5 work is done in 3 days Therefore, Whole work will be done in (3*5)=15 days.

21)Seven men can complete a work in 12 days. They started the work and after 5 days, two men left. In how many days will the work be completed by the remaining men?

(A) 5 (B) 6 (C ) 7 (D) 8 (E) none

Sol: 7*12 men complete the work in 1 day. Therefore, 1 man's 1 day's work=1/84 7 men's 5 days work = 5/12 =>remaining work = 1-5/12 = 7/12

5 men's 1 day's work = 5/84 5/84 work is don by them in 1 day 7/12 work is done by them in ((84/5) * (7/12)) = 49/5 days = 9 4/5 days. Ans: E

22).12 men complete a work in 9 days. After they have worked for 6 days, 6 more men joined them. How many days will they take to complete the remaining work?

(a) 2 days (b) 3 days (c) 4 days (d) 5days

Sol : 1 man's 1 day work = 1/108 12 men's 6 days work = 6/9 = 2/3 remaining work = 1 – 2/3 = 1/3 18 men's 1 days work = 18/108 = 1/6 1/6 work is done by them in 1 day therefore, 1/3 work is done by them in 6/3 = 2 days. Ans : A

23).A man, a woman and a boy can complete a job in 3,4 and 12 days respectively.How many boys must assist 1 man and 1 woman to complete the job in ¼ of a day?

(a). 1 (b). 4 (c). 19 (d). 41

Sol : (1 man + 1 woman)'s 1 days work = 1/3+1/4=7/12 Work done by 1 man and 1 women n 1/4 day=((7/12)*(1/4))=7/48 Remaining work= 1- 7/48= 41/48 Work done by 1 boy in ¼ day= ((1/12)*(1/4)) =1/48 Therefore, Number of boys required= ((41/48)*48)= 41 days So,Answer: D

24)12 men can complete a piece of work in 4 days, while 15 women can complete the same work in 4 days. 6 men start working on the job and after working for2 days, all of them stopped working. How many women should be put on the job to complete the remaining work, if it is to be completed in 3 days.

(A) 15 (B) 18 (C) 22 (D) data inadequate

Sol: one man's one day's work= 1/48 one woman's one day's work=1/60 6 men's 2 day's work=((6/48)*2)= ¼ Remaining work=3/4 Now, 1/60 work s done in 1 day by 1 woman. So, ¾ work will be done in 3 days by (60*(3/4)*(1/3))= 15 woman. So, Answer: A

25)Twelve children take sixteen days to complete a work which can be completed

by 8 adults in 12 days. Sixteen adults left and four children joined them. Howmany days will they take to complete the remaining work?

(A) 3 (B) 4 ( C) 6 (D) 8

Sol: one child's one day work= 1/192; one adult's one day's work= 1/96; work done in 3 days=((1/96)*16*3)= 1/2 Remaining work= 1 – ½=1/2 (6 adults+ 4 children)'s 1 day's work= 6/96+4/192= 1/12 1/12 work is done by them in 1 day. ½ work is done by them 12*(1/2)= 6 days So, Answer= C

26)Sixteen men can complete a work in twelve days. Twenty four children can complete the same work in 18 days. 12 men and 8 children started working and after eight days three more children joined them. How many days will they nowtake to complete the remaining work?

(A) 2 days (B) 4 days ( C) 6 days (D) 8 days

ol: one man's one day's work= 1/192 one child's one day's work= 1/432 Work done in 8 days=8*(12/192+ 8/432)=8*(1/16+1/54) =35/54 Remaining work= 1 -35/54= 19/54 (12 men+11 children)'s 1 day's work= 12/192 + 11/432 = 19/216 Now, 19/216 work is done by them in 1 day. Therefore, 19/54 work will be done by them in ((216/19)*(19/54))= 4 days So,Answer: B

27)Twenty-four men can complete a work in 16 days. Thirty- two women can complete the same work in twenty-four days. Sixteen men and sixteen women started working and worked for 12 days. How many more men are to be added to complete the remaining work in 2 days?

(A) 16 men (B) 24 men ( C) 36 men (D) 48 men

Sol: one man's one day's work= 1/384 one woman's one day's work=1/768 Work done in 12 days= 12*( 16/384 + 16/768) = 12*(3/48)=3/4 Remaining work=1 – ¾=1/4 (16 men+16 women)'s two day's work =12*( 16/384+16/768)=2/16=1/8 Remaining work = 1/4-1/8 =1/8 1/384 work is done n 1 day by 1 man. Therefore, 1/8 work will be done in 2 days in 384*(1/8)*(1/2)=24men

28)4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?

(A) 35 days (B) 40 days ( C) 45 days (D) 50 days

Sol: Let 1 man's 1 day's work =x days and 1 woman's 1 day's work=y Then, 4x+6y=1/8 and 3x+7y=1/10. Solving these two equations, we get: x=11/400 and y= 1/400 Therefore, 1 woman's 1 day's work=1/400 => 10 women will complete the work in 40 days. Answer: B

29)One man,3 women and 4 boys can do a piece of work in 96hrs, 2 men and 8 boyscan do it in 80 hrs, 2 men & 3 women can do it in 120hr. 5Men & 12 boys can do it in?

(A) 39 1/11 hrs (B) 42 7/11 hrs ( C) 43 7/11 days (D) 44hrs

Sol: Let 1 man's 1 hour's work=x 1 woman's 1 hour's work=y 1 boy's 1 hour's work=z Then, x+3y+4z=1/96 -----------(1) 2x+8z= 1/80 ----------(2) adding (2) & (3) and subtracting (1) 3x+4z=1/96 ---------(4) From (2) and (4), we get x=1/480 Substituting, we get : y=1/720 and z= 1/960 (5 men+ 12 boy)'s 1 hour's work=5/480+12/960 =1/96 + 1/80=11/480 Therefore, 5 men and 12 boys can do the work in 480/11 or 43 7/11hours. So,Answer: C

Pipes and Cisterns

Important Facts:

1.INLET:A pipe connected with a tank or cistern or a reservoir, that fills it, it is known as Inlet.

OUTLET:A pipe connected with a tank or a cistern or a reservoir, emptying it, is known as Outlet.

2. i) If a pipe can fill a tank in x hours, then : part filled in 1 hour=1/x.

ii)If a pipe can empty a tank in y hours, then :

part emptied in 1 hour=1/y.

iii)If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours( where y>x), then on opening both the pipes, the net part filled in 1 hour=(1/x -1/y).

iv)If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours( where x>y), then on opening both the pipes, the net part filled in 1 hour=(1/y -1/x).

v) If two pipes A and B can fill a tank in x hours and y hours respectively. If both the pipes are opened simultaneously, part filled by A+B in 1 hour= 1/x +1/y.

Simple Problems

1)Two pipes A& B can fill a tank in 36 hours and 45 hours respectively.If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Sol: Part filled by A in 1 hour=1/36 Part filled by B in 1 hour= 1/45; Part filled by (A+B)'s in 1 hour=1/36 +1/45= 9/180 =1/20 Hence, both the pipes together will fill the tank in 20 hours.

2)Two pipes can fill a tank in 10 hours & 12 hours respectively. While 3rd pipe empties the full tank n 20 hours. If all the three pipesoperate simultaneously,in how much time will the tank be filled?

Sol: Net part filled in 1 hour=1/10 +1/12 -1/20 =8/60=2/15 The tank be filled in 15/2hours= 7 hrs 30 min

3)A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously,then after how much time will the cistern get filled?

Sol: Net part filled in 1 hour= 1/4 -1/9= 5/36 Therefore the cistern will be filled in 36/5 hours or 7.2 hours.

4)If two pipes function simultaneously, the reservoir will be filled in 12 days.One pipe fills the reservoir 10 hours faster than the other. How many hours does it take the second pipe to fill the reservoir.

Sol: Let the reservoir be filled by the 1st pipe in x hours. The second pipe will fill it in (x+10) hours 1/x + (1/(x+10))= 1/12 => (2x+10)/((x)*(x+10))= 1/12 => x=20 So, the second pipe will take 30 hours to fill the reservoir.

5)A cistern has two taps which fill it in 12 min and 15 min respectively.

There is also a waste pipe in the cistern. When all the three are opened,the empty cistern is full in 20 min. How long will the waste pipe take to empty the full cistern?

Sol: Work done by a waste pipe in 1 min =1/20 -(1/12+1/15)= -1/10 (-ve means emptying)

6)A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

Sol: Time taken by one tap to fill the half of the tank =3 hours Part filled by the four taps in 1 hour=4/6=2/3 Remaining part=1 -1/2=1/2 Therefore, 2/3:1/2::1:x or x=(1/2)*1*(3/2)=3/4 hours. i.e 45 min So, total time taken= 3hrs 45min.

7)A water tank is two-fifth full. Pipe A can fill a tank in 10 min. And B can empty it in 6 min. If both pipes are open, how long will it take to empty or fill the tank completely ?

Sol: Clearly, pipe B is faster than A and So, the tank will be emptied. Part to be emptied=2/5. Part emptied by (A+B) in 1 min= 1/6 -1/10=1/15 Therefore, 1/15:2/5::1:x or x=((2/5)*1*15)=6 min. So, the tank be emptied in 6 min.

8)Bucket P has thrice the capacity as Bucket Q. It takes 60 turns for Bucket P to fill the empty drum. How many turns it will take for both the buckets P&Q, having each turn together to fill the empty drum?

Sol: Let the capacity of P be x lit. Then capacity of Q=x/3 lit Capacity of the drum=60x lit Required number of turns= 60x/(x+(x/3))= 60x*3/4x=45

Complex Problems

1)Two pipes can fill a cistern in 14 hours and 16 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in thebottom it took 32min more to fill the cistern. When the cistern is full,in what time will the leak empty it?

Sol: Work done by the two pipes in 1 hour= 1/14+1/16=15/112 Time taken by these two pipes to fill the tank=112/15 hrs. Due to leakage, time taken = 7 hrs 28 min+ 32 min= 8 hours Therefore, work done by (two pipes + leak) in 1 hr= 1/8

work done by leak n 1 hour=15/112 -1/8=1/112 Leak will empty full cistern n 112 hours.

2)Two pipes A&B can fill a tank in 30 min. First, A&B are opened. After7 min, C also opened. In how much time, the tank s full.

Sol: Part filled n 7 min = 7*(1/36+1/45)=7/20 Remaining part= 1-7/20=13/20 Net part filled in 1 min when A,B and C are opened=1/36 +1/45- 1/30=1/60 Now, 1/60 part is filled in 1 min. 13/20 part is filled n (60*13/20)=39 min Total time taken to fill the tank=39+7=46 min

3)Two pipes A&B can fill a tank in 24 min and 32 min respectively. If both the pipes are opened simultaneously, after how much time B shouldbe closed so that the tank is full in 18 min.

Sol: Let B be closed after x min, then part filled by (A+B) in x min+ part filled by A in (18-x) min=1 x(1/24+1/32) +(18-x)1/24 =1 => x=8 Hence B must be closed after 8 min.

4)Two pipes A& B together can fill a cistern in 4 hours. Had they been opened separately, then B would have taken 6 hours more than A to fill the cistern. How much time will be taken by A to fill the cistern separately?

Sol: Let the cistern be filled by pipe A alone in x hours. Pipe B will fill it in x+6 hours 1/x + 1/x+6=1/4 Solving this we get x=6. Hence, A takes 6 hours to fill the cistern separately.

5)A tank is filled by 3 pipes with uniform flow. The first two pipes operating simultaneously fill the tan in the same time during whichthe tank is filled by the third pipe alone. The 2nd pipe fills the tank5 hours faster than first pipe and 4 hours slower than third pipe. Thetime required by first pipe is :

Sol: Suppose, first pipe take x hours to fill the tank then B & C will take (x-5) and (x-9) hours respectively. Therefore, 1/x +1/(x-5) =1/(x-9) On solving, x=15 Hence, time required by first pipe is 15 hours. 6)A large tanker can be filled by two pipes A& B in 60min and 40 min respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time & A and B fill it together for the other half?

Sol: Part filled by (A+B) n 1 min=(1/60 +1/40)=1/24 Suppose the tank is filled in x minutes Then, x/2(1/24+1/40)=1

=> (x/2)*(1/15)=1 => x=30 min.

7)Two pipes A and B can fill a tank in 6 hours and 4 hours respectively.If they are opened on alternate hours and if pipe A s opened first, in how many hours, the tank shall be full.

Sol: (A+B)'s 2 hours work when opened alternatively =1/6+1/4 =5/12 (A+B)'s 4 hours work when opened alternatively=10/12=5/6 Remaining part=1 -5/6=1/6. Now, it is A's turn and 1/6 part is filled by A in 1 hour. So, total time taken to fill the tank=(4+1)= 5 hours.

8)Three taps A,B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternatively, the tank will be full in.

Sol: (A+B)'s 1 hour's work=1/12+1/15=9/60=3/20 (A+C)'s 1 hour's work=1/20+1/12=8/60=2/15 Part filled in 2 hours=3/20+2/15=17/60 Part filled in 2 hours=3/20+2/15= 17/60 Part filled in 6 hours=3*17/60 =17/20 Remaining part=1 -17/20=3/20 Now, it is the turn of A & B and 3/20 part is filled by A& B in 1 hour. Therefore, total time taken to fill the tank=6+1=7 hours

9)A Booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m3. The emptying capacity of the tank is10 m3 per minute higher than its filling capacity and the pump needs 8minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

Sol: Let, the filling capacity of the pump be x m3/min Then, emptying capacity of the pump=(x+10) m3/min. So,2400/x – 2400/(x+10) = 8 on solving x=50.

10)A leak in the bottom of a tank can empty the full tan in 8 hr. An inletpipe fills water at the rate of 6 lits a minute. When the tank is full,the inlet is opened and due to the leak, the tank is empty in 12 hrs.How many liters does the cistern hold?

Sol: Work done by the inlet in 1 hr= 1/8 -1/12=1/24 Work done by the inlet in i min= (1/24)*(1/60)=1/1440 Therefore, Volume of 1/1440 part=6 lit Volume of whole=(1440*6) lit=8640 lit.

11)Two pipes A and B can fill a cistern in 37 ½ min and 45 minutes respectively. Both the pipes are opened. The cistern will be filled in just half an hour, if the pipe B is turned off after:

sol: Let B be turned off after x min. Then, Part filled by (A+B) in x min+ part filled by A in (30-x)min=1

Therefore, x(2/75+1/45)+(30-x)(2/75)=1 11x/225 + (60-2x)/75=1 11x+ 180-6x=225 x=9. So, B must be turned off after 9 minutes.

Time and Distance

Formulae:

I)Speed = Distance/Time

II)Time = Distance/speed

III) Distance = speed*time

IV) 1km/hr = 5/18 m/s

V)1 m/s = 18/5 Km/hr

VI)If the ratio of the speed of A and B is a:b,then the ratio ofthe time taken by them to cover the same distance is 1/a : 1/bor b:a

VII) suppose a man covers a distance at x kmph and an equal distance at y kmph.then the average speed during the whole journey is (2xy/x+y)kmph

Problems

1)A person covers a certain distance at 7kmph .How many meters does he cover in 2 minutes.

Solution::speed=72kmph=72*5/18 = 20m/sdistance covered in 2min =20*2*60 = 2400m

2)If a man runs at 3m/s. How many km does he run in 1hr 40min

Solution::speed of the man = 3*18/5 kmph = 54/5kmphDistance covered in 5/3 hrs=54/5*5/3 = 18km

3)Walking at the rate of 4knph a man covers certain distance in 2hr 45 min. Running at a speed of 16.5 kmph the man will cover the same distance in.

Solution::Distance=Speed* time4*11/4=11kmNew speed =16.5kmph

therefore Time=D/S=11/16.5 = 40min Complex Problems

1)A train covers a distance in 50 min ,if it runs at a speedof 48kmph on an average.The speed at which the train must runto reduce the time of journey to 40min will be.

Solution::Time=50/60 hr=5/6hrSpeed=48mphdistance=S*T=48*5/6=40kmtime=40/60hr=2/3hrNew speed = 40* 3/2 kmph= 60kmph

2)Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?

Solution::Let total distance be Stotal time=1hr24minA to T :: speed=4kmphdiistance=2/3ST to S :: speed=5kmdistance=1-2/3S=1/3S21/15 hr=2/3 S/4 + 1/3s /584=14/3S*3S=84*3/14*3= 6km

3)walking at ¾ of his usual speed ,a man is late by 2 ½ hr.the usual time is.

Solution::Usual speed = SUsual time = TDistance = DNew Speed is ¾ SNew time is 4/3 T4/3 T – T = 5/2T=15/2 = 7 ½

4)A man covers a distance on scooter .had he moved 3kmph faster he would have taken 40 min less. If he had moved 2kmph slower he would have taken 40min more.the distance is.

Solution::Let distance = x mUsual rate = y kmphx/y – x/y+3 = 40/60 hr2y(y+3) = 9x --------------1x/y-2 – x/y = 40/60 hr y(y-2) = 3x -----------------2

divide 1 & 2 equations

by solving we get x = 40

5)Excluding stoppages,the speed of the bus is 54kmph and including stoppages,it is 45kmph.for how many min does the busstop per hr.

Solution::Due to stoppages,it covers 9km less.time taken to cover 9 km is [9/54 *60] min = 10min

6)Two boys starting from the same place walk at a rate of5kmph and 5.5kmph respectively.wht time will they take to be 8.5km apart, if they walk in the same direction

Solution::The relative speed of the boys = 5.5kmph – 5kmph = 0.5 kmphDistance between them is 8.5 kmTime= 8.5km / 0.5 kmph = 17 hrs

7)2 trains starting at the same time from 2 stations 200km apart and going in opposite direction cross each other ata distance of 110km from one of the stations.what is the ratio oftheir speeds.

Solution::In same time ,they cover 110km & 90 km respectivelyso ratio of their speed =110:90 = 11:9

8)Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph resp. At the time of the meeting thesecond train has traveled 120km more than the first.the distance between them.

Solution::Let the distance traveled by the first train be x kmthen distance covered by the second train is x + 120kmx/50 = x+120 / 60x= 600so the distance between A & B is x + x + 120 = 1320 km

9)A thief steals a ca r at 2.30pm and drives it at 60kmph.the theft is discovered at 3pm and the owner sets off in another car at 75kmph when will he overtake the thief

Solution::Let the thief is overtaken x hrs after 2.30pmdistance covered by the thief in x hrs = distance covered by the owner in x-1/2 hr60x = 75 ( x- ½)x= 5/2 hrthief is overtaken at 2.30 pm + 2 ½ hr = 5 pm

10)In covering distance,the speed of A & B are in the ratio of 3:4.A takes 30min more than B to reach the destion.The time taken by A to reach the destinstion is.

Solution::

Ratio of speed = 3:4Ratio of time = 4:3let A takes 4x hrs,B takes 3x hrsthen 4x-3x = 30/60 hrx = ½ hrTime taken by A to reach the destination is 4x = 4 * ½ = 2 hr

11)A motorist covers a distance of 39km in 45min by moving at a speed of xkmph for the first 15min.then moving at double the speed for the next 20 min and then again moving at his original speed for the rest of the journey .then x=?

Solution::Total distance = 39 kmTotal time = 45 minD = S*Tx * 15/60 + 2x * 20/60 + x * 10/60 = 39 kmx = 36 kmph

12)A & B are two towns.Mr.Fara covers the distance from A t0 Bon cycle at 17kmph and returns to A by a tonga running at a uniform speed of 8kmph.his average speed during the whole journey is.

Solution::When same distance is covered with different speeds,then the average speed = 2xy / x+y=10.88kmph

13)A car covers 4 successive 3km stretches at speed of10kmph,20kmph,30kmph&:60kmph resp. Its average speed is.

Solution::Average speed = total distance / total time total distance = 4 * 3 = 12 kmtotal time = 3/10 + 3/20 + 3/30 + 3/60= 36/60 hrspeed =12/36 * 60 = 20 kmph 14)A person walks at 5kmph for 6hr and at 4kmph for 12hr.The average speed is.

Solution::avg speed = total distance/total time = 5*6 + 4*12 / 18=4 1/3 mph

15)A bullock cart has to cover a distance of 80km in 10hrs.If it covers half of the journeyin 3/5th time.wht should be its speed to cover the remaining distance in the time left.

Solution::Time left = 10 - 3/5*10= 4 hrspeed =40 km /4 hr =10 kmph

16)The ratio between the speeds of the A& B is 2:3 an therefore A takes 10 min more than the time taken by B to reach the destination.If A had walked at double the speed ,he would have covered the distance in ?

Solution::Ratio of speed = 2:3Ratio of time = 3:2A takes 10 min more3x-2x = 10 min A's time=30 min --->A covers the distance in 30 min ,if its speed is x-> He will cover the same distance in 15 min,if its speed doubles (i.e 2x)

17)A is twice as fast as B and B is thrice as fast as C is.The journey covered by B in?

Solution::speed's ratio a : b = 2: 1b : c = 3:1Time's ratiob : c = 1:3b : c = 18:54(if c covers in 54 min i..e twice to 18 min )

18)A man performed 3/5 of the total journey by ratio 17/20 bybus and the remaining 65km on foot.wht is his total journey.

Solution:: Let total distance is xx-(3/5x + 17/20 x) =6.5x- 19x/20 = 6.5x=20 * 6.5=130 km

19)A train M leaves Meerat at 5 am and reaches Delhi at 9am .Another train N leaves Delhi at 7am and reaches Meerut at 1030amAt what time do the 2 trains cross one another

Solution::Let the distance between Meerut & Delhi be xthey meet after y hr after 7amM covers x in 4hrN covers x in 3 ½ i.e 7/2 hrspeed of M =x/4speed of N = 2x/7Distance covered by M in y+2 hr + Distance covered by N in y hr is xx/4 (y+2) +2x/7(y)=xy=14/15hr or 56 min

20)A man takes 5hr 45min in walking to certain place and riding back. He would have gained 2hrs by riding both ways.The time he would take to walk both ways is?

Solution::Let x be the speed of walkedLet y be the speed of rideLet D be the distance

Then D/x + D/y = 23/4 hr -------1D/y + D/y = 23/4 – 2 hrD/y = 15/8 --------2substitute 2 in 1D/x + 15/8 = 23/4D/x = 23/4 -15/8 =46-15/8 =31/8Time taken for walk one way is 31/8 hrtime taken to walk to and fro is 2*31/8 = 31/4 hr=7 hr 45 min

Trains

General Concept:

(1) Time taken by a train x mt long in passing a signal postor a pole or a standing man = time taken by the train to cover x mt

(2) Time taken by a train x mt long in passing a stationaryobject of length y mt = time taken by the train to cover x+y mt

(3) Suppose two trains or two bodies are moving in the same direction at u kmph and v kmph such that u > v then their relative speed is u-v kmph

(4)If two trains of length x km and y km are moving in oppositediredtions at u kmph and vmph,then time taken by the train tocross each other = (x+y)/(u+v) hr

(5) Suppose two trains or two bdies are moving in opposite directionat u kmph and v kmph then,their relative speed = (u+v) kmph

(6)If two train start at the same time from 2 points A & B towardseach other and after crossing they take a & b hours in reaching B & A respectively then A's speed : B's speed = (b^1/2 : a^1/2 )

Problems (1)Find the time taken by a train 180m long,running at 72kmph in crossing an electric pole

Solution: Speed of the train =72*5/18m/s =20 m/s Distance move din passing the pole = 180m Requiredtime = 180/20 = 9 seconds

(2)A train 140 m long running at 60kmph.In how much time will it

pass a platform 260m long.

Solution: Distance travelled =140 + 260 m =400 m, speed = 60 * 5/18 = 50//3 m time=400*3 / 50 = 24 Seconds

(3)A man is standing on a railway bridge which is 180 m.He finds that a train crosses the bridge in 20 seconds but himself in8 sec. Find the length of the train and its sppeed

Solution: i)D=180+x T = 20 seconds S= 180+x / 20 ------------ 1 ii)D=x T=8 seconds D=ST x=8S ------------- 2 Substitute 2 in 1 S=180 + 8 S / 20 S=15 m/s Length of the train,x is 8 *15 = 120 m

(4)A train 150m long is running with a speed of 68 mphIn wht time will it pass a man who is running at a speed of 8kmph in the same direction in which the train is going

Solution: Relative Speed = 68-8=60kmph*5/18 = 50/3 m/s time= 150 * 3 /50 =9sec

5)A train 220m long is running with a speed of 59 k mph /..In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which train is going.

Solution: Relative Speed = 59+7=66kmph*5/18 = 55/3 m/s time= 220/55 * 3 =12sec

(6)Two trains 137m and 163m in length are running towards each other on parallel lines,one at the rate of 42kmph & another at 48 mph.In wht time will they be clear of each other from the moment they meet.

Solution: Relative speed =42+48 = 90 *5/18 = 25m/s time taken by the train to pass each other = time taken to cover (137+163)m at 25 m/s = 300 /25 s =12 s

(7)A train running at 54 kmph takes 20 sec to pass a platform. Next it takes 12 sec to pass a man walking at 6kmph in the same direction in which the train is going.Find length of the train

and length of platform

Solution: Relative speed w.r.t man = 54-6=48kmph the length of the train is 48 * 5/18 * 12 =160m time taken to pass platform =20 sec Speed of the train = 54 * 5/18 =15m/s 160+x =20 *15 x=140m length of the platform is 140m

(8)A man sitting in a train which is travelling at 50mph observesthat a goods train travelling in opposite irection takes 9 secto pass him .If the goos train is 150m long fin its speed

Solution: Relative speed =150/9 m/s =60 mph speed of the train = 60-50 =10kmph

(9)Two trains are moving in the sam e direction at 65kmph and 45kmph. The faster train crosses a man in slower train in18sec.thelength of the faster train is

Solution: Relative speed =65-45 kmph = 50/9 m/s Distancce covered in18 s =50/9 * 18 = 100m the length of the train is 100m

(10)Atrain overtakes two persons who are walking in the same direction in which the train is going at the rate of 2kmph an 4kmph and passes them completely in 9 sec an 10 sec respectively.The length of train is

Solution: 2kmph = 5/9 m/s 4 mph =10/9 m/s Let the length of the trainbe x meters and its speed is y m/s then x / (y- 5/9) = 9 and x / (y- 10/9) = 10 9y-5 =x and 10(9y-10)=9x 9y-x=5 and 90y-9x=100 on solving we get x=50,lenght of trains

(11) Two stations A & B are 110 km apart on a straight line.One train starts from A at 7am and travels towards B at 20kmph.Another train starts from B at 8am an travels toward A at a speedof 25kmph.At what time will they meet

Solution: Suppose the train meet x hr after 7am Distance covered by A in x hr=20x km 20x+25(x-1) = 110 45x=135 x=3 So they meet at 10 am

(12)A traintravelling at 48kmph completely crosses another train having half its length an travelling inopposite direction at 42kmph

in12 sec.It also passes a railway platform in 45sec.the length of platform is

Solution: Let the length of the first train be x mt then,the length of second train is x/2 mt relative speed = 48+42 kmph =90 * 5/18 m/s = 25m/s (x+ x/2)/25 =12 x=200 Length of the train is 200m Let the length of the platform be y mt speed f the first train = 48*5/18 m/s = 40/3 m/s 200+y * 3/40 = 45 y=400m

(13)The length of a running trsain in 30% more than the length ofanother train B runnng in the opposite direction.To find out the speed of trtain B,which of the following information given in the statements P & Q is sufficient P : The speed of train A is 80 kmph Q : They too 90 sec to cross each other(a) Either P & Q is sufficient(b)Both P & Q are not sufficient(c)only Q is sufficient(d)Both P & Q are neeed Ans: B

Solution: Let the length of th e train A be x mt Length of the train B = 130/100 x mt =13x/10 mt Let the speed of B be y mph,speed of the train A=80mph relative speed= y+80 * 5/18 m/s time taken by the trains t cross each other is gven by 90 = (x + 13x/10)/ (5y+400 / 18) to find y,clearly xis also needed so,both P & Q are not sufficient

(14)The speed of a train A,100m long is 40% more than then the speedof another train B,180m long running in opposite direction.To fin outthe speed of B,which of the information given in statements P & Q issufficient P :The two trains crossed each other in 6 seconds Q : The difference between the spee of the trains is 26kmph(a)Only P is sufficient(b)Only Q is sufficient(c)Both P & Q are needed(d)Both P & Q are not sufficientAns : A

Solution: Let speed of B be x kmph then,speed of A =140x/100 kmph =7x/5 mph relative speed = x + 7x/5 =2x/3 m/stime taken to cross each other = (100+180)*3/2x s =420/x snow,420/x = 6

x=70 mphthus,only P is sufficient

(15)The train running at certain speed crosses astationary enginein20 seconds.to find out the sped of the train,which of the following information is necessary(a)Only the length of the train(b)only the length of the engine(c)Either the length of the train or length of engine(d)Both the length of the train or length of engineAns : D

Solution:

Since the sum of lengths of the tran and the engine is needed, so both the length must be known

BOATS AND STREAMS

Important facts:

1)In water, the direction along the stream is called down stream.

2)Direction against the stream is called upstream.

3)The speed of boat in still water is U km/hr and the speed of stream is Vkm/hr then

speed down stream =U + V km/hrspeed up stream = U â€“ V km/hr

Formulas:

If the speed down stream is A km/hr and the speed up stream is B km/hr then

speed in still water = Â½(A+B) km/hr

rate of stream =1/2(A-B) km/hrPROBLEMS: 1. In one hour a boat goes 11 km long the stream and 5 km against the stream. The speed

of the boat in still water is?Sol:Speed in still water = Â½ ( 11+5) km/hr= 8 kmph

2.A man can row 18 kmph in still water. It takes him thrice as long as row up as to rowdown the river. find the rate of stream.Sol:Let man's rate up stream be xkmphthen, in still water =1/2[3x+x]=2x kmphso, 2x= 18, x=9rate upstream =9kmphrate downstream =27 kmphrate of stream = Â½ [27-9]= 9kmph

3.A man can row 71/2kmph in still watre . if in a river running at 1.5 km an hour, iftakes him 50 min to row to place and back. how far off is the place?Sol:speed down stream =7.5+1.5=9kmphspeed upstream =7.5-1.5=6kmphlet the required distence x km. then ,x/9+x/6=50/60 = 2x+3x= 5/6*185x=15, x=3Hence, the required distence is 3 km

4.A man can row 3 quarters of a km aganist the stream is 111/4 min. the speed of theman in still water is ?Sol:rate upstream = 750/625 m/sec =10/9 m/secrate downstream =750/450 m/sec = 5/3 m/secrate in still water =1/2[10/9+5/3] = 25/18 m/sec= 25/18*18/5=5 kmph

5.A boat can travel with a speed of 13 kmph in still water. if the speed of stream is 4 kmph,find the time taken by the boat to go 68 km downstream?

Sol: Speed down stream = 13+4= 17 kmphtime taken to travel 68km downstream =68/17 hrs = 4 hrs

6.A boat takes 90 min less to travel 36 miles downstream then to travel the samedistence upstream. if the speed of the boat in still water is 10 mph . the speed of the stream is :Sol:Let the speed of the stream be x mph .then, speed downstream = [10+x]mphspeed upstream =[10-x] mph36/[10+x] - 36/[10-x] = 90/60 =72x*60= 90[100-x2](x+50)(x-2) =0x=2 kmph

7.At his usual rowing rate, Rahul 12 miles down stream in a certain river in 6 hrs less than it takes him to travel the same distence upstream. but if he could double his usual rowing rate for his 24 miles roundthe down stream 12 miles would thentake only one hour less than the up stream 12 miles. what is the speed of the current in miles per hours?Sol:Let the speed in still water be x mph and the speed of the curren be y mph.then, speed upstream = (x-y)speed downstream =(x+y)12/(x-y) - 12/(x+y) = 66(x2 â€“ y2) m= 2xy => x2 â€“ y2 =4y -(1)and 12/(2x-y) - 12/(2x+y) =1 => 4x2 â€“ y2 = 24yx2= ( 24y + y2)/4 -->(2)from 1 and 2 we have4y+ y2 =( 24y+y2)/4 y=8/3 mphy= 22/3 mph

8.There is a road beside a river. two friends started from a place A, moved to a temple situated at another place B and then returned to A again. one of them moves on a cycle at a speed of 12 kmph, while the other sails on a boat at aspeed of 10 kmph . if the river flows at the speedof 4 kmph , which of the two friends will return to place A first ?Sol:Clearly, The cyclist moves both ways at a speed of 12 kmphso, average speed of the cyclist = 12 kmph the boat sailor moves downstream = (10+4) = 14 kmphupstream =(10-4) = 6 kmphSo, average speed of the boat sailor =[ 2*14*6]/[14+6] kmph=42/5 kmph =8.4 kmphSince, the average speed of the cyclist is greater, he will return to A first.

9.A boat takes 19 hrs for travelling downstream from point A to point B. and coming back to a point C midway between A and B. if the velocity of the sream is 4 kmph .and the speed of the boat in still water is 14 kmph. what is the distence between A and B?Sol:speed downstream =14+4 =18 kmphspeed upstream = 14 -4 = 10 kmphlet the distence between A and B be x km. then,x/18 + (x/2)/10 = 19x/18 + x/20 =1919x/180 =19 =>x = 180kmHence, the distence between A and B bw 180 km

ALLIGATION OR MIXTURES

Important Facts and Formula:1.Allegation:It is the rule that enables us to find the ratio in which two of moreingredients at the given price must be mixed to produce a mixture of a desired price.

2.Mean Price:The cost price of a unit quantity of the mixtureis called the mean price.

3.Rule of Allegation:If two ingredients are mixed then Quantity of Cheaper / Quantity of Dearer = (C.P of Dearer â€“ Mean Price) /(Mean Priceâ€“C.P of Cheaper).

C.P of a unit quantity of cheaper(c) C.P of unit quantity of dearer(d)

Mean Price(m)

(d-m) (m-c)

Cheaper quantity:Dearer quantity = (d-m):(m-c)

4.Suppose a container contains x units of liquid from which y unitsare taken out and replaced by water. After n operations the quantity of pure liquid = x (1 â€“ y/x)n units.

SOLVED PROBLEMS

Simple problems:

1.In what ratio must rice at Rs 9.30 per Kg be mixed with riceat Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg?

Solution: C.P of 1 Kg rice of 1st kind 930 p C.P of 1 Kg rice of 2n d kind

1080p

Mean Price 1000p

80 70

Required ratio=80:70 = 8:7

2.How much water must be added to 60 liters of milk at 11/2 litersfor Rs 20 so as to have a mixture worth Rs 10 2/3 a liter?

Solution:C.P of 1 lit of milk = 20*2/3 = 40/3

C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3

Mean Price 32/3

8/3 32/3

Ratio of water and milk =8/3 : 32/3 = 1:4Quantity of water to be added to 60 lit of milk =1/4*60=15 liters.

3.In what ratio must water to be mixed with milk to gain20% by selling the mixture at cost price?

Solution:Let the C.P of milk be Re 1 per literThen S.P of 1 liter of mixture = Re.1

Gain obtained =20%.Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6

C.P of 1 liter of water 0 C.P of 1 liter of milk1

Mean Price 5/6

1/6 5/6

Ratio of water and milk =1/6 : 5/6 = 1:5.

4.In what ratio must a grocer mix two varieties of pulsescosting Rs 15 and Rs 20 per Kg respectively so as to get

a mixture worth Rs 16.50 per Kg?

Solution:Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of 2nd kind 20

Mean Price Rs 16.50

3.50 1.50

Required ratio =3.50 : 1.50 = 35:15 = 7:3.

5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice at Rs 6 per Kg .Find the average price of the mixture?

Solution:

rice of 5 Rs per Kg rice of 6 Rs per Kg

Average price Aw

6-Aw Aw-5

(6-Aw)/(Aw-5) = 4/8 =1/212-2Aw = Aw-5

3Aw = 17Aw = 5.66 per Kg.

6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice toget a mixture costing Rs 7 per Kg. Find the price

of the costlier rice?

Solution:Using the cross method:

rice at Rs 6 per Kg rice at Rs x per Kg

Mean price Rs 7 per Kg

5 4

x-7:1=5:44x-28 = 54x=33=>x=Rs 8.25.Therefore price of costlier rice is Rs 8.25 per KgMedium Problems:

1.A butler stole wine from a butt of sherry which contained 40% of spirit and he replaced,what he had stolen by winecontaining only 16% spirit. The butt was then of 24% strength only. How much of the butt did he steal?

Solution:

Wine containing 40%spirit Wine containing 16% spirit

Wine containing 24% spirit

8 16

They must be mixed in the ratio of =1:2.

Thus 1/3 of the butt of sherry was left and hence the butler drew out 2/3 of the butt.

2.The average weekly salary per head of the entire staff of a factory consisting of supervisors and the laborers is Rs 60.The average salary per head of the supervisors is Rs 400 and that of the laborers is Rs 56.Given that the number of supervisors is 12.Find the number oflaborers in the factory.Solution:Average salary of laborer Rs 56 Average salary of supervisors Rs 400

Average salary of entire staff Rs 60

340 4

Number of laborer / Number of Supervisors = 340 / 4=85/1Thus,if the number of supervisors is 1,number of laborers =85.Therefore if the number of supervisors is 12 number of laborers 85*12=1020.

3.The cost of type 1 rice is Rs 15 per Kg and type 2 rice is Rs 20 per Kg. If both type1 and type 2 are mixed in theratio of 2:3,then the price per Kg of the mixed variety of rice is?

Solution:Let the price of the mixed variety be Rs x per Kg.

Cost of 1 Kg of type 1 rice Rs 15 Cost of 1 Kg of type 2 rice Rs 20 Mean Price Rs x

20-x x-15

(20-x) /( x-15) = 2/3

=> 60-3x = 2x-305x = 90=>x=18.

4.In what ratio must a grocer mix two varieties of tea worthRs 60 a Kg and Rs 65 a Kg so that by selling the mixtureat Rs 68.20 a Kg he may gain 10%?

Solution:S.P of 1 Kg of the mixture = Rs 68.20,gain =10%S.P of 1 Kg of the mixture = Rs (100/110*68.20)=Rs 62.

Cost of 1 Kg tea of 1st kind 60 Cost of 1 Kg tea of 2nd kind 65

Mean Price Rs 62 3 2

Required ratio =3:2.

5.A dishonest milkman professes to sell his milk at cost price but he mixes t with water and there by gains 25% .The percentage of water in the mixture is?

Solution:Let C. P of 1 liter milk be Re 1.Then S.P of 1 liter mixture=Re 1. Gain=25%C.P of 1 liter mixture =Re(100/125*1) = Re 4/5.

C.P of 1 liter milk Re 1 C.P of 1 liter of water 0

Mean Price 4/5

4/5 1/5

Ratio of milk to water =4/5 : 1/5 = 4:1Hence percentage of water n the mixture=1/5*100=20%.

12.A merchant has 1000Kg of sugar,part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole.The quantity sold at 18% profit is?

Solution:

Profit on 1st part 8% Profit on 2nd part 18%

Mean Profit 14%

4 6

Ratio of 1st and 2nd parts =4:6 =2:3.Quantity of 2nd ind =3/5*1000Kg =600 Kg.

6.A jar full of whiskey contains 40% alcohol. A part of this whiskey is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%.The quantity of whiskey replaced is?

Solution:Strength of first jar 40% Strength of 2nd jar 19% Mean Strength 26%

7 14

So,ratio of 1st and 2nd quantities =7:14 =1:2Therefore required quantity replaced =2/3.

7.A container contains 40lit of milk. From this container 4 lit of milk was taken out and replaced by water.This process was repeated further two times. How much milk is now contained by the container?

Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit=(40*9/10*9/10*9/10) = 29.16 litComplex Problems:

1.Tea worth Rs 126 per Kg are mixed with a third variety in the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,the price of the third variety per Kg will be?

Solution:Since First and second varieties are mixed in equal proportionsso their average price =Rs (126+135)/2 = 130.50.

So the mixture is formed by mixing two varieties ,one atRs 130.50 per Kg and the other at say Rs x per Kg in theratio 2:2 i e,1:1 we have to find x.

Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d kind Rs x.

Mean Price Rs 153

x-153 22.50

(x=153)/22.5 = 1 =>x-153 = 22.5 x = 175.50.Price of the third variety =Rs 175.50 per Kg.

2.The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

Solution:Let the C.P of milk be Re 1 per liter.Milk in 1 liter mixture of A = 4/7 liter.Milk in 1 liter mixture of B = 2/5 liter.Milk in 1 liter mixture of C = 1/2 liter.C.P of 1 liter mixture in A=Re 4/7C.P of 1 liter mixture in B=Re 2/5.Mean Price = Re 1/2.By rule of allegation we have:

C.P of 1 liter mixture in A C.P of 1 liter mixture in B 4/7 2/5

Mean Price Â½

1/10 1/14

Required ratio = 1/10 : 1/14 = 7:5.

3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44 and Rs 1.74 per Kg so that the mixture may be worth Rs 1.41 per Kg?

Solution:Step1:Mix wheat of first and third kind to get a mixtureworth Rs 1.41 per Kg.

C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 3rd kind 174p Mean Price 141p

33 21

They must be mixed in the ratio =33:21 = 11:7

Step2:Mix wheats of 1st and 2n d kind to obtain a mixtureworth of 1.41.per Kg.

C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d kind 144p Mean Price 141p 3 21

They must be mixed in the ratio = 3:21=1:7.

Thus,Quantity of 2n d kind of wheat / Quantity of3rd kind of wheat = 7/1*11/7= 11/1Quantities of wheat of 1st :2n d:3rd = 11:77:7.

4.Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio n whichthese mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8:5?

Solution:Let the C.P of spirit be Re 1 per liter.Spirit in 1 liter mix of A = 5/7 liter.C.P of 1 liter mix in A =5/7.Spirit in 1 liter mix of B = 7/13 liter.C.P of 1 liter mix in B =7/13.Spirit in 1 liter mix of C = 8/13 liter.C.P of 1 liter mix in C =8/13.

C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture in B 7/13

Mean Price 8/13

1/13 9/91

Therefore required ratio = 1/13 : 9/91 = 7:9.

5.A milk vendor has 2 cans of milk .The first contains 5% waterand the rest milk. The second contains 50% water. How much milkshould he mix from each of the container so as to get 12 litersof milk such that the ratio of water to milk is 3:5?

Solution:Let cost of 1 liter milk be Re 1.Milk in 1 liter mixture in 1st can = 3/4 lit.C.P of 1 liter mixture in 1st can =Re 3/4Milk in 1 liter mixture in 2n d can = 1/2 lit.C.P of 1 liter mixture in 2n d can =Re 1/2Milk in 1 liter final mixture = 5/8 lit.Mean Price = Re 5/8.

C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2

Mean Price 5/8

1/8 1/8

There ratio of two mixtures =1/8 :1/8 = 1:1.So,quantity of mixture taken from each can=1/2*12 = 6 liters.

6.One quantity of wheat at Rs 9.30 per Kg are mixedwith another quality at a certain rate in the ratio 8:7. If the mixture so formed be worth Rs 10 per Kg ,what is the rate per Kg of the second quality of wheat?

Solution:Let the rate of second quality be Rs x per Kg.

C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p

Mean Price 1000p

100x-1000p 70 p

(100x-1000) / 70 = 8/7700x -7000 = 560700x = 7560 =>x = Rs 10.80.Therefore the rate of second quality is Rs10.80

7.8lit are drawn from a wine and is then filled with water.This operation is performed three more times.The ratio ofthe quantity of wine now left in cask to that of the wateris 16:81. How much wine did the cask hold originally?

Solution:Let the quantity of the wine in the cask originally be x liters.Then quantity of wine left in cask after4 operations = x(1- 8/x)4lit.Therefore x((1-(8/x))4)/x = 16/81.(1- 8/x)4=(2/3) 4(x- 8)/x=2/33x-24 =2xx=24.

8.A can contains a mixture of two liquids A and B in theratio 7:5 when 9 liters of mixture are drawn off and the can is filled with B,the ratio of A and B becomes 7:9. How many liters of liquid A was contained by the can initially?

Solution:Suppose the can initially contains 7x and 5x liters of mixtures A and B respectively .

Quantity of A in mixture left = (7x- (7/12)*9 )lit= 7x - (21/4) liters.Quantity of B in mixture left = 5x - 5/12*9= 5x - (15/4) liters

Therefore (7x â€“ 21/4)/ (5x â€“ 15/4+9)=7/9(28x-21)/(20x +21)= 7/9(252x -189)= 140x +147 112x = 336=> x=3.So the can contains 21 liters of A.

9.A vessel is filled with liquid,3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn offand replaced with water so that the mixture may be half water and half syrup?

Solution:Suppose the vessal initially contains 8 liters of liquid.Let x liters of this liquid be replaced with water then quantity of water in new mixture= 3-(3x/8)+x liters.Quantity of syrup in new mixture = 5 - 5x/8 liters.Therefore 3 - 3x/8 +x = 5 - 5x/8 5x+24 = 40-5x10x = 16.x= 8/5.So part of the mixture replaced = 8/5*1/8 =1/5.

SIMPLE INTEREST

CONCEPT

->Principal or Sum:- The money borrowed or lent out for a certain period is called Principal or the Sum.

->Interest:- Extra money paid for using others money is called Interest.

->Simple Interest:- If the interest on a sum borrowed for a certain

period is

reckoned uniformly,then it is called Simple Interest.

->Formulae:-

Principal = P Rate = R% per annum Time = T years. Then,

(i)Simple Interest(S.I)= (P*T*R)/100

(ii) Principal(P) = (100*S.I)/(R*T) Rate(R) = (100*S.I)/(P*T) Time(T) = (100*S.I)/(P*R)

SIMPLE PROBLEMS 1.Find S.I on Rs68000 at 16 2/3% per annum for 9months.

Sol:- P=68000 R=50/3% p.a

T=9/12 years=4/3 years

S.I=(P*R*T)/100 =(68000*(50/3)*(3/4)*(1/100))

=Rs 8500

Note:- If months are given we have to converted into years by dividing 12 ie., no.of months/12=years

(2)Find S.I on Rs3000 at 18% per annum for the period from 4th Feb to

18th April 1995 Sol:- Time=(24+31+18)days =73 days

=73/365=1/5 years

P= Rs 3000 R= 18% p.a

S.I = (P*R*T)/100

=(3000*18*1/5*1/100) =Rs 108

Remark:- The day on which money is deposited is not counted while the

day on which money is withdrawn is counted.

3. In how many years will a sum of money becomes triple at 10% per annum.

Sol:- Let principal =P S.I = 2P

S.I = (P*T*R)/100 2P = (P*T*10)/100 T = 20 years

Note:-(I) Total amount = Principal + S.I (2) If sum of money becomes double means Total amount or Sum

= Principal + S.I = P + P = 2P

MEDIUM PROBLEMS

1. A sum at Simple interest at 13 1/2% per annum amounts to Rs 2502.50 after 4 years.Find the sum.

Sol:- Let Sum be x. then, S.I = (P*T*R)/100

= ((x*4*27)/(100*2)) = 27x/100 Amount = (x+(27x)/100) = 77x/50

77x/50 = 2502.50 x = (2502.50*50)/77

= 1625

Sum = 1625

2. A some of money becomes double of itself in 4 years in 12 years it will become how many times at the same rate.

Sol:- 4 yrs - - - - - - - - - P 12 yrs - - - - - - - - - ?

(12/4)* P =3P Amount or Sum = P+3P = 4 times

3. A Sum was put at S.I at a certain rate for 3 years.Had it been put at

2% higher rate ,it would have fetched Rs 360 more .Find the Sum. Sol:- Let Sum =P

original rate = R T = 3 years

If 2% is more than the original rate ,it would have fetched 360 more

ie., R+2

(P*(R+2)*3/100) - (P*R*3)/100 = 360 3PR+ 6P-3PR = 36000

6P = 36000 P = 6000 Sum = 6000.

4.Rs 800 amounts to Rs 920 in 3yrs at S.I.If the interest rate is

increased by 3%, it would amount to how much? Sol:- S.I = 920 - 800 = 120

Rate = (100*120)/(800*3) = 5% New Rate = 5 + 3 = 8% , Principal = 800 , Time = 3 yrs

S.I = (800*8*3)/100 = 192

New Amount = 800 + 192

= 992

5. Prabhat took a certain amount as a loan from bank at the rate of 8% p.a S.I and gave the same amount to Ashish as a loan at the rate of 12%

p.a . If at the end of 12 yrs, he made a profit of Rs. 320 in the deal,What was the original amount?

Sol:- Let the original amount be Rs x. T = 12

R1 = 8% R2 = 12% Profit = 320 P = x

(P*T*R2)/100 - (P*T*R1)/100 =320 (x*12*12)/100 - (x*8*12)/100 = 320

x = 2000/3 x = Rs.666.67

6. Simple Interest on a certail sum at a certain rate is 9/16 of the sum . if the

number representing rate percent and time in years be equal ,then the rate is.

Sol:- Let Sum = x .Then, S.I = 9x/16

Let time = n years & rate = n% n = 100 * 9x/16 * 1/x * 1/n

n * n = 900/16 n = 30/4 = 7 1/2%

COMPLEX PROBLEMS

1. A certain sum of money amounts t 1680 in 3yrs & it becomes 1920 in 7 yrs .What is the sum.

Sol:- 3 yrs - - - - - - - - - - - - - 1680 7 yrs - - - - - - - - - - - - -

1920 then, 4 yrs - - - - - - - - - - - - - 240 1 yr - - - - - - - - - - - - -

? (1/4) * 240 = 60 S.I in 3 yrs = 3*60 = 18012 Sum = Amount - S.I

= 1680 - 180 = 1500

we get the same amount if we take S.I in 7 yrs I.e., 7*60 =420

Sum = Amount - S.I = 1920 - 420

= 1500

2. A Person takes a loan of Rs 200 at 5% simple Interest. He returns Rs.100 at the end of 1 yr. In order to clear his dues at the end of

2yrs ,he would pay: Sol:- Amount to be paid

= Rs(100 + (200*5*1)/100 + (100*5*1)/100) = Rs 115

3. A Man borrowed Rs 24000 from two money lenders.For one loan, he paid 15% per annum and for other 18% per annum .At the end of one

year,he paid Rs.4050.How much did he borrowed at each rate?

Sol:- Let the Sum at 15% be Rs.x & then at 18% be Rs (24000-x)

P1 = x R1 = 15 P2 = (24000-x) R2 = 18

At the end of ine year T = 1

(P1*T*R1)/100 + (P2*T*R2)/100 = 4050 (x*1*15)/100 + ((24000-x)*1*18)/100 = 4050 15x + 432000 - 18x = 405000

x = 9000 Money borrowed at 15% = 9000

Money borrowed at 18% = (24000 - 9000) = 15000

4. What annual instalment will discharge a debt of Rs. 1092 due in 3

years at 12% Simple Interest ?

Sol:- Let each instalment be Rs x

(x + (x * 12 * 1)/100) + (x + (x * 12 * 2)/100) + x = 1092

28x/25 + 31x/25 + x =1092 (28x +31x + 25x) = (1092 * 25)

84x = 1092 * 25 x = (1092*25)/84 = 325

Each instalement = 325

5. If x,y,z are three sums of money such that y is the simple interest on x,z

is the simple interest on y for the same time and at the same rate of

interest ,then we have: Sol:- y is simple interest on x, means

y = (x*R*T)/100 RT = 100y/x z is simple interest on y,

z = (y*R*T)/100 RT = 100z/y 100y/x = 100z/y y * y = xz

6.A Sum of Rs.1550 was lent partly at 5% and partly at 5% and partly at 8% p.a Simple interest .The total interest received after 3 years was

Rs.300.The ratio of the money lent at 5% to that lent at 8% is: Sol:- Let the Sum at 5% be Rs x

at 8% be Rs(1550-x) (x*5*3)/100 + ((1500-x)*8*3)/100 = 300 15x + 1500 * 24 - 24x = 30000 x = 800

Money at 5%/ Money at 8% = 800/(1550 - 800) = 800/750 = 16/15

7. A Man invests a certain sum of money at 6% p.a Simple interest and

another sum at 7% p.a Simple interest. His income from interest after 2 years was Rs 354 .one fourth of the first sum is equal to one fifth of

the second sum.The total sum invested was:

Sol:- Let the sums be x & y R1 = 6 R2 = 7

T = 2 (P1*R1*T)/100 + (P2*R2*T)/100 = 354

(x * 6 * 2)/100 + (y * 7 * 2)/100 = 354 6x + 7y = 17700

â€”â€”â€”(1) also one fourth of the first sum is equal to one

fifth of the second sum x/4 = y/5 => 5x - 4y = 0 â€”â€” (2)

By solving 1 & 2 we get, x = 1200 y = 1500 Total sum = 1200 +1500

= 2700

8. Rs 2189 are divided into three parts such that their amounts after 1,2& 3 years respectively may be equal ,the rate of

S.I being 4% p.a in all cases. The Smallest part is: Sol:- Let these parts be x,y and[2189-(x+y)] then,

(x*1*4)/100 = (y*2*4)/100 = (2189-(x+y))*3*4/100

4x/100 = 8y/100 x = 2y

By substituting values (2y*1*4)/100 = (2189-3y)*3*4/100 44y = 2189 *12 y = 597

Smallest Part = 597

9. A man invested 3/3 of his capital at 7% , 1/4 at 8% and the remainder at 10%

If his annual income is Rs.561. The capital is: Sol:- Let the capital be Rs.x

Then, (x/3 * 7/100 * 1) + ( x/4 * 8/100 * 1)

+ (5x/12 * 10/100 * 1) = 561

7x/300 + x/50 + x/24 = 561 51x = 561 * 600

x = 6600

COMPOUND INTEREST

CONCEPT

Compound Interest:Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time ,say yearly or half-yearly or quarterly to settle the previous account.In such cases ,the amount after the first unit of time becomes the principalfor the 2nd unit ,the amount after second unit becomes the principal for the

3rd unit and so on.After a specified period ,the difference between the amount and the moneyborrowed is called Compound Interest for that period.

Formulae:

Let principal=p,Rate=R% per annum Time=nyears

1.When interest is compounded Annually,Amount=P[1+(R/100)]n2.When interest is compounded Halfyearly,Amount=P[1+((R/2)100)]2n3.When interest is compounded Quaterly,Amount=P[1+((R/4)100)]4n4.When interest is compounded Annually,but time in fractions say 3 2/5 yrsAmount=P[1+(R/100)]3[1+((2R/5)/100)]5.When rates are different for different years R1%,R2%,R3% for 1st ,2nd ,3rd yrs respectivelyAmount=P[1+(R1/100)][1+(R2/100)][1+(R3/100)]6.Present Worth of Rs.X due n years hence is given by Present Worth=X/[1+(R/100)]n

SIMPLE PROBLEMS

1.Find CI on Rs.6250 at 16% per annum for 2yrs ,compounded annually.Sol:

Rate R=16,n=2,Principle=Rs.6250Method1:

Amount=P[1+(R/100)]n=6250[1+(16/100)]2

=Rs.8410C.I=Amount-P

=8410-6250=Rs.2160Method2:

Iyear------------------6250+1000\\Interest for 1st yr on 6250

II yr---------------6250+1000+160\\Interest for I1yr on 1000

C.I.=1000+1000+160=Rs.2160

2.Find C.I on Rs.16000 at 20% per annum for 9 months compounded quaterly

Sol:MethodI:

R=20%12months------------------------20%

=> 3 months------------------------5%For 9 months,there are '3' 3months

--------16000+800--------16000+800+40

--------16000+800+40+10+2=>Rs.2522MethodII:

Amount=P[1+(R/100)]n=16000[1+(5/100)]3

=Rs.18522C.I=18522-16000

=Rs.2522MEDIUM PROBLEMS

1.The difference between C.I and S.I. on a certain sum at 10% per annum for 2 yrs is Rs.631.find the sumSol:MethodI:NOTE:a) For 2 yrs -------->sum=(1002D/R2)b) For 3 yrs -------->sum=(1003D/R2(300+R))Sum=1002*631/102 =Rs.63100MethodII:Let the sum be Rs.X,Then C.I.=X[1+(10/100)]2-XS.I=(X*10*2)/100=X/5C.I-S.I.=21X/100-X/5

=X/100X/100=631X=Rs.63100

2.If C.I on a certain sum for 2 yrs at 12% per annum is Rs.1590.What would be S.I?sol:C.I.=Amount-PrincipleLet P be XC.I=X[1+(12/100)]2-X=>784X/625-X=1590=>X=Rs.6250S.I=(6250*12*2)/100=Rs.1500

3.A sum of money amounts to Rs.6690 after 3 yrs and to Rs.10035bafter 6 yrs on C.I .find the sumsol:For 3 yrs,Amount=P[1+(R/100)]3=6690-----------------------(1)For 6 yrs,Amount=P[1+(R/100)]6=10035----------------------(2)(1)/(2)------------[1+(R/100)]3=10035/6690=3/2[1+(R/100)]3=3/2-----------------(3)substitue (3) in (1)p*(3/2)=6690=>p=Rs.4460sum=Rs.4460

4.A sum of money doubles itself at C.I in 15yrs.In how many yrs will it become 8 times?sol: Compound Interest for 15yrs p[1+(R/100)]15 p[1+(R/100)]15=2P=>p[1+(R/100)]n=8P=>[1+(R/100)]n=8=>[1+(R/100)]n=23=>[1+(R/100)]n=[1+(R/100)]15*3

since [1+(R/100)] =2n=45yrs

5.The amount of Rs.7500 at C.I at 4% per annum for 2yrs issol:Iyear------------------7500+300(300------Interest on 7500)IIyear ----------------7500+300+12(12------------4% interest on 300)Amount=7500+300+300+12=Rs.8112

6.The difference between C.I and S.I on a sum of money for 2 yrs at121/2% per annum is Rs.150.the sum issol:Sum=1002D/R2=( 1002*150) /(25/2)2=Rs.9600

7.If the S.I on sum of money at 15% per annum for 3yrs is Rs.1200,the C.I on the same sum for the same period at same rate is---------sol:S.I=1200P*T*R/100=1200P*3*5/100=1200=>P=Rs.8000C.I for Rs.8000 at 5% for 3 yrs is-------------8000+400-------------8000+400+20-------------8000+400+20+20+1C.I =400+400+20+400+20+20+1=Rs.1261

COMPLEX PROBLEMS

1.A certain sum amounts to Rs.7350 in 2 yrs and to Rs.8575in 3 yrs.Find the sum and rate%?sol:S.I. on Rs.7350 for 1yr =Rs.(8575-7350)=Rs.1225S.I. on Rs.7350 for 2yrs=Rs.2*1225=Rs.2450PTR/100=2450

=>P*2*R/100=2450Since S.I. on Rs.7350 for 1yr =Rs.(8575-7350)=Rs.1225 Rate R=100*1225/(7350*10=16 2/3%Since it is C.I ,Let sum be Rs.XThen X[1+(R/100)]2=7350=>X[1+(50/100)]2=7350=>X=7350*(36/49)Sum=Rs.5400

2.If the difference between C.I compounded halfyearly and S.I ona sum at 10% per annum for one yr is Rs.25 the sum issol:p[1+((R/2)/100)]2n-PTR/100=25P[1+((10/2)/100)]2n-P*1*10/100=25=>P=Rs.400

3.A man borrowed Rs.800 at 10 % per annum S.I and immediately lent the whole sum at 10% per annum C.I What does he gain at the end of 2yrs?sol:C.I.=Rs.[800[1+(10/100)2]-800]=Rs.168S.I=Rs.[800*10*2/100]=Rs.160Gain=C.I-S.I=Rs(168-160)=Rs.8

4.On what sum of money will be S.I for 3 yrs at 8% per annum be halfof C.I on Rs.400 for 2 yrs at 10% per annum?sol:C.I on Rs.400 for 2yrs at 10%=Rs.[400*[1+(10/100)]2-400]=Rs.84Required S.I =1/2*84=42/-New S.I=Rs.42,Time=3yrs Rate=8%Sum=Rs.[100*42/(3*8)]=Rs.175

5.A sum of money placed at C.I doubles itself in 5yrs .It will amount to8 times itself in-------------sol:p[1+(R/100)]5=2P=>[1+(R/100)]5=2To become 8 times =>8Pp[1+(R/100)]5=2^3P=[1+(R/100)]^(5*3)=[1+(R/100)]^15n=15years

Areas

Important Facts and Formulae:

Results On Triangle

1.Sum of the angles of a triangle is 180 degrees.

2.The sum of any two sides of a triangle is greaterthan third side.

3.Pythagoras Theorem:

In a right angled triangle (Hypotenuse)2 = (Base)2 +(Height)2

4.The line joining the mid point of a side of a triangleto the opposite vertex is called the MEDIAN.

5.The point where the three medians of a triangle meet,is called CENTROID. The centroid divides each of the medians in the ratio 2:1

6.In an isosceles triangle, the altitude from the vertex bisects the base

7.The median of a triangle divides it into two triangles of the same area.

8.The area of the triangle formed by joining the mid pointsof the sides of a given triangle is one-fourth of the areaof the given triangle.

Results On Quadrilaterals

1.The diagonals of a Parallelogram bisect each other.

2.Each diagonal of a Parallelogram divides it into twotriangles of the same area.

3.The diagonals of a Rectangle are equal and bisect

each other

4.The diagonals of a Square are equal and bisect each other at right angles.

5.The diagonals of a Rhombus are unequal and bisect each other at right angles.

6.A Parallelogram and a Rectangle on the same baseand between the same parallels are equal in area.

7.Of all he parallelogram of given sides the parallelogramwhich is a rectangle has the greatest area.

Formulae

1.Area of a RECTANGLE = length * breadth

Length = (Area/Breadth) and Breadth = (Area/Length)

2.Perimeter of a RECTANGLE = 2(Length + Breadth)

3.Area of a SQUARE = (side)2 = Â½ ( Diagonal)2

4.Area of four walls of a room = 2(length + breadth) * height

5.Area of a TRIANGLE = Â½ * base * height

6.Area of a TRIANGLE = âˆš[s * (s-a) * (s-b) * (s-c)],where a,b,c are the sides of the triangle and s = 1/2(a+b+c)

7.Area of EQUILATERAL TRIANGLE = âˆš(3/4)* (side)2

8.Radius of in circle of an EQUILATERAL TRIANGLE of side a = r / 2âˆš3

9.Radius of circumcircle of an EQUILATERAL TRIANGLE of side a = r / âˆš3

10.Radius of incircle of a triangle of area âˆ† and semi perimeter S = âˆ† / s

11.Area of a PARALLELOGRAM = (base * height)

12.Area of RHOMBUS = 1/2 (product of diagonals)

13.Area of TRAPEZIUM = =1/2 * (sum of parallel sides)* (distance between them)

14.Area of a CIRCLE = ï›‡ r2 where r is the radius

15.Circumference of a CIRCLE = 2ï›‡r

16.Length of an arc = 2 ï›‡rÃ¸ / 360, where Ã¸ is central angle

17.Area of a SECTOR = Â½ (arc * r) = ï›‡r2Ã¸ / 360

18.Area of a SEMICIRCLE = ï›‡r2 / 2

19.Circumference of a SEMICIRCLE = ï›‡r Simple Problems

1.One side of a rectangular field is 15m and one of its diagonalis 17m. Find the area of field?

Sol: Other side = âˆš[(17*17) â€“ (15*15)] = âˆš(289-225) = 8mArea = 15 * 8 =120 sq. m

2.A lawn is in the form of a rectangle having its sides in the ratio 2:3 The area of the lawn is 1/6 hectares. Find the lengthand breadth of the lawn.

Sol: let length = 2x meters and breadth = 3x mtNow area = (1/6 * 1000)sq m = 5000/3 sq m

2x * 3x = 5000/3 =>x * x =2500 / 9x = 50/3

length = 2x = 100/3 m and breadth = 3x = 3*(50/3) = 50m

3.Find the cost of carpeting a room 13m long and 9m broad with a carpet 75cm wide at the rate of Rs 12.40 per sq meter

Sol: Area of the carpet = Area of the room = 13* 9 =117 sq m length of the carpet = (Area/width) = 117 * (4/3) = 156 m

Cost of carpeting = Rs (156 * 12.40) = Rs 1934.40

4.The length of a rectangle is twice its breadth if its length is decreased by 5cm and breadth is increased by 5cm, the area of the rectangle is increased by 75 sq cm. Find the length of the rectangle.

Sol: let length = 2x and breadth = x then(2x-5) (x+5) â€“ (2x*x)=755x-25 = 75 => x=20length of the rectangle = 40 cm

5.In measuring the sides of a rectangle, one side is taken 5% in excess and the other 4% in deficit. Find the error percent in the area, calculate from the those measurements.

Sol: let x and y be the sides of the rectangle then correct area = (105/100 * x) * (96 / 100 *y)

=(504/500 xy) â€“ xy = 4/500 xyError% = 4/500 xy*(1/xy)*100 % = 4/5% = 0.8%

6.A room is half as long again as it is broad. The cost of carpeting the room at Rs 5 per sq m is Rs 2.70 and the cost ofpapering the four walls at Rs 10 per sq m is Rs 1720. If a doorand 2 windows occupy 8 sq cm. Find the dimensions of the room?

Sol: let breadth=x mt ,length= 3x/2 mt and height=h mt Area of the floor = (total cost of carpeting /rate) = 270/5 sq m = 54 sq m x * 3x/2=54 => x*x= 54*(2/3)=36 => x = 6m so breadth = 6m and length=3/2*6 = 9m now papered area = 1720 /10 = 172 sq m Area of one door and 2 windows =8 sq m

total area of 4 walls = 172+8 = 180 sq m2(9+6)*h = 180 => h=180/30 = 6m

7.The altitude drawn to the base of an isosceles triangle is 8cmand the perimeter is 32cm. Find the area of the triangle?

Sol: let ABC be the isosceles triangle, the AD be the altitudelet AB = AC=x then BC= 32-2x

since in an isoceles triange the altitude bisects the base so BD=DC=16-x in âˆ†ADC,(AC) 2 = (AD) 2 + (DC) 2

x*x=(8*8) + (16-x)*(16-x)32x =320 => x = 10BC = 32-2x = 32-20 = 12 cmHence, required area = Â½ * BC * AD

= Â½ * 12 * 10 = 60 sq cm

8.If each side of a square is increased by 25%, find the percentage change in its area?

Sol: let each side of the square be a , then area = a * aNew side = 125a / 100 = 5a / 4 New area =(5a * 5a)/(4*4) = (25aÂ²/16) â€“ aÂ²

= 9aÂ²/16Increase %= 9aÂ²/16 * 1/aÂ² * 100% = 56.25%

9.Find the area of a Rhombus one side of which measures 20cm and one diagonal 24cm.

Sol: Let other diagonal = 2x cm since diagonals of a rhombus bisect each other at right angles, we have 20Â² = 12Â² + xÂ² => x = âˆš[20Â² -12Â²]= âˆš256 = 16cm

so the diagonal = 32 cmArea of rhombus = Â½ * product of diagonals

= Â½ * 24 * 32= 384 sq cm

10. The area of a circular field is 13.86 hectares. Find the costof fencing it at the rate of Rs. 4.40 per meter.

Sol: Area = 13.86 * 10000 sq m = 138600 sq mï›‡rÂ²= 138600 => rÂ² = 138600 * 7/22 => 210 mcircumference = 2ï›‡r = 2 * 22/7 * 210m = 1320 mcost of fencing = Rs 1320 * 4.40 = Rs. 5808

Medium Problems:

11.Find the ratio of the areas of the incircle and circumcircle ofa square.

Sol: let the side of the square be x, then its diagonal = âˆš2 xradius of incircle = x/2 andradius of circmcircle =âˆš2 x /2 = x/âˆš2required ratio = ï›‡xÂ²/4 : ï›‡xÂ²/2 = Â¼ : Â½ = 1:2

12.If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

Sol: let original radius = r and new radius = 50/100 r = r/2original area = ï›‡rÂ² and new area = ï›‡(r/2)Â²decrease in area = 3 ï›‡rÂ²/4 * 1/ ï›‡rÂ² * 100 = 75%

13.Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively. Find the width of the ring.

sol: let the inner and outer radii be r and R metersthen, 2ï›‡r = 352/7 => r = 352/7 * 7/22 * Â½ = 8m2ï›‡R = 528/7 => R= 528/7 * 7/22 * Â½ = 12mwidth of the ring = R-r = 12-8 = 4m

14.If the diagonal of a rectangle is 17cm long and its perimeteris 46 cm. Find the area of the rectangle.

sol: let length = x and breadth = y then2(x+y) = 46 => x+y = 23xÂ²+yÂ² = 17Â² = 289now (x+y)Â² = 23Â² =>xÂ²+yÂ²+2xy= 529289+ 2xy = 529 => xy = 120area =xy=120 sq. cm

15.A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt

sol: area of theplot = 110 * 65 = 7150 sq m area of the plot excluding the path = (110-5)* (65-5) = 6300 sq marea of the path = 7150- 6300 =850 sq mcost of gravelling the path = 850 * 80/100 = 680 Rs

16. The perimeters of ttwo squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the differenceof the areas of the two squares.

sol: side of first square = 40/4 =10cmside of second square = 32/4 = 8cmarea of third squre = 10Â² â€“ 8Â² = 36 sq cmside of third square = âˆš36 = 6 cmrequired perimeter = 6*4 = 24cm

17. A room 5m 44cm long and 3m 74cm broad is to be paved with squretiles. Find the least number of squre tiles required to cover the floor.

sol: area of the room = 544 * 374 sq cmsize of largest square tile = H.C.F of 544cm and 374cm= 34cmarea of 1 tile = 34*34 sq cmno. of tiles required = (544*374) / (34 * 34) = 176

18. The diagonals of two squares are in the ratio of 2:5. Find the ratio of their areas.

sol: let the diagonals of the squares be 2x and 5x respectivelyratio of their areas = Â½ * (2x)Â² : Â½*(5x)Â² = 4:25

19.If each side of a square is increased by 25%. Find the percentage change in its area.

sol: let each side of the square be a then area = a Â²new side = 125a/100 = 5a/4new area = (5a/4)Â² = 25/16 aÂ²increase in area = (25/16)aÂ² - aÂ² = (9/16)aÂ²increase % = (9/16)aÂ² * (1/aÂ²) * 100 = 56.25%

20.The base of triangular field os three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18.Find its base and height.

sol: area of the field = total cost/ rate = 333.18 /24.68 = 13.5 hectares=> = 13.5 * 10000 = 135000 sq mlet the altitude = x mt and base = 3x mtthen Â½ *3x * x = 135000 => xÂ² = 90000 => x = 300base= 900m and altitude = 300m

21.In two triangles the ratio of the areas is 4:3 and the ratio of their heights is 3:4. Find the ratio of their bases?

Sol: let the bases of the two triangles be x &y and their heightsbe 3h and 4h respectively.(1/2*x*3h)/(1/2*y*4h) =4/3 => x/y = 4/3 *4/3 = 16/9

22.Find the length of a rope by which a cow must be tethered in orderthat it may be able to graze an area of 9856 sq meters.

Sol:clearly the cow will graze a circular field of area 9856 sq m and radius equal to the length of the rope.

Let the length of the rope be r mtsthen ï›‡rÂ²=9856 => rÂ²=9856*7/22 = 3136 => r=56m

23.The diameter of the driving wheel of a bus is 140cm. How manyrevolutions per minute must the wheel make inorder to keep a speed of66 kmph?

Sol: Distance to be covered in 1min = (66*1000)/60 m =1100mdiameter = 140cm => radius = r =0.7mcircumference of the wheel = 2*22/7*0.7 = 4.4mno of revolutions per minute = 1100/4.4 = 250

24.The inner circumference of a circular race track, 14m wide is 440m.Find the radius of the outer circle.

Sol: let inner radius be r meters.Then 2ï›‡r =440 => r=440*7/22*1/2 = 70mradius of outer circle = 70+4 =84m

25.A sector of 120 degrees, cut out from a circle, has an area of 66/7 sq cm. Find the radius of the circle.

Sol: let the radius of the circle be r cm. Thenï›‡rÂ²Ã¸/360 =66/7=> 22/7*rÂ²*120/360 = 66/7 =>rÂ² = 66/7 *7/22*3 =9radius = 3cm

26.The length of the room is 5.5m and width is 3.75m. Find the cost of paving the floor by slabs at the rate of Rs.800 per sq meter.

Sol: l=5.5m w=3.75marea of the floor = 5.5 * 3.75 = 20.625 sq mcost of paving = 800 *20.625 =Rs. 16500

27.A rectangular plot measuring 90 meters by 50 meters is to be enclosed by wire fencing. If the poles of the fence are kept 5 metersapart. How many poles will be needed?

Sol: perimeter of the plot = 2(90+50) = 280mno of poles =280/5 =56m

28.The length of a rectangular plot is 20 meters more than its breadth.If the cost of fencing the plot @ 26.50 per meter is Rs. 5300. What isthe length of the plot in meter?

Sol: let breadth =x then length = x+20perimeter = 5300/26.50 =200m2(x+20+x) =200 => 4x+40 =200x = 40 and length = 40+20 = 60m

29.A rectangular field is to be fenced on three sides leaving a side of20 feet uncovered. If the area of the field is 680 sq feet, how many

feet of fencing will be required?

Sol: l=20feet and l*b=680 => b= 680/20 = 34feetlength of fencing = l+2b = 20+68 =88 feet

30.A rectangular paper when folded into two congruent parts had a perimeter of 34cm foer each part folded along one set of sides andthe same is 38cm. When folded along the other set of sides. What isthe area of the paper?

Sol: when folded along the breadth we have 2(l/2 +b) = 34 or l+2b = 34...........(1) when folded along the length, we have 2(l+b/2)=38 or 2l+b =38.....(2)from 1 &2 we get l=14 and b=10Area of the paper = 14*10 = 140 sq cm

31.A took 15 seconds to cross a rectangular field diagonally walking atthe rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?

Sol: length of the diagonal = 52*15/60 =13msum of length and breadth = 68*15/60 = 17mâˆš(lÂ²+bÂ²)=13 or l+b = 17area =lb = Â½ (2lb) = Â½[(l+b)Â² â€“ (lÂ²+bÂ²)] = Â½[17Â² -169]=1/2*120 = 60 sq meter

32 . A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

sol: area of cross roads = 55*4 +35*4-4*4 = 344sq mcost of graveling = 344 *75/100 =Rs. 258

33.The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m

sol: perimeter = total cost / cost per m = 10080 /20 = 504mside of the square = 504/4 = 126mbreadth of the pavement = 3mside of inner square = 126-6 = 120marea of the pavement = (126*126)-(120*120)=246â€*6 sq m�cost of pavement = 246*6*50 = Rs. 73800

34.Amanwalked diagonally across a square plot. Approximately what was the percent saved by not walking along the edges?

Sol: let the side of the square be x meterslength of two sides = 2x meters diagonal = âˆš2 x = 1.414x msaving on 2x meters = .59x msaving % = 0.59x /2x *100%= 30% (approx)

36.A man walking at the speed of 4 kmph crosses a square field diagonally in 3 meters.The area of the field is

sol: speed of the man = 4*5/18 m/sec = 10/9 m/sectime taken = 3*60 sec = 180 seclength of diagonal = speed * time = 10/9 * 180 = 200mArea of the field = Â½ *(dioagonal)Â²= Â½ * 200*200 sq m = 20000sq m

37.A square and rectangle have equal areas. If their perimetersare p and q respectively. Then

sol: A square and a rectangle with equal areas will satisfy the relation p < q 38.If the perimeters of a square and a rectangle are the same,then the area a & b enclosed by them would satisfy the condition:

sol: Take a square of side 4cm and a rectangle having l=6cm and b=2cmthen perimeter of square = perimeter of rectanglearea of square = 16 sq cm area of rectangle = 12 sq cmHence a >b

39.An error of 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is

sol: 100cm is read as 102 cma = 100*100 sq cm and b = 102 *102 sq cmthen a-b = 404 sq cmpercentage error = 404/(100*100) = 4.04%

40.A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is

sol: area to be plastered = [2(l+b)*h]+(l*b)= 2(25+12)*6 + (25*12)= 744 sq mcost of plastering = Rs . 744*75/100 = Rs. 5581

41.The dimensions of a room are 10m*7m*5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m*3m. One window is of size 2m*1.5m and the other 2 windows are of size 1m*1.5m.The cost of painting the walls at Rs. 3 per sq m is

sol: Area of 4 walls = 2(l+b)*h=2(10+7)*5 = 170 sq mArea of 2 doors and 3 windows = 2(1*3)+(2*1.5)+2(1*1.5) = 12 sq marea to be planted = 170 -12 = 158 sq mcost of painting = Rs. 158 *3 = Rs. 474

42.The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is

sol: a = Â½ *15*12 = 90 sq cmb = 2a = 2 * 90 = Â½ * 20 *h => h= 18cm

43.The sides of a triangle are in the ratio of Â½:1/3:1/4. If the perimeter is 52cm, then the length of the smallest side is

sol: ratio of sides = Â½ :1/3 :1/4 = 6:4:3perimeter = 52 cm, so sides are 52*6/13 =24cm52*4/13 = 16cm52 *3/13 = 12cmlength of smallest side = 12cm

44.The height of an equilateral triangle is 10cm. Its area is

sol: aÂ² = (a/2)Â² +(10)Â² aÂ² â€“ aÂ²/4 = 100 =>3aÂ² = 100*4area = âˆš3/4 *aÂ² = âˆš3/4*400/3 = 100/âˆš3 sq cm

45.From a point in the interior of an equilateral triangle, the perpendicular distance of the sides are âˆš3 cm, 2âˆš3cm and 5âˆš3cm. The perimeter of the triangle is

sol: let each side of the triangle be â€˜aâ€™ cmthen area(AOB) +area(BOC)+area(AOC) = area(ABC)Â½ * a *âˆš3 +1/2 *a *2âˆš3 +1/2 * a*5âˆš3 = âˆš3/4 a Â² a/2âˆš3(1+2+5) = âˆš3/4 a Â² => a=16perimeter = 3*16 = 48cm Complex Probems:

1.If the area of a square with side a s equal to the area of a triangle with base a, then the altitude of the triangle is

sol: area of a square with side a = a Â² sq untsarea of a triangle with base a = Â½ * a*h sq untsa Â² =1/2 *a *h => h = 2aaltitude of the triangle is 2a

2.An equilateral triangle is described on the diagonal of a square. What is the ratio of the area of the triangle to that ofthe square?

Sol: area of a square = a Â² sq cmlength of the diagonal = âˆš2a cmarea of equilateral triangle with side âˆš2a = âˆš3/4 * (âˆš2a) Â²required ratio = âˆš3aÂ² : a Â² = âˆš3 : 2

3.The ratio of bases of two triangles is x:y and that of their areas is a:b. Then the ratio of their corresponding altitudes wll be

sol: a/b =(Â½ * x*H) /(1/2 * y * h)bxH = ayh =>H/h =ay/bxHence H:h = ay:bx

4 .A parallelogram has sides 30m and 14m and one of its diagonals is 40m long. Then its area is

sol: let ABCD be the given parallelogram area of parallelogram ABCD = 2* (area of triangle ABC)now a = 30m, b = 14m and c = 40ms = Â½(30+14+40) = 42mArea of triangle ABC = âˆš[ s(s-a)(s-b)(s-c) = âˆš(42*12*28*2 = 168sq marea of parallelogram ABCD = 2 *168 =336 sq m

5.If a parallelogram with area p, a triangle with area R and a triangle with area T are all constructed on the same base and allhave the same altitude, then which of the following statements is false?

Sol: let each have base = b and height = hthen p = b*h, R = b*h and T = Â½ * b*hso P = R, P = 2T and T = Â½ R are all correct statements

6.If the diagonals of a rhombus are 24cm and 10cm the area and the perimeter of the rhombus are respectively.Sol: area = Â½*diagonal 1 *diagonal 2= Â½ * 24 * 10= 120 sq cmÂ½ * diagonal 1 = Â½ * 24 = 12cmÂ½ * diagonal 2 = Â½ *10 =5 cmside of a rhombus = (12) Â² + (5) Â² = 169 => AB = 13cm

7.If a square and a rhombus stand on the same base, then the ratio of the areas of the square and the rhombus is:

sol: A square and a rhombus on the same base are equal in area

8.The area of a field in the shape of a trapezium measures 1440sq m. The perpendicular distance between its parallel sides is 24cm. If the ratio of the sides is 5:3, the length of the longer parallel side is:

sol: area of field =1/2 *(5x+3x) *24 = 96x sq m96x = 1440 => x = 1440 /96 = 15hence, the length of longer parallel side = 5x = 75m

9.The area of a circle of radius 5 is numerically what percent itscircumference?

Sol: required percentage = ï›‡(5)Â²/(2ï›‡*5) *100 = 250%

10.A man runs round a circular field of radius 50m at the speed of

12m/hr. What is the time taken by the man to take twenty rounds ofthe field?

Sol: speed = 12 k/h = 12 * 5/18 = 10/3 m/sdistance covered = 20 * 2*22/7*50 = 44000/7mtime taken = distance /speed = 44000/7 * 3/10 = 220/7min

11.A cow s tethered in the middle of a field with a 14feet long rope.If the cow grazes 100 sq feet per day, then approximately what time will be taken by the cow to graze the whole field?

Sol: area of the field grazed = 22/7 * 14 * 14 = 616 sq feet

12.A wire can be bent in the form of a circle of radius 56cm. If it is bent in the form of a square, then its area will besol: length of wire = 2ï›‡ r = 2 *22/7 *56 = 352 cmside of the square = 352/4 = 88cmarea of the square = 88*88 = 7744sq cm

13.The no of revolutions a wheel of diameter 40cm makes intraveling a distance of 176m is

sol: distance covered in 1 revolution = 2ï›‡ r = 2 *22/7 *20 = 880/7 cmrequired no of revolutions = 17600 *7/880 = 140

14.The wheel of a motorcycle 70cm in diameter makes 40revolutions in every 10sec.What is the speed of motorcyclen km/hr?

Sol: distance covered in 10sec = 2 *22/7 *35/100 *40 =88mdistance covered in 1 sec =88/10m = 8.8mspeed =8.8m/s = 8.8 * 18/5 *k/h = 31.68 k/h

15.Wheels of diameters 7cm and 14cm start rolling simultaneouslyfrom x & y which are 1980 cm apart towards each other in oppositedirections. Both of them make the same number of revolutions per second. If both of them meet after 10seconds.The speed of the smaller wheel is

sol: let each wheel make x revolutions per sec. Then(2ï›‡ *7/2 *x)+(2ï›‡ * 7*x)*10 = 1980(22/7 *7 * x) + (2 * 22/7 *7 *x) = 19866x = 198 => x = 3distance moved by smaller wheel in 3 revolutions = 2 *22/7 *7/2 *3 = 66cmspeed of smaller wheel = 66/3 m/s = 22m/s

16.A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the poolis 11/25 that of the pool, then the radius of the pool is?

Sol: let the radius of the pool be R ftradius of the pool including the wall = (R+4)ftarea of the concrete wall = ï›‡ [(R+4)2 - R2 ]=> = ï›‡[R+4+R][R+4-R]= 8ï›‡(R+2) sq feet8ï›‡(R+2) = 11/25 ï›‡ R2 => 11 R2 = 200 (R+2)Radius of the pool R = 20ft

17.A semicircular shaped window has diameter of 63cm. Itsperimeter equals

sol: perimeter of window = ï›‡r +2r = [22/7 * 63/2 +63] = 99+63 = 162 cm

18.Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

sol: required area = (area of an equilateral triangle of side 7 cm) - (3 * area of sector with Ã˜ = 6o degrees and r = 3.5cm)= ( âˆš Â¾ * 7 * 7) â€“ (3* 22/7 *3.5 *3.5*60/360 ) sq cm= 49âˆš3/4 â€“ 11*0.5*3.5 sq cm = 1.967 sq cm

19. Four circular cardboard pieces, each of radius 7cm are placedin such a way that each piece touches two other pieces. The area of the space encosed by the four pieces is

sol: required area = 14*14 â€“ (4 * Â¼ * 22/7 * 7 *7) sq cm= 196 â€“ 154 = 42 sq cm

RACES AND GAMES OF SKILL


Races :- A contest of speed in running ,riding,driving,sailing or rowing is called a Race

Race Course :-The ground or path on which contests are made is called

a race course

STARTING POINT :-The point from which a race begins is called starting point.

Winning point or goal:-The point set to bound a race is called a winning point.

Dead Heat Race:-If all the persons contesting a race reach the goal exactly

at the same time,then the race is called a dead heat race.

Start:-suppose A and B are two contestants in a race .If before the start of

the race,A is at the satrtint point and B is ahead of A by 12 metres.

Then we say that "A gives B a start 12 metres.

->To cover a race of 100metres in this case,A will have to cover 100m while B

will have to cover 88m=(100-12)

->In a100m race 'A can give B 12m' or 'A can give B a start of 12m' or 'A beats

B by 12m'means that while A runs 100m B runs 88m.

GAMES:- A game of 100m,means that the person among the contestants who

scores 100 points first is the winner.

If A scores 100 points while B scores only 80 points then we say that 'A can give

B 20 points'.


PROBLEMS :- 1) In a 1 km race,A beats B by 28 m or 7sec.Find A's time over the course? Sol: B covers 28 m in 7sec so,B's time over the course = 7/28 *1000 =250 sec A's time over the course =250-7 =243 sec = 4min ,3 sec.

2) A runs 1 3/4 times as fast as B.If A gives B a start of 84 m,how far must be winning post be so that A and B might reach it at the same time ? Sol: Ratio of rates of A:B =7/4 :1 =7 :4 In a game of 7 m A gains 3m over B. 3m are gained by A in a race of 7m 84 m are gained by A in a race of 7*84/3 =196 m Winning post must be 196m away from the starting point.

3)A can run 1km in 3 min ,10sec and B can cover same distance in 3 min 20 sec. By what distance A beat B? Sol: clearly A beats B by 10 sec. Distance covered by B in 10 sec =1000/200 *10 =50 m A beats B by 50 metres.

4) In a 100m race,A runs at 8km per hour.If A gives b a start of 4 m and still beats him by 15 sec,what is the speed of B? Sol: 8000 m -------60*60 sec 100m ------- 60*60*1000/8000 =45 sec. Time taken by A to cover 100m =45 sec. B covers 100-4 m =96 min 45 sec =60 sec B's speed =96 *60*60/60*1000 =5.76 km/hr

5) A and B take part in 100m race .A runs at 5km per hour. A gives B a start of 8 m and still beats him by 8 sec. What is the speed of B? Sol : A'speed = 5km/hr =5*5/18 =25/18 m/s Time taken by A to cover 100m =100*18/25 =7.2 sec Time taken by A to cover 92m = 72+8 =80 sec B's speed =92*18/80*5 = 4.14 kmph.

6) A runs 1 2/3 times as fast as B.If A gives B a start of 80 m, how far must the winning post be so that A and B might reach if at the same time? Sol: Ratio of the speed of A and B =5/3 :1 Thus in a race of 5m ,A gains 2m over B 2m are gained by in a race of 5m 80 m will be gained by A in a race of 5/2* 80 =200 m

7) A,B and C are three contestans in a kmrace .If A can give B a start of 40 m and A can give C a start of 64 m,how amny metres start can b give C? Sol: A covers 100m,B covers (1000-40) =960 m C covers 1000-64 m or 936m when B covers 960 m,C covers 936 m when B covers 1000 m,C covers 936*1000/960 m =975 m B can give C a start of 1000-975 or 25 m.

8) In a 100m race,A covers the distance in 36 sec and B in 45 sec.In this race A beats B by? Sol: Distance covered by B in 9secs =100*9/45 =20 m A beats B by 20m

9) In a 200 m race A beats B by 35m or 7 sec.What is the A's time over the course? Sol: B runs 35 m in 7sec. B covers 200m in =7*200/35 = 40 sec. B's time over the course =40 sec A's time over the course =40-7 =33 sec.

10) In a 300 m race A beats B by 22.5 m or 6 sec.What is the B's time over the course? Sol: B runs 22.5 m in 6sec. B runs 300m in =6*300*2/45 =80 sec. B's time over the course =80 sec.

11) A can run 22.5 m while B runs 25 m.In a kilometre race B beats A by? Sol: B runs 25 m ,A runs 45/2 m B runs 1000 A runs = 1000*45/2*25 =900m B beats A by 100m

12)In a 500 m race, the ratio of the speeds of two contestants A and B is 3:4.A has a start of 140 m.Then ,A win by B? Sol: The speeds of A and B =3:4 To reach the winning post A will have to cover a distance of 500-140 m i.e 360 m while A covers 3m ,b covers 4m A covers 360m B covers 4/3*360=480m Thus when A reaches the winning post, b covers 480m and therefore remains 20m behind. A wins by 20m.

13) In a 100m race,A can beat B by 25 m and B can beat c by 4m.In the same race A can beat C by/ Sol: If A:B =100 :75 B :C=100 :96 then A :C =A/B*B/C=100/75* 100/96 = 100/72 A beats C by 100 -72 m=28 m

14) In a 100 race,A can give B 10 mand C 28 m.In the same race B can give C? Sol: A:B =100 :90 A :C=100 :72 B:C =B/A *A/C =90/100*100/72 =90/72 When B runs 90 C runs 72 when B runs 100 C runs =72*100/90 B beats C by 20m

15) In a 100m race ,A beats B by 10 m and C by 13 m.In the race of 180m. B will beat c by? Sol :A : B =100 :90 A/B =100/90 A/C =100/87 B/C =B/A *A/C =90 /87 When B runs 90m C runs 87 When B runs 180 m then C

runs =87 *180/90 B beats C by 180-174= 6m

16) In a race of 200 m, A can beat B by 31 m and C by 18m.In arace of 350 m, C will beat B by? Sol : In a race of 200 m A :B =200:169 A :C =200 :182 C/B =C/A*A/B =182/200*200/169 When C runs 182 m B runs 169 when C runs 350 m B runs =350*169/182 =325m

17) In agame of 100 points A can give B 20 points and C 28 points then B can give C? In a game of 100 points. A :B =100 :80 A :C =100 :72 B/C=B/A*A/C =80/100*100/72 = 80/72 when B runs 80m C runs 72 when B runs 100m C runs =100*72/80 =90 B can give C 10 points in agame of 100.

18)At a game of billiards,A can give B 15 points in 60 and A can give C 20 points in60.How amny points can B give C in a game of 90? Sol: A:B =60:45 A:C =60:40 B/C =B/A*A/C =45/60*60/40 =90/80 B can give C 10 points in agame of 90.

19) in agame of 80 points,A can give B 5 points and C 15 points.then how many points B can give C in agame of 60? Sol: A :B =80 :75 A :C =80:65 B/C=B/A*A/C = 75/80*80/65 =15/13 B:C =60:52

Clocks

General Concepts:

The face or dial of a watch is a circle whose circumference is divided into 60 equal parts,called minute spaces.

A clock has two hands, the Smaller one is called the hour hand or short hand while the larger one is called the minute hand orlong hand.

Important points:

a) In every 60 minutes, the minute hand gains 55 minutes on thehour hand

b)In every hour, both the hands coincide once ,i.e 0 degrees.

c)the hands are in the same straight line when they are coincidentor opposite to each other. i.e 0 degrees or 180 degrees.

d)when the two hands are at right angles, they are 15 minute spacesapart,i.e 90 degrees.

e)when the hands are in the opposite directions,they are 30 minutespaces apart,i.e 180 degrees.

f)Angle traced by hour hand in 12hrs = 360 degrees.

g)Angle traced by minute hand in 60 min = 360 degrees. If a watchor a clock indicated 8.15,when the correct time is 8, it is said to be 15 minutes too fast. On the other hand, if it indicates 7.45,when the correct time is 8,it is said to be 15 minutes slow.

h)60 min --> 360 degrees 1 min --> 60

i)the hands of a clock coincide in a day or 24 hours is 22 times,in 12hours 11minutes.

j)the hands of clock are straight in a day is 44 times .

k)the hands of a clock at right angle in a day is 44 times .

l)the hands of a clock in straight line but opposite in direction is22 times per day Simple Problems:

Type1:Find the angle between the hour hand and the minute hand of a clock when the time is 3.25

solution : In this type of problems the formulae is as follows30*[hrs-(min/5)]+(min/2)In the above problem the given data is time is 3.25. that isapplied in the formulae 30*[3-(25/5)]+(25/2)30*(15-25)/5+25/2 = 30*(-10/5)+25/2 = -300/5+25/2 = -600+(25/2)=-475/10=-47.5 i.e 47 1/20therefore the required angle is 47 1/20.

Note:The -sign must be neglected.Another shortcut for type1 is :The formulae is 6*x-(hrs*60+X)/2Here x is the given minutes,so in the given problem the minutes is 25 minutes,that is applied in the given formulae 6*25-(3*60+25)/2 150-205/2 (300-205)/2=95/2 =47 1/20.therefore the required angle is 47 1/20.

Type2:At what time between 2 and 3 o' clock will be the hands of aclock be together?

Solution : In this type of problems the formulae is 5*x*(12/11)Here x is replaced by the first interval of given time. here i.e 2. In the above problem the given data is between 2 and 3 o' clock 5*2*12/11 =10*12/11=120/11=10 10/11min.Therefore the hands will coincide at 10 10/11 min.past2.

Another shortcut for type2 is:Here the clocks be together but not opposite to each other so the angle is 0 degrees. so the formulae is 6*x-(2*60+x)/2=06*x-(120+x)/2=012x-120-x=0 11x=120 x=120/11=10 10/11therefore the hands will be coincide at 10 10/11 min.past2. Medium Problems

Type3:At what time between 4 and 5 o'clock will the hands of a clockbe at rightangle?

Solution : In this type of problems the formulae is(5*x + or -15)*(12/11)Here x is replaced by the first interval of given time here i.e 4

Case 1 : (5*x + 15)*(12/11)(5*4 +15)*(12/11)(20+15)*(12/11)35*12/11=420/11=38 2/11 min.Therefore they are right angles at 38 2/11 min .past4

Case 2 : (5*x-15)*(12/11)(5*4-15)*(12/11)(20-15)*(12/11)5*12/11=60/11 min=5 5/11minTherefore they are right angles at 5 5/11 min.past4.

Another shortcut for type 3 is:Here the given angle is right angle i.e 900.

Case 1 : The formulae is 6*x-(hrs*60+x)/2=Given angle6*x-(4*60+x)/2=906*x-(240+x)/2=9012x-240-x=18011x=180+24011x=420x=420/11= 38 2/11 min

Therefore they are at right angles at 38 2/11 min. past4.

Case 2 : The formulae is (hrs*60+x)/2-(6*x)=Given angle(4*60+x)/2-(6*x)=90(240+x)/2-(6*x)=90240+x-12x=180

-11x+240=180240-180=11x x=60/11= 5 5/11 min

Therefore they art right angles at 5 5/11 min past4.

Type 4:Find at what time between 8 and 9 o'clock will the hands of a clock be in the same straight line but not together ?

Solution : In this type of problems the formulae is

(5*x-30)*12/11x is replaced by the first interval of given time Here i.e 8

(5*8-30)*12/11(40-30)*12/1110*12/11=120/11 min=10 10/11 min.

Therefore the hands will be in the same straight line but nottogether at 10 10/11 min.past 8.

Another shortcut for type 4 is:Here the hands of a clock be in the samestraight line but not together the angle is 180 degrees.The formulae is (hrs*60+x)/2-(6*x)=Given angle(8*60+x)/2-6*x=180(480+x)/2-(6*x)=180480+x-12*x=36011x=480-360x=120/11=10 10/11 min.therefore the hands will be in the same straight line but nottogether at 10 10/11 min. past8. Type 5:At what time between 5 and 6 oâ€™ clock are the hands of a 3 minutes apart ?

Solution : In this type of problems the formuae is (5*x+ or - t)*12/11

Here x is replaced by the first interval of given time here xis 5.t is spaces apart

Case 1 : (5*x+t)*12/11 (5*5+3)*12/1128*12/11 = 336/11=31 5/11 mintherefore the hands will be 3 min .apart at 31 5/11 min.past5.

Case 2 : (5*x-t)*12/11(5*5-3)*12/11(25-3)*12/11=24 mintherefore the hands wi be 3 in apart at 24 min past 5.

Typicalproblems

problems:A watch which gains uniformly ,is 5 min,slow at 8 o'clock inthe morning on sunday and it is 5 min.48 sec.fast at 8 p.m on

following sunday. when was it correct?

Solution : Time from 8 am on sunday to 8 p.m on following sunday = 7 days 12 hours = 180 hoursthe watch gains (5+(5 4/5))min .or 54/5 min. in 180 hoursNow 54/5 minare gained in 180 hours.Therefore 5 minutes are gained in(180*5/54*5)hours=83 hours20 min. =3 days11hrs20min.therefore watch is correct at 3 days 11 hours 20 minutes after 8 a.mof sundaytherefore it wil be correct at 20 min.past 7 p.m on wednesday

TRUE DISCOUNT

CONCEPT

Let rate=R%per annum, Time= T years

1.Present worth (PW) = (100*Amount)/(100+(R*T))

= (100* TrueDiscount)/(R*T)

2.TrueDiscount (TD) = (P.W*R*T)/100

= (Amount*R*T)/(100+(R*T))

3.Sum =(SimpeInterest*TrueDiscount)/(SimpleInterest-TrueDiscount)

4.SimpleInterest-TrueDiscount=SimpeInterest on TrueDiscount

5.When the sum is put at CompoundInterest,then

PresentWorth=Amount/(1+(R/100))^T

GeneralConcept: Suppose a man has to pay Rs.156 after 4 years and the rate of interest is 14%per annum Clearly ,Rs.100 at 14% will amount to Rs156 in 4 years So,the payment of Rs.100 now wil cear off the debtof Rs156 due 4 years henceWe say that Sum due = Rs156 due 4 years hence

Present Worth =Rs100True Discount=Rs.156-Rs100=Rs56=Sumdue-PWWe define TrueDiscount= Interest on Present Worth(PW)Amount = PresentWorth+TrueDiscount

SIMPLE PROBLEMS

1.Find the present worth of Rs.930 due 3 years hence at 8% per

annum.Aso find the discount?

Sol: Amount=RS.930,Time=3years,Rate=8%

TrueDiscount = (Presentworth*Time*Rate)/100

Presentworth = (Amount*100) /(100+(R*T))

Presentworth = (930*100)/(100+(8*3)

=Rs.750

TrueDiscount = (930*3*8)/(100+(8*3)) =Rs.180 (or)

TrueDiscount = Amount-Presentworth

=Rs.930-Rs.750

=Rs.180

2.The truediscount on a bill due 9 months hence at 12% per

annum is Rs540.Find the amount of the bill and its presentworth?

Sol: Time=9months=9/12years=3/4years

Rate=12%

TrueDiscount=Rs540

TrueDiscount=(Amount*100)/(100+(R*T))

=>Amount=(TrueDiscount*(100+(R*T))) /100

=(540*(100+12*(3/4))/100

=Rs.6540

Presentworth=Amount-TrueDiscount

=Rs.6540-Rs540

=Rs.6000

3.The TrueDiscount on a certain sum of money due 3 years hence is Rs.250

and SimpeInterest on the same sum for same time and same rate is Rs375

find sum and rate%?

Sol: Time=3 years

Truediscount=Rs.250

SimpeInterest=Rs375

Sum = (SimpeInterest*TrueDiscount)/(SimpeInterest-TrueDiscount)

= (375*250)/(375-250)=Rs.750SimpeInterest=(Principle*Time*Rate)/100375 = ( 750*3R)/100R=50/3%=> Rate=162/3%

4.The difference between SimpeInterest and TrueDiscount on a certain sum

of money for 6 months at 121/2% per annum is Rs25.find the sum.

Sol: Let amount be Rs.xSimpleInterest=(Amount*T*R)/100TrueDiscount=(Amount*T*R)/(100+(R*T))SI-T.D=Rs25=>((x*6/12*25/2)/100)-((x*6/12*25/2)/(100+(6/12*25/2))=25=>x=Rs.6800

5.The Present worth of Rs.2310 due 21/2yrs henceThe interest being 15%per annum issol: Amount=Rs2310,Time=21/2yrs,Rate=15%

Presentworth = (Amount*100) /(100+(R*T))

=2310*100/(100+15*5/2)

=Rs1680.

MEDIUM PROBLEMS

1.A bill falls due in 1 year.The creditor agrees to acceptimmediate payment of the half and to defer the payment of the other half for 2 years .By this arrangement he gains Rs.40.what is the amount of bill,if the money be worth 121/2%?

sol: Let the amount be Rs.xIf the bill falls due for 1 yr,Truediscount=(Amount*100)/(100+R*T)=(x*100)/(100+(25/2*1)If Accepting immediate payment of the half and to defer the payment of the other half for 2 years, Truediscount=(Amount*100)/(100+R*T)=[x/2+((x/2*100)/(100+(25/2*2))]x/2 for immediate payment,((x/2*100)/(100+(25/2*2)) for paying after 2 yrsHe gains Rs40=>[x/2+((x/2*100)/(100+(25/2*2))] â€“ (x*100)/(100+(25/2*1) = Rs40=> x/2+2x/5-8x/9=40=>x=Rs.3600

Amount of the bill=Rs.3600

2.If the truediscount on a sum due 2yrs hence at 14% per annum be Rs168.The sum due is?Sol: Time=2yrs,Rate=14%Truediscount=Rs168,Amount=xTrueDiscount=(Amount*R*T)/(100+R*T)168 =(x*14*2)/(100+(14*2))=>x=Rs.768

3.The truediscount on Rs.2562 due 4 months hence is Rs.122.The rate % is?Sol: Amount=Rs.2562 Time=4/12yrsTrueDiscount=Rs122Rate=X%

TrueDiscount=(Amount*R*T)/(100+R*T)=>122=(2562*x*(4/12))/(100+(x*(4/12))=>36600+122x=2562x=>Rate=15%

4.The Truediscount on Rs1760 due after a certain time at12% per annum is Rs160The time after which it is due is:Sol: Amount=Rs.1760Rate=12%Truediscount=Rs160Time=xTrueDiscount=(Amount*R*T)/(100+R*T)=>x*11*12=100+12x=>x=5/6yrTime=10months

5.The interest on Rs.760 for 2yrs is the same as theTruediscount on Rs960 due 2yrs hence .If the rate of interest is same in both cases,it is

sol: Principal=Rs.760Time=2yrsamount=Rs960P*T*R/100=(Amount*R*T)/(100+R*T)=>(750*2*r)/100=(960*r*2)/(100+2r)=>150r=2100rate=14%

6.The SimpleInterest & TrueDiscount on a certain sum ofmoney for a given time & at a given rate are Rs85 & Rs.80.The sum is.............sol:S.I= Rs.85TrueDiscount=Rs80Sum=(S.I.*TD)/(SI-T.D)=Rs.1360

7.A trader owes a maerchant Rs10028 due 1yr hence The trader wants to settle the account after 3 months .If the rate of interest is 12%per annum How much cash should he pay?Sol:Time=1yr but he settles account after 3monthsso time = 9monthsCash to pay=(Amount*R*T)/(100+R*T)=(10028*100)/(100+12*(9/12))=Rs.9200

8.A man buys a watch for Rs1950 in cash and sells it for Rs2200 at a credit of 1yrThe rate of interest is 10% per annum Then he gains / loose---------------amount?Sol:If he sells it for Rs2200 at a credit of 1yr then thepresent worth of that amount isPresentworth = (Amount*100) /(100+(R*T))=(2200*100)/(100+10*1)=2000/-So he gains 2000-1950=Rs.50

9.A owes Rs.1573 payable 11/2yrs hence ,Aso B owes A Rs.144450 payabe 6months hence If they want to settle the accunt forth with ,keeping 14% as the rate of interest ,then who should pay& how much?Sol:B->A-----------Amount=Rs.1573Rate=14%Time=3/2Presentworth = (Amount*100) /(100+(R*T))=(1573*100)/(100+(3/2)*14)=Rs1300A->B------------Amount=Rs.1444.50Rate=14%Time=1/2Presentworth = (Amount*100) /(100+(R*T))=(1444.50*100)/(100+(1/2)*14)=Rs1350so B should pay Rs.1350

10.If Rs20 is the TrueDiscount on Rs260due after a certain time.What wil be trueDiscount on same sum due after Â½ of former time,the rate of interest beng the same?Sol:

Simple Interest on (260-20) for a gven time=Rs20Simple Interest on (260-20) for a half time=Rs20*(1/2)=Rs10True Discount on Rs.250=Rs10True Discount on Rs.260=Rs10*260/250=Rs10.40

11.A has to pay Rs.220 to B after 1yr.B asks A to pay Rs.110in cash and defer the payment of Rs,.110 for 2 yrs.A agrees to it.If the rate of interest be 10% per annum in this mode of payment:............... sol:A has to pay =Present worth of Rs220 due 1yr hence=Rs.(220*100)/(100+(10*1))

=Rs.200 A actually pays=Rs.110+Presentworth of Rs110 due 2 yrs hence=[110+ (Amount*100) /(100+(R*T))]=[110+((110*100)/(100+(10*2)))]=Rs192.66So A gains Rs(200-192.66)=Rs.7.34

COMPLEX PROBLEMS

1.If Rs.10 be allowed as truedscount on a bill of Rs.110 at the end of a certaimn time ,then discount allowed on the same sum due at the end of double the time..sol:Amount=Rs110TrueDiscount=Rs10Present worth=Amount-TrueDiscount=Rs110-10=Rs.100SI on Rs.100 for a certain time =Rs.10SI on Rs.100 for doube the time =Rs.20TrueDiscount on (100+20)=120-100=Rs20TrueDiscount on Rs.110 =(110*20)/120=Rs18.33

2.A man wants to se hs scooter .There are two offers one atRs12000 cash and other at acredit of Rs12880 to be paid after 8 months ,money being at 18% per annum which is better offer?Sol:offer1=Rs12000offer2 Present worth= (Amount*100) /(100+(R*T))=Rs.12880*100/(100+(18*(8/12))=Rs11,500The first offer is better as if he gains 500/- if he sells at Rs.12000 is better

3.Goods were bought for Rs.600 and sold the same day for Rs.688.80 at a credit of 9 months and thus gaining 2% rate of interest per annum issol:

MethodI:Amount=Rs.688.50Time=9.12Gaining 2%----------->Presentworth=102%of 600=(102*600)/100=Rs.612Presentworth=(Amount*100)/(100+(T*R))612=(688.80*100)/(100+((9/2)*R))=>R=162/3%MethodII:TrueDiscount=688.50-612=Rs.76.50Rate=TrueDiscount*100/(P.W*T)=(76.50*100)/(612*(9/12))=162/3%

4.The present worth of Rs.1404 due in 2 equal halfyearly instalments at 8% per annu8m S.I is SOl:Presentworth=(Amount*100)/(100+(T*R))PresentWorth=Presentworth of Rs.702 6 months hence +Presentworth of Rs.702 1yr hence= (702*100)/(100+(1/2*8)) + (702t*100)/(100+(1*8))=Rs.1325

BANKER'S DISCOUNT

CONCEPT-> Suppose a merchant A buys goods worth Rs.10000 from another merchant B at a credit of say 5 months-> Then,B prepares a bill , called the bill of exchange -> A signs this bill & allows B to withdraw the amount from his bank account after exactly 5 months,the date exactly after 5 months is called Nominally due date-> Three days (grace days) are added to it get a date known as legally due date-> Suppose B wants to have money before legally due date then he can have the money from banker or a broker who deducts S.I on the face value (i.e., 10000) for the period from the date on which the bill was discounted (i.e paied by the banker) & the legally due date this amount is known as Bankers Discount-> Thus , B.D is the S.I on the face for the period from the date on which the bill was discounted and the legally due date-> Bankers Gain (B.G) = (B.D) â€“ (T.D) for the unexpired time

Note:- When the date of the bill is not given,grace days are not to be added

Formulae:-

(1)B.D = S.I on bill for unexpired time(2)B.G = (B.D) â€“ (T.D) = S.I on T.D = (T.D)^2 /P.W(3)T.D = sqrt(P.W * B.G)(4)B.D = (Amount * Rate * Time)/100(5)T.D = (Amount * Rate * Time)/(100+(Rate * time)(6)Amount = (B.D * T.D)/(B.D â€“ T.D)(7)T.D = (B.G * 100)/(Rate * Time)

SIMPLE PROBLEM1.If the true discount on a certain sum due 6 months hence at 15% isRs 120.What is the bankers discount on the same for same time and the same rate.Sol:- B.G = S.I on T.D = RS (120 * 15 * Â½ * 1/100) = 9 (B.D) â€“ (T.D) = 9 B.D = 120 + 9 =1292.The bankers discount on Rs 1800 at 12 % per annum is equal to the true discount on Rs 1872 for the same time at the same rate .Find the time.Sol:- S.I on Rs 1800 = T.D on Rs 1872 P.W of Rs 1872 is Rs 1800 Rs 72 is S.I on Rs 1800 at 12% Time = (100 * 72)/(12 * 1800) = 1/3 years = 4 months

3.The bankers discount and true discount on a sum of money due 8 months hence are Rs.120 & Rs.110 resp. Find the sum & the rate per centSol:- Sum = (B.D * T.D) / (B.D) â€“ (T.D) = (120 * 110) / (120 â€“ 110) = 1320 Since B.D is S.I on sum due, so S.I on Rs 1320 for 8 months is Rs 120 Rate = (100 * 120) / (1320 * 2/3) = 13 7/11%

MEDIUM PROBLEMS1.The Bankers discount on Rs 1650 due a certain time hence is Rs 165.find the true discount and the bankers gain.Sol :- Sum = (B.D * T.D) / (B.D â€“ T.D) = (B.D * T.D) / B.G T.D/B.G = Sum/B.D =650/165 =10/1 Thus if B.G is Rs 1 ,T.D = Rs 10 if B.D is Rs 11 ,T.D = Rs 10 if B.D is Rs 165,T.D = Rs (10/11 * 165) = 150 B.G =Rs(165 â€“ 150) = Rs 152.The Present worth of a bill due something hence is Rs 1110 and the true discount on the bill is Rs.110 . Find the bankes discount & the bankers gain.Sol:- T.D = sqrt (P.W * B.G)

B.G = (T.D)^2 /P.W = (110 * 110) / 1110 = 11 B.D = (T.D + B.G) = (110 + 11) = Rs 121

3.What rate percent does a man get for his money when in discountingSol:- Let the amount of the bill =100 Money deducted = 10 Money received by the holder of the bill = (100 â€“ 10) = 90 S.I on Rs 90 for 10 months = 10 Rate = (100 * 10) / (90 * 10/12) = 13 1/3%

COMPLEX PROBLEMS 1.A bill for Rs.6000 is drawn on July 14 at 5 months . It is

discounted on 5th October at 10%.Find the bankers discount

true discount, bankers gain and the money that the holder of

the bill receives.

Sol:- Face value of the bill = Rs.6000

date on which the bill was drawn = July 14 at 5 months

nominally due date = December 14

legally due date = December 17

Date on which the bill was discounted = October 5 Unexpired time : Oct Nov Dec 26 + 30 + 17 = 73days =1/5 years B.D = S.I on Rs 6000 for 1/5 year = Rs (6000 * 10 * 1/5 * 1/100) = Rs 120 T.D = Rs(6000 * 10 *1/5)/(100 + (10 * 1/5)) = Rs. 117.64 B.G = (B.D) â€“ (T.D) = Rs(120 -117.64) = Rs 2.36 Money received by the holder of the bill = Rs(6000 â€“ 120) = 5880

2. The bankers gain on a certain sum due 1 Â½ year hence

is 3/25 of the bankers discount .The rate percent is

Sol:- Let B.D = 1 then B.G = 3/25 T.D = (B.D â€“ B.G) = (1 â€“ 3/25) = 22/25 sum = (1 * 22/25) / (1 â€“ 22/25) = 22/3 S.I on Rs 22/3 for 1 Â½ year is 1. Rate = (100 * 1) / (22/3 * 3/2) = 9 1/9%

3. The bankers gain of a certain sum due 2 years hence

at 10% per annum is Rs 24 .The percent worth is

Sol:- T.D = (B.G * 100) / (Rate * Time) 00) / (10 * 2) = 120. P.W = (100 *T.D) / (Rate * Time) = (100 * 120) /(10 * 2) = 600

ODDMAN OUT AND SERIES

In any type of problems,a set of numbers is given in such a way that each one except one satiesfies a particular definite property.The one which does not satisfy that characteristic is to be

taken out.

Some important properties of numbers are given below :

1.Prime Number Series Example:

2,3,5,7,11,.................2.Even Number Series Example:

2,4,6,8,10,12,............3.Odd Number Series: Example:

1,3,5,7,9,11,...................4.Perfect Squares:

Example: 1,4,9,16,25,..............

5.Perfect Cubes:

Example: 1,8,27,64,125,.................

6.Multiples of Number Series: Example:

3,6,9,12,15,..............are multiples of 3

7.Numbers in Arthimetic Progression(A.P): Example:

13,11,9,7................8.Numbers in G.P:

Example: 48,12,3,.....

SOME MORE PROPERTIES:1. If any series starts with 0,3,.....,generally the relation will be

(n2-1).2. If any series starts with 0,2,.....,generally the relation will be

(n2-n).3. If any series starts with 0,6,.....,generally the relation will be

(n3-n).4. If 36 is found in the series then the series will be in n2 relation.

5. If 35 is found in the series then the series will be in n2-1 relation.

6. If 37 is found in the series then the series will be in n2+1 relation.

7. If 125 is found in the series then the series will be in n3 relation.

8. If 124 is found in the series then the series will be in n3-1 relation.

9. If 126 is found in the series then the series will be in n3+1 relation.

10. If 20,30 found in the series then the series will be in n2-n relation.

11. If 60,120,210,........... is found as series then the series will be in n3-n relation.

12. If 222,............ is found then relation is n3+n13. If 21,31,.......... is series then the relation is n2-n+1.14. If 19,29,.......... is series then the relation is n2-n-1.

15. If series starts with 0,3,............ the series will be on n2-1 relation.

EXAMPLE PROBLEMS: 1.Find the odd one out. 3,5,7,12,17,19

SOLUTION: In the above series except 12 all elements are

odd numbers.so 12 is the odd one.

2. Find the odd one out. 1,4,9,16,23,25,36

SOLUTION: In the above series all elements except 23 are

perfect sqares.so 23 is odd one.

3. Find the odd one out. 41,43,47,53,61,71,73,81

SOLUTION: In the above series all elements except 81

are prime numbers.so 81 is odd one.

4.Find the odd one out. 1,4,9,16,20,36,49

SOLUTION:

In the above series all elements except 20 are perfect squares.So 20 is odd one.


SOLUTION: In the above series all elements except 100

are perfect cubes.so 100 is odd one.


SOLUTION: In the above series all elements in the pattern

like 12, 12+22,12+22+32,................. But 50 is not in this pattern,so odd one.

7.Find the odd one out.

835,734,642,751,853,981,532

SOLUTION: In the above series,the difference between third and first digit of each element is equal to its middle digit.But

751 is not in this pattern,so odd one.


SOLUTION: In the above series,the sum of first and

third digit of each element is equal to its middle

digit.But 427is not in this pattern,so odd one.


SOLUTION:

In the above series,the product of first and third digit of each element is equal to its middle digit.

But 383 is not in this pattern,so odd one.

10. Find the odd one out. 2,5,10,17,26,37,50,64

SOLUTION: In the above series,the elements are in the pattern of x2+1,Where x is 1,2,3,4,5,6,7.but 82+1 is not equal to

64.It is 65.64 is odd one.


SOLUTION:

In the above series,the elements are in the pattern of x2+3,Where x is 4,5,6,7,8,9,10.but 102+3 is not equal to

102.It is 103.so 102 is odd one.


SOLUTION:

In the above series,the elements are in the pattern of x2+3,Where x is 4,5,6,7,8,9,10.but 102+3 is not equal to

102.It is 103.so 102 is odd one. 13.Find the odd one out.

2,5,10,50,500,5000

SOLUTION: In the above series,the pattern as

follows: 1st term * 2nd term = 3rd term 2nd term * 3rd term = 4th term 3rd term * 4th term = 5th term

But 50*500=25000 which is not equal to 5000.

so 5000 is odd one.


SOLUTION:

In the above series, alternatively 23 is added and 17 is subtracted from the terms.

So 634 is odd one.


SOLUTION:

In the above series,the terms are successively divided by

12,10,8,6,..... so 24 is not in this pattern.

so 24 is odd one.


SOLUTION:

In the above series,the terms at odd places are 5,6,7,8.......and at even places is 16. So 9 is odd one.


SOLUTION:

In the above series,the difference between two successive terms from the beginning are 7,5,7,5.........

so 40 is odd one.

18.Find the odd one out. 56,72,90,110,132,150

SOLUTION:

The above series as follows: 7*8,8*9,9*10,10*11,11*12,12*13. So it will be 56,72,90,110,132,156

so 150 is wrong.


SOLUTION:

Add 1square ,2square ,....,6square to the terms.

so 91 is wrong.

20.Find the odd one out. 105,85,60,30,0,-45,-90

SOLUTION:

Subtract 20,25,30,35,40,45 from the terms.

So 0 is odd one.

21.Find out the odd one out. 3,10,21,36,55,70,105

SOLUTION: The pattern in the series is

1*3, 2*5, 3*7, 4*9, 5*11, 6*13, 7*15. So the series will be 3,10,21,36,55,78,105.

So 70 is wrong term in the series.

22.Find out the odd one out. 4,9,19,39,79,160,319

SOLUTION: Double the number and add 1 to it.

So the series will be 4,9,39,79,159,319. So 160 is wrong.

23.Find out the odd one out. 10,14,28,32,64,68,132.

SOLUTION: Alternatively add 4 and double the next term.

So 132 is wrong.

1.Find the missing term in the series: 4,-8,16,-32,64,( )

SOLUTION: The terms are doubled and change the sign.

So the next term is -128

2.16,33,65,131,261,( )

SOUTION: The terms are doubled and 1 is added.

So 261*1+1=522+1=523 So the missing term is 523.

3.2,6,12,20,30,42,56,( )

SOLUTION:

The pattern is 1 * 2 , 2 * 3 , 3 * 4 , 4 * 5 , 5 * 6 , 6 * 7 , 7 * 8 , 8 * 9 .

So the series is 2,6,12,20,30,42,56,72. So 72 is the missing term.

4. 8,24,12,36,18,54,( )

SOLUTION:

Numbers are alternatively multiplied by 3 and divided by 2. So the next term is 54 / 2 = 27.

5. 165,195,255,285,345,( )

SOLUTION: Each number is 15 multiplied by a prime number.

i.e the series is 15*11,15*13,15*17,15*19,15*23,15*29. So series is 165,195,255,285,345,435.

So 435 is the missing term.

6. 7,16,63,124,215,342,( ).

SOLUTION: Numbers are 23 -1,33-1,43-1,....................so 83-

1=511. So 511 is the missing term.

7.2,4,12,48,240,( )

SOLUTION: Go on multiplying by 2,3,4,5,6.

So the last term in the series is 240*6=1440.

8.8,7,11,12,14,17,17,22,( )

SOLUTION: There are two series 8,11,14,17,20 and 7,12,17,22

So increasing by 3 and 5.So 20 the missing term.

9.71,76,69,74,67,72,( )

SOLUTION: Alternately add 5 and subtract 7.

so the series is 71+5=76

10.2,5,9,19,37

SOLUTION: Second number is one more than twice the first,Third number is one less than twice the second,Forth is one

more than twice the third and so on. So the next number is 2 * 37 + 1 = 74+1 = 75.

11. Find the wrong number in the given series.

3,8,15,24,34,48,63

SOLUTION: The difference between consecutive terms are

respectively 5,7,9,11,13. So 34 is the wrong number in the series.

12. 125,106,88,76,65,58,53 SOLUTION:

Subtract 24,21,18,15,12,9 from the numbers to get the next number.

So 128 is wrong.

13. 1,1,2,6,24,96,720

SOLUTION: Multiply with 1,2,3,4,5,6 to get the next number.

So 96 is wrong.

14 . 32,36,41,61,86,122,171,235

SOLUTION: Second term = First term + 22 Third term = Second term + 32 Fourth term = Third term + 42 Fifth term = Forth term + 52 Sixth term = Fifth term + 62 Seventh term = Sixth term + 72

So the third term should be 45 instead of 41.

15 . 15,16,34,105,424,2124,12576

SOLUTION: Second term = First term * 1 + 1 = 16 Third term = Second term * 2 + 2 = 34 Forth term = Third Term * 3 + 3 = 105 Fifth term = Forth term * 4 + 4 = 424 Sixth term = Fifth term * 5 + 5 =2125

Seventh term = Sixth term * 6 + 6 = 12576. So 2124 is wrong.

16 . 40960,10240,2560,640,200,40,10

SOLUTION: Go on dividing by 4 ,the series will be

40960,10240,2560,640,160,40,10. So 200 is wrong.

17.7,8,18,57,228,1165,6996

SOLUTION: Let the numbers be A,B,C,D,E,F,G then

A,A*1+1,B*2+2,C*3+3,............ so 288 is wrong.

18. 19,26,33,46,59,74,91

SOLUTION: Go on adding 7,9,11,13,15,17.

So 33 is wrong.

19 . 10,26,74,218,654,1946,5834.

SOLUTION: Second term = first term * 3 â€“ 4 = 26. Third term = Second term * 3 â€“ 4 =74 Forth term = Third term * 3 â€“ 4 =218 Fifth term = Forth term * 3 â€“ 4 =650

So 654 is wrong .

Probability

Introduction:

Experiment:An operation which can produce some well-defined outcome iscalled an experiment.

Random Experiment:An experiment in which all possible out comes are known andthe exact output cannot be predicted in advance is called a random experiment.EX:1) Rolling an unbiased dice.2) Tossing a fair coin.3) Drawing a card from a pack of well-shuffled cards .4)Picking up a ball of certain colour from a bag containing balls of different colours.

Details:

1) When we thrown a coin ,then either a Head(H) or a Tail(T)appears.

2)A dice is a solid cube ,having 6 faces,marked 1,2,3,4,5,6respectively. When we throw a die ,the outcome is the number that appears on its upper face.3)A pack of cards has 52 cards.It has 13 cards of each suit,namely spades,clubs,hearts anddiamonds. Cards of spades and clubs are balck cards.Cards of hearts and diamonds are red cards.There are four honours of each suit.These are Aces,Kings,queens and Jacks.These are called Face cards.

Sample Space:When we perform an experiment ,then the set of S of all possible outcomes is called the Sample space .

EX:1)In tossing a coin S= {H,T}.2)If two coins are tossed then S= {HH,HT,TH,TT}.3)In rolling a dice ,we have S={1,2,3,4,5,6}.

Event:Any subset of a sample space is called an Event.Probability of occurrence of an Event:Let S be the sample space.Let E be the Event.Then E cS i.e E is subset of S thenprobability of E p(E) =n(E)/n(S).

Reults on Probability:1)P(S) =1.2)0 < P(E) < 1probability of an event lies between 0 and 1.Max value of probability of an event is one.3)P(Ð¤)=0.4)For any events A and B we have .P(AUB) =P(A) +P(B) -P(AnB).5)If A denotes (not -A) thenP(A) =1-P(A)P(A)+P(A) =1. Problems:

1)An biased die is tossed.Find the probability of getting a multiple of 3?

Sol: Here we have sample space S={1,2,3,4,5,6}.Let E be the event of getting a multiple of 3.Then E={3,6}.P(E) =n(E)/n(S).n(E) =2,n(S) =6.P(E) =2/6P(E) =1/3.

2)In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7?

Sol: Here we have sample space n(S) =6*6 =36.Let E be the event of getting a total more than 7.={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),(5,3),(6,2),(4,5),(5,4), (5,5),(4,6),(6,4)}n(E) =15P(E) = n(E)/n(S)= 15/36.P(E) = 5/12.

3)A bag contains 6 white and 4 black balls .Two balls aredrawn at random .Find the probability that they are of thesame colour?

Sol: Let S be the sample space.Number of ways for drawing two balls out of 6 white and 4 red balls = 10C2=10!/(8!*2!)= 45.n(S) =45.Let E =event of getting both balls of the same colour.Thenn(E) =number of ways of drawing ( 2balls out of 6) or (2 balls out of 4).= 6C2 +4C2= 6!/(4!*2!) + 4!/(2! *2!)= 6*5/2 +4 *3/2=15+6 =21.P(E) =n(E)/n(S) =21/45 =7/45.

4)Two dice are thrown together .What is the probability thatthe sum of the number onthe two faces is divisible by 4 or 6?Sol: n(S) = 6*6 =36.E be the event for getting the sum of the number on the two faces is divisible by 4 or 6.E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)(5,1)(5,3)(6,2)(6,6)}n(E) =14.Hence P(E) =n(E)/n(S)= 14/36.P(E) = 7/18

5)Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens? Sol: total number of ways for choosing 2 cards from52 cards is =52C2 =52 !/(50!*2!)= 1326.Let A= event of getting bothe black cards.Let B= event of getting bothe queensAnB=Event of getting queens of black cardsn(A) =26C2.We have 26 black cards from that we have to choose 2 cards.n(A) =26C2=26!/(24!*2!)= 26*25/2=325from 52 cards we have 4 queens.

n(B) = 4C2= 4!/(2!* 2!) =6n(AnB) =2C2. =1P(A) = n(A) /n(S) =325/1326P(B) = n(B)/n(S) = 6/1326P(A n B) = n(A n B)/n(S) = 1/1326P(A u B) = P(A) +P(B) -P(AnB)= 325/1326 + 6/1326 -1/1326= 330/1326P(AuB) = 55/221

6)Two diced are tossed the probability that the total score is a prime number?

Number of total ways n(S) =6 * 6 =36E =event that the sum is a prime number.Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)}n(E) =15P(E) =n(E)/n(S)= 15/36P(E) = 5/12 7)Two dice are thrown simultaneously .what is the probabilityof getting two numbers whose product is even?

Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6= 36E=Event of getting two numbers whose product is evenE={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)(3,4)(3,6)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}n(E) = 27P(E) = n(E)/n(S)= 27 /36P(E) =3/4probability of getting two numbers whose product is even isequals to 3/4.

8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is drawn at random. what is the probability of getting a prize ?

Sol: By drawing lottery at random ,we have n(S) =10C1+25C1 = 10+25 = 35.E =event of getting a prize.n(E) =10C1 =10out of 10 prozes we have to get into one prize .The number of ways 10C1.n(E) =10 n(S) =35 P(E) =n(E)/n(S) =10/35 = 2/7Probability is 2/7.

9)In a class ,30 % of the students offered English,20 % offeredHindi and 10 %offered Both.If a student is offered at random,what is the probability that he has offered English or Hindi?

Sol:English offered students =30 %.Hindi offered students =20%Both offered students =10 %Then only english offered students E =30 -10=20 %only Hindi offered students S =20 -10 % = 10 %All the students =100% =E +S +E or S 100 =20 +10 + E or S +E and SHindi or English offered students =100 -20-10-10 =60 %Probability that he has offered English or Hindi =60/100= 2/5

10) A box contains 20 electricbulbs ,out of which 4 are defective,two bulbs are chosen at random from this box.What is theprobability that at least one of these is defective ?

Sol: out of 20 bulbs ,4 bulbs are defective.16 bulbs are favourable bulbs.E = event for getting no bulb is defective.n(E) =16 C 2out of 16 bulbs we have to choose 2 bulbs randomly .so the numberof ways =16 C 2n(E) =16 C2n(S) =20 C 2P(E) =16 C2/20C2= 12/19probability of at least one is defective + probability of oneis non defective =1P(E) + P(E) =112/19 +P(E) =1P(Eâ€™) =7/19

11)A box contains 10 block and 10 white balls.What is the probability of drawing two balls of the same colour?

Sol: Total number of balls =10 +10=20 ballsLet S be the sample space.n(S) =number of ways drawing 2 balls out of 20= 20 C2= 20 !/(18! *2!)= 190.Let E =event of drawing 2 balls of the same colour.n(E) =10C2+ 10C2= 2(10 C2)= 90P(E) =n(E)/n(S)P(E) =90/190= 9/19

12) A bag contains 4 white balls ,5 red and 6 blue balls .Threeballs are drawn at random from the bag.What is the probability

that all of them are red ?

Sol: Let S be the sample space.Then n(S) =number of ways drawing 3 balls out of 15.=15 C3.=455Let E =event of getting all the 3 red balls.n(E) = 5 C3 =5C2 = 10P(E) =n(E) /n(S) =10/455 =2/91.

13)From a pack of 52 cards,one card is drawn at random.What is theprobability that the card is a 10 or a spade?Sol: Total no of cards are 52.These are 13 spades including tne and there are 3 more tens.n(E) =13+3 = 16P(E) =n(E)/n(S).=16/52P(E) =4/13.

14) A man and his wife appear in an interview for two vacancies inthe same post.The probability of husband's selection is 1/7 and theprobabililty of wife's selection is 1/5.What is the probabililty that only one of them is selected?

Sol: let A =event that the husband is selected.B = event that the wife is selected.E = Event for only one of them is selected.P(A) =1/7andp(B) =1/5.P(A') =Probability of husband is not selected is =1-1/7=6/7P(B') =Probaility of wife is not selected =1-1/5=4/7P(E) =P[(A and B') or (B and A')]= P(A and B') +P(B and A')= P(A)P(B') + P(B)P(A')= 1/7*4/5 + 1/5 *6/7P(E) =4/35 +6/35=10/35 =2/7

15)one card is drawn at random from a pack of 52 cards.What is theprobability that the card drawn is a face card?

Sol: There are 52 cards,out of which there 16 face cards.P(getting a face card) =16/52= 4/13

16) The probability that a card drawn from a pack of 52 cards willbe a diamond or a king?

Sol: In 52 cards 13 cards are diamond including one king there are3 more kings. E event of getting a diamond or a king.n(E) =13 +3= 16P(E) =n(E) /n(S) =16/52

=4/13

17) Two cards are drawn together from apack of 52 cards.What is theprobability that one is a spade and one is a heart ?

Sol: S be the sample space the n (S) =52C2 =52*51/2=1326 let E =event of getting 2 kings out of 4 kings n(E) =4C2= 6P(E) =n(E)/n(S)=6/1326=1/221

18) Two cards are drawn together from a pack of 52 cards.What is theprobability that one is a spade and one is a heart?

Sol: Let S be the sample space thenn(S) =52C2=1326E = Event of getting 1 spade and 1 heart.n(E) =number of ways of choosing 1 spade out of 13 and 1 heart outof 13.= 13C1*13C1 =169P(E)= n(E)/n(S)=169/1326 =13/102.

19) Two cards are drawn from a pack of 52 cards .What is the probability that either both are Red or both are Kings?

Sol: S be the sample space.n(S) =The number of ways for drawing 2 cards from 52 cards.n(S) =52C2=1326E1 be the event of getting bothe red cards.E2 be the event of getting both are kings.E1nE2 =Event of getting 2 kings of red cards.We have 26 red balls.From 26 balls we have to choose 2 balls.n(E1) =26C2= 26*25/2=325We have 4 kings .out of 4 kings,we have to choosed 2 balls.n(E2) =4C2=6n(E1nE2) =2C2 =1P(E1) = n(E1)/n(S)=325/1326P(E2) =n(E2)/n(S)=6/1326P(E1nE2) =n(E1nE2)/n(S) =1/1326P(both red or both kings) = P(E1UE2)= P(E1) +P(E2)-P(E1nE2)=325/1326 +6/1326 -1/1326=330/1326 =55/221

PERMUTATIONS AND COMBINATIONS

CONCEPT

Formulae:-

-> Factorial Notation :- Let n be positive integer.Then ,factorial n dentoed by n! is defined as

n! = n(n-1)(n-2). . . . . . . .3.2.1 eg:- 5! = (5 * 4* 3 * 2 * 1)

= 120 0! = 1

->Permutations :- The different arrangements of a given number of things by taking some or all at a time,are called permutations.

eg:- All permutations( or arrangements)made with the letters a,b,c by

taking two at a time are (ab,ba,ac,ca,bc,cb)

->Numbers of permutations :- Number of all permutations of n things , taken r at a time is given by

nPr = n(n-1)(n-2). . .. . . (n-r+1) = n! / (n-r)!

->An Important Result :- If there are n objects of which p1 are alike of one

kind ; p2 are alike of another kind ; p3 are alike of third kind and so on

and pr are alike of rth kind, such that (p1+p2+. . . . . . . . pr) = n

Then,number of permutations of these n objects is: n! / (p1!).(p2!). . . . .(pr!)

->Combinations :- Each of different groups or selections which can be formed by taking some or all of a number of objects,is called a

combination. eg:- Suppose we want to select two out of three boys

A,B,C . then ,possible selection are AB,BC & CA.

Note that AB and BA represent the same selection.

-> Number of Combination :- The number of all combination of n things taken r at atime is:

nCr = n! / (r!)(n-r)! = n(n-1)(n-2). . . . . . . tor

factors / r!

Note that : nCn = 1 and nC0 =1

An Important Result : nCr = nC(n-r)

PROBLEMS

1.Evaluate 30!/28!Sol:- 30!/28! = 30 * 29 * (28!) / (28!)

= 30 * 29 =870

2.Find the value of 60P3Sol:- 60P3 = 60! / (60 â€“ 3)! = 60! / 57!

= (60 * 59 *58 * (57!) )/ 57!

= 60 * 59 *58 = 205320

3. Find the value of 100C98 50C 50Sol:- 100C98 = 100C100-98)

= 100 * 99 / 2 *1 = 4950 50C50 = 1

4.How many words can be formed by using all the letters of the word â€œDAUGHTRâ€ so that vowels always come together & vowels are never�

together?Sol:- (i) Given word contains 8 different letters

When the vowels AUE are always together we may suppose them to form an entity ,treated as one letter then the letter to be arranged are DAHTR(AUE) these 6 letters can be arranged in 6p6 = 6!

= 720 ways The vowels in the group (AUE) may be arranged in 3! = 6

ways Required number of words = 760 * 6 =4320

(ii)Total number of words formed by using all the letters of the given words

8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320

Number of words each having vowels together = 760 * 6 = 4320

Number of words each having vowels never together = 40320 â€“ 4320

= 36000

5.In how many ways can a cricket eleven be chosen out of a batchof 15 players.

Sol:- Required number of ways = 15C 11 = 15C (15-11)

= 15 C 4 15C4 = 15 * 14 * 13 * 12 / 4 * 3 * 2 *1

= 1365

6.In how many a committee of 5 members can be selected from 6men

5 ladies consisting of 3 men and 2 ladiesSol:- (3 men out of 6) and (2 ladies out of 5) are to be

chosen Required number of ways

=(6C3 * 5C2) = 200

7.How many 4-letter word with or without meaning can be formed outof the letters of the word 'LOGARITHMS' if repetition of letters is

not allowedSol:- 'LOGARITHMS' contains 10 different letters

Required number of words = Number of arrangements of 100 letters

taking 4 at a time

= 10P4 = 10 * 9 * 8 * 7

= 5040

8.In how many ways can the letter of word 'LEADER' be arrangedSol:- The word 'LEADER' contains 6 letters namely

1L,2E,1A,1D and 1R Required number of ways

= 6! / (1!)(2!)(1!)(1!)(1!) = 6 * 5 * 4 * 3 * 2 *1 / 2 * 1

=360

9.How many arrangements can be made out of the letters of the word 'MATHEMATICS' be arranged so that the vowels always come

togetherSol:- In the word ' MATHEMATICS' we treat vowels AEAI as one letter thus we have MTHMTCS(AEAI)

now we have to arrange 8 letters out of which M occurs twice ,T occurs twice & the rest are different

Number of ways of arranging these letters = 8! / (2!)(2!)

= 10080now AEAI has 4 letters in which A occurs 2 times and the rest

are different Number of ways of arranging these letters

= 4! / 2! = 12 Required number of words = (10080 * 12) = 120960

10.In how many different ways can the letter of the word 'DETAIL' bearranged in such a way that the vowels occupy only the odd positions

Sol:- These are 6 letters in the given word , out of which there are 3 vowels and 3 consonants Let us mark these positions as under

(1)(2) (3) (4)(5)(6) now 3 vowels can be placed at any of the three places out

of 4

marked 1,3,5 Number of ways of arranging the vowels = 3P3 = 3! =6

Also,the 3 consonants can be arranged at the remaining 3 positions Number of arrangements = 3P3 = 6 Total number of ways = (6 * 6) =36

11.How many 3 digit numbers can be formed from the digits 2,3,5,6,7 and 9 which are divisible by 5 and none of the digits is repeated?

Sol:- Since each desired number is divisible by 5, so we much have 5 at the unit place. The hundreds place

can now be filled by any of the remaining 4 digits .so, there 4 ways of filling it.

Required number of numbers = (1 * 5 * 4) = 20

12.In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a self so that two books on Hindi may not

be together?Sol:- In order that two books on Hindi are never together,

we must place all these books as under: X E X E X . . . . . . . . . . X E X

Where E denotes the position of an English and X that of a Hindi book.

Since there are 21 books on English,the number of places marked X are therefore 22.

Now, 19 places out of 22 can be chosen in 22 C 19 = 22 C 3 =22 * 21 * 20 / 3 * 2 *1

Hence the required number of ways = 1540

13.Out of 7 constants and 4 vowels how many words of 3 consonants and 2 vowels can be formed?

Sol:- Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

= 7C3 * 4C2 = 210

Number of groups each having 3 consonants and 2 vowels = 210

Each group contains 5 letters Number of ways of arranging 5 letters among themselves

= 5! = (5 * 4 * 3 * 2 * 1) = 210

Required number of words = (210 * 210) = 25200

PUZZLES

Puzzles are dealt in a detailed manner with certain solutions.

Different puzzles are gathered from ShakuntalaDeviâ€™s puzzle books.

Keeping in mind certain puzzles for Infosys some reasoning problems are also dealt.

Puzzle name at the top of each problem will give a brief idea regarding the mode of application.

SELECTING A CANDIDATE

For an advertisement of six local posts,twelve persons applied for the job.Can you tell in how many different ways the selection can be made?

Solution:

6^12

SET OF BAT AND BALL

When I wanted to buy a bat and ball ,the shopkeeper said they would together cost Rs.3.75.But I did not want to buy a ball.The shopkeeper said that bat would cost 75paise more than the ball.What was the cost of

bat and the ball?

Soluton:

Given that bat and ball together cost Rs.3.75 = 375paise

Let cost of the ball alone be x.

Given cost of the bat is 75p greater than cost of the ball.

So cost of the bat = x+75

x+x+75 = 375

2x = 375 â€“ 75

2x = 300

x = 150p

Hence cost of the ball = Rs.1.50

=>Cost of the bat = 1.50 + 75 = Rs.2.25

PLAYING CHILDREN

A group of boys and girls are playing.15 boys leave.There remain 2 girls for each boy.Then 45 girls leave.There remain 5 boys for each girl.How

many boys were in the orginal group?

Solution:

Let B and G represent no.of boys and girls in the original group respectively.

G ---------> 2

B-15 ----------> 1

G/B-15 = 2/1

i.e., 2 girls are left for 15 boys who are alone.

G-45 -------------------->1

B-15 ------------------------->5

5 boys are left out when 15 girls are alone.

=>G/B-15=2/1 --------------------------------(1)

=>G-45/B-15 =1/5 ----------------------------(2)

(1) & (2) =>

G = 2B-30

5G â€“ 225 = B - 15

5 ( 2B â€“ 30 ) = B â€“ 15 + 225

10B = B â€“ 15 + 225 + 150

9B = 360

B = 40

(1)=> G/40-15 = 2

G=50 girls.

SOLVNG PROBLEMS

Reshma appeared for a maths exam.She was given 100 problems to solve.She tried to solve all of them correctly but some went wrong.But she scored

85.

Her score was calculated by subtracting two times th no.of wrong answers from the no.of correct answers.How many problems did Reshma do

correctly?

Soluton:

Assume W as wrong answers and R as correct answers

Given total no.of questions as 100

R+W=100 ---------------------------(1)

Score is calculated by subtracting 2 times wrong answers(2W)

from right answers(R) and given as 85

R-2W=85 -------------------------------(2)

(2)-(1)

R-2W=85

R+W=100

---------------------W=5

Hence,100-5=95 is the no.of correct answers of Reshma.

A RUNNNG RACE

Sneha,Shilpa,Sushma join a running race.The distance is 1500 metres.Sneha beats Shilpa by 30 metres and Sushma by 100 metres.By how much could Shilpa beat Sushma over the full distance if they both ran as

before?

Solution:

Total distance covered by Sneha=1500m

Shilpa=1500-30=1470

Sneha =1500-100=1400

Distance covered by Shilpa=1500*1400/1470=1428.6

Distance to be covered by Shilpa to beat Sushma over full distance 1500-1428.6=71.4m

FILLING A CISTERN

Pipe S1 can fill a cistern in 2 hours and pipe S2 in 3 hours.Pipe S3 can empty it in 5 hours.Supposing all the pipes are turned on when the

cistern is competely empty,how long will it take to fill?

Solution:

S1 fills cistern in 1/2 hours

S2 fills cistern in 1/3 hours

S3 empties it in 1/5 hours

A the pipes S1,S2,S3 working i.e.,filling the cistern

1/2+1/3-1/5=15+10-6/30=25-6/30=19/30

No.of hours to fill=30/19=1 11\19hours.

SEQUENCE PROBLEMS

What are the next two terms in the sequence?1,1,5,17,61,217.................

Solution:

The order in this cases is

Tn=3*Tn-1 +2*Tn-2

= 3(217)+2(61)

= 773

Tn+1=3(773)+2(271)

=2319+542

=2753

SEQUENCES

What are the next two terms in the sequence?

1,1,5,17,61,217.................

Solution:

Tn=3Tn-1+2Tn-2 =3(217)+2(61)

=773Tn+1=3(773)+2(217)=2753

What are the next three terms to the series?

1+3+7+15+31+63...........

Solution:

Actual term is 2exp n-1.

The next three terms are:

2exp7-1=1272exp8-1=2552exp9-1=511

A PROBLEM OF SHOPPING

Samsrita went out for shopping by taking with her Rs.15/- in one rupee notes and 20p coins.On return she had as many one rupee notes as she

originally had and as many 20p coins as she had one rupee notes.She came back with 1/3rd with what she had.How much did Samsrita spend and how

much did she take?

Solution:

Let x be no.of rupee notes y be no.of 20pcoins.

So,when going for shopping 100x+20y paise were there with Samsrita.

On return she had 100y+20x paise.

Also it is given that she had 1/3 rd of the orginal amount.

1/3(10x+20y)=100y+20x

=>4x=280y

=>x=7y

y=1 => x=7 total =7.20 <15

y=2 =>x=14 total=14.40~=15

y=3 =>x=21 total=21.60 >>15

Hence the suitable value nearer to the amunt is 14.40 and so is the amount Samsrita carried with her.

1/3(1440)=480paise.

Rs.4.80/- is amount spent by Samsrita.

A PUZZLE OF CULTURAL GROUPS

Literary,Dramatic,Musical,Dancing and Painting are the 5 groups of a club.Literary group meets every other day,dramatic every third

day,musical every fourth day,dancing every fifth day,painting every sixth day.Five groups meet on NewYears day of 1975 and starting from

that day they met regularly on schedule.

How many times did all the 5 groups meet on same day in first quarter excluding Jan1,1975.How many days did none of them met?

Solution:

LCM of 2,3,4,5,6 is 60.

Hence excluding Jan1,1975 they met on every 61st day.

60/2=30 60/3=20 60/4=15 60/5=12 60/6=10

Literary meet for 30 2 day intervals.

Dramatic meet for 20 3 day intervals.

Musical meet for 15 4 day intervals.

Dancing meet for 12 5 day intervals.

Paintng meet for 10 6 day intervals.

First quartr implies 3 months with 90 days.

So inorder to a nswer that how many days do they don/t meet atleast once in first quarter is got by rounding all other categories.

By counting all the intervals for other groups no.of days in Jan 8,Feb 7,Mar 9.

Total is 24.

STOLEN MANGOES

Three naughty boys stole some mangoes from a garden.Among them one counted and ate one.From the remainder he took precise third and went

back to sleep.

After sometime second boy woke up,counted the mangoes,ate one,took an exact third of the remaining and went back to sleep.

After sometime third boy also did the same.In the morning they found one which was rotten and hence threw it away from the remainder,they made an

exact division.How many mangoes did they steal?

Solution:

Let the noof mangoes be x

After the first boy had eaten noof mangoes =x-1

After taking 1/3 rd of remaining it is 2x-2/3

Second boy ate one and tok 1/3 then it is 2(2x-2/3 -1)=4x-10/9

Third boy ate and tok 1/3 as 2(4x-19)/27=8x-38/27

Deducting the rotten one from remaining noogf mangoes left=8x-38/27=8x-765/27

This is divided among the three equally 8x-65/27=3n

8x=81n+65

Let n be equal to odd number 2b+1

8x=81(2b+1)+65

4x=81b+73

Let b=2c+1

4x=81(2c+1)+73

2x=81c+77

Let c=2d+1

x=81d+79

Least value of x for d=0 is 79

for d=1 is 160

for d=3 is 241

On verfication,79-1=78/3=26

Hence 79 is the correct answer.

AN ELECTION PROBLEM

My club had a problem recently.They had to appoint a Ssecretary from among the men and a joint secretary from among the women.

We have a membership of 12 men and 10 women.In how many ways can the selection be made?

Solution:

As per the permutations and combinatins concept of mathematics,

out of 12 men one selected as secretary can be done in 12c1 ways

out of 10 women one selected as joint secretary can be done in 10c1 ways

Hence one secrtary and one joint secretary is 12*10=120

SNAPPING A PLANE

A plane has a span of 12 metres.It was photographed as it was flying directly overhead with a camera with a depth of 12cm.In the photo the

span of the plane was same.Can you tell how higher was the plane when it was snapped?

Solution:

Actual span of the plane was 12m

Span of the plane n photograph was 800m

Depth of the plane is 12000m=12cm

Hence,height of the plane when photographed be x

12000:800 = x:12

x=180m

A THRST PROBLEM

Pramatha and Pranathi went camping.They took their own water in bg plastic bottles.

Pramatha got thirsty and drank half the water in her bottle.A little later on she drank 1/3 f what was left.Sometime afterwards she drank 1/4

of what remained and so onPranathi also had a bottle of the same size.She drank half the bottle at the first instance ,half of what remained when she drank next and so on.

Aftr each took 10 drins ,the water Pramatha left was how many times greatr than the water Pranathi had left?

Soltuion:

Pramatha for the first drink 1/2

for the second drink 1/3

for the third drink 1/4

She drank 10 times and hence by the end of the 10th drink 1/11 of water she had in the bottle was over.

Pranathi for the first drink 1/2

for the second drin 1/4

for the third drink 1/8

So Pranathi as per the given information has drunk 1/1024 of water she had in the bottle.

Water left for Pramatha/Water left for Pranathi=1/11 / 1/1024 =1024/11

NAME OF THE EXCHANGE

In GreatBritain some years ago the first threeletters of a telephone number usd to indicate the name of the exchangeHow many such

arrangements of 3 letters is it possible to devise from the 26 letters of the alphabet?

Solution:

For permutations the no.of ways to select is npr=n!/(n-r)!

Hence out of 26 letters the possible outcomes are 26p3=26!/23!=15600

VALUE OF THE SERIES

Take a good look at the following series.

1-1/3+1/5-1/7+1/9-1/11+1/13....................

Find the value of the series and multiply the answer by 4.You will notce that a well-known vale approximates this product.Even more interestng is

that as you add more terms the approximation becomes closer.

Solution:

Ths is an Arithmetic progresson with value .76 when two terms adde becomes .77 and multiplidd by 4 it becomes 3.04 and 3.08 repeatition it is 3.14 which pi value adjusted to 2 decimals.

PLANTING TREES

If you wished to plant some trees so that each was equidistant from every pther tree,what is the largest number you would plant?

Solution:

From the above informatin,as per equidstant formula of triangle,it is an equilateral triangle.

Planting at all the three corners only 3 can be planted.

The centroid is the middle point placed exactly equidistant from all the corners.

Hence 4 plants can be planted at euqidistant.

LENGTH OF A TRAIN

A train is travelling at the speed of 96 kmph.It takes 3 seconds to enter a tunnel and 30 seconds more to pass thorugh it completely.What is

the length of the train and the tunnel?

Solution:

Speed of the train=95*5/18 m/sec

Time taken=3ssec

Length of the train=96*5/18*3=80m

Length of the tunnel=96*5/18*30=800m

A GAME OF BILLIARDS

Rajv,Sanjiv,Vinay were playing a game of Billiards.Rajiv can give Sanjiv 10 points in 50 and Sanjiv can give 10 points in 50.How many points in

50 must Rajiv gve Sanjiv to make an even game?

Solution:

Rajiv 50 Sanjiv 40

Sanjiv 50 Vinay 40

Sanjiv 40 Vinay 40*40/50=32

Rajiv 50 Sanjiv 40 Vinay 32

Rajiv gains 18 points than Vinay (50-32=18)

WOMEN AT CLUB SOCIALS

Women outnumbered men by 16 at a club social.Seventimes the no.of women exceeds nine times the no.of men by 32.What was the number of men and

women at club?

Solution:

Let W and M be the no.of women and men respectively.

Given W=M+16.................(1)

7W=9M+32................(2)

7*(1)=>7W=7M+112........(3)

(3)-(2) =>M=40

W=56

FILLING WINE IN BARRELS

A friend of mine in London has a very nice cellar.He has two large barrels in the cellar.The larger barrel is mostly empty.But the smaller

barrel is only 5/6 th full f wine while it can hold 536 litres.Supposing he empties the smaller barrel and fills the bigger barrel to find that the wine fills only 4/9 of it.How much wine would the larger barrel hold

when full?

Solution:

5/6-------------------536

4/9-------------------?

5/6*536=4/9*x

=>x=1005 litrs

WEIGHT OF A BRICK

We have a brick of regular size.It weighs 4 kilograms.How much do you think asmaller brick four times small, but made of the same material

weigh?

Solution:

The weight of the given brick = 4 Kilograms = 4000 grams

It is given that the smaller brick's volume is 4 times smaller than the given one.

The volume of smaller brick = 4 * 4 * 4 = 64 times smaller

The smaller brick's weight = 4000/64 = 62.5 grams

A JUMPING FROG

A frog starts climbing a 30 ft wall. Each hour it climbs 3 ft and slips back 2. How many days does it take him to reach the top and get out?

Solution:

Hours Feets 1 3ft - 2ft = 1ft

2 3ft - 2ft = 1ft

3 3ft - 2ft = 1ft

: : : : 27 3ft - 2ft = 1 ft

At the end of 27th hour the frog climbs 27fts and on 28th hour it climbs the remaining 3fts and comes out.

QUESTION OF PROBABILITY

My friend Parveen teaches at a school. One day she conducted a test for 3 of her students and when they handed back the test papers, they had

forgotten to write their names.

Parveen returned the papers to the students at random.What is the probability that none of the 3 students will get the right paper?

Solution:

The possible combinations for the given condition are:

1 2 3 condition met

1 2 3 No

1 3 2 Yes

2 1 3 No

2 3 1 Yes

3 1 2 Yes

3 2 1 No

The required probability = 2/6 = 1/3

MATHEMATICAL ODDITY

In the 20th century there are only seven years whose numbers are a mathematical oddity because their numbers signify a prime number.The first one of its kind was the year 1951.Can you name the other sin?

Solution:

1973,1979,1987,1993,1997,1999.

DOWN THE ESCALATOR

Recently,while in London,I decided to walk down the escalator of a tube station.I did some quick calculation in my mind.I found that if I wa

down 26 steps, I require 30 seconds to reach the bottom.However ,if I am able to step down 34 stairs I would only require 18 secionds to get to

the bottom.

If the time is measured from the moment the top step begins to descend to the time I step off the last step at the bottom , can you yel the

height of the stairway in steps?

Solution:

Given that after walking 26 steps time needed to reach the bottom is

30 seconds ---(1)

Similarly after walking 34 steps, it needs

18 seconds more to reach the bottom ---(2)

from (1) & (2) it is clear that

for (34-26) steps it took (30-18) seconds.

i.e; 12 seconds for 8 steps

in 1 second --------?

(1*8)/12=2/3 steps/sec.

i.e; 2 steps in 3 seconds

for 30 seconds -----------how many steps?

=> (30*2)/3=20 steps.

Finally Total number of steps = 26 + (steps covered in 30 seconds)=26+20=46 steps.

A COMPUTING PROBLEM

Compute (100-1)(100-2)(100-3).........................(100+1)(100+2)(100+3)=?

Solution:

=>(100-1)(100-2)(100-3).........(100-100)(100+1)(100+2)(100+3)

=>(100-1)(100-2)(100-3).........(0)(100+1)(100+2)(100+3)

=0

A CIRCLE AND A TRIANGLE

What do you call a circle which passes through the vertices of of a triangle?

Solution:

Circumscribed.The meaning to circumscribe is to describe a figure round another so as to touch it at points without cutting.This is exactly

what takes place with the circumscribed circle.

To find the center of such a circle,we have to bisect the sides of a triangle and errect perpendiculars which are concurrent at the

circumcentre.

The radius r of the circumscribed circle of the triangle ABC is given by

R=a/2SinA=b/2SinB=c/2Sinc

MISSING TERMS

48,60,58,72,68,104.......

Here is a sequenc.Can you find the two missing terms?

Solution:

The odd terms are in the decimal system and differ by 10.And each even term is the preceeding odd term expressed in the octonamy system.78-

8=70,remainder 6:9:8=1,remainder 1.Therefore the next two terms are: 78,116

PACKETS OF CANDY

If 6 men can pack 6 packets of candy in 6 minutes.How many are required to pack 60 packets in 60 minutes?

Solution:

Given that for 6 men to pack 6packets of candy it takes 6 miutes.i.e., for 1 man to pack 1 packet it takes 1 miute.

Hence,for 60 packets to be packed in 60 miutes we need 60 men.

A PROBLEM OF WEIGHT

In my neighbourhood lives a man who weighs 200 pounds.He has two sons.They both weigh 100 pounds each.On a festival day they decide to go across the river on a boat to vissit some relations.But the boat could carry a maximum load at only 200 pounds.Yes they managed to come across

the river by boat.How did they?

Solution:

Let us assume that c1 ----------------->first son

c2 ---------------->second son

f ------------------>father

First the two sons c1,c2rowed across the river and c1 stayed behind while c2 returned in the boat to his father.

The son remained behind while the father crossed the river.Then the other son brought back and the two brothers c1,c2 rowed over

together.

A PROBLEM OF CANDY BARS

Recently I attended a birthday party.All the children in the party were given candy bars.All the children got 3 candy bars each except the child

sitting in the end.She got only 2 candy bars.If only child had been given 2 candy bars there would have been 8 candy bars remaining.How many

candy bars were there altogether to begin with?

Solution:

Suppose that there were x children at the party.

If we distribute the candies in the above mentioned ways,then the resulting expressions

3(x-1)+2---------------------(1)

2x+8-------------------------(2)

3(x-1)+2=2x+8

3x-3+2=2x+8

3x-1=2x+8

x=9

Therefore the no.of candies for distribution 2x+8=2*9+8=18+8=26

FIND OUT THE SUM

What is the sum of all numbers between 100 and 1000 which are divisible by 14?

Solution:

Let us assume that the sum is S

s=112+126+........+994

s=14(8+9+10+.....................+71)

s=14(8+71)(71-8+1)/2=7(79)(64)=35392

WALKING ALL THE WAY

One day I decided to walk all the way from Banglore to Tumkur.I started exactly at noon and some one I know in Tumkur decided to walk all the

way from Tumkur to Banglore and she started exactly at 2 PM on the same day.

We met on the Banglore - Tumkur road at 5 past four and we both reached our destination at exactly the same time.At what time did we

both arrive?

Solution:

There fore Total time = 2:00 PM + 3:55 PM + 1:55 PM = 7:50PM

THE TRAINS AND THE FALCON

Two trains start from two opposite directions towards each other.The stations from which they start are 50 miles apart.Both the trains start at the same time towards the other train. As soon as it reaches the

second one, it fies back to the first train and so on and so forth. It continues to do so, flying bacwards and forwards from one train to other

until the trains meet.Both the trains travel at a speed of 25 miles per hour,and the bird

flies at 100 miles per hour. How many miles will the falcon have flown before the trains meet?

Solution:

The trains travel at 25 miles per hour.

Hence they will meet after travelling for one hour and the falcon also must have been flyingfor one hour. Since it travels at 100 miles per

hour the bird must have flown 100 miles

VALUE OF 'S'

S434S0, what number must be substituted with to make it divisible by 36?

Solution:

To be divisible by 36, the number has to be divided by 4 and 9

To be divisible by 4 , the number 'S' must be an even number and to be divisible by 9, the sum of all the digits of the number must be either

equal to 9 or a multiple of 9

i.e; S + 4 + 3 + 4 + S + 0 = 9n

The only digit that meets these two condition is 8

HEIGHT OF A ROOM

Given the floor area of a room as 24 feet by 48 feet,and the space diagonal of the room as 56 feet,Can you find the height of the room?

Solution:

We know that,

Volume of a cube =l pow(2) + b pow(2) + h pow(2).

Here the values of l & b are given.

We also know that

(diagonal)pow(2) = (length)pow(2) + (breadth)pow(2).

=> (x)pow(2) = 24 pow(2) + 48 pow(2)

=> x = 24squareroot(5);

Therefore,volume= h pow(2) + x pow(2) =56 pow(2)

=> h=16

Therefore,height of the room=16 ft.

A QUESTION OF DISTANCE

It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen I was driving from Banglore to Brindavan

Gardens. It took me 1 hour and 30 miutes to complete the journey.

After lunch I returned to Banglore. I drove for 90 minutes.How do you explain it?

Solution:

90 minutes = 1 hour 30 minutes. Hence,the driving time there and back is absolutely the same

because 90 minutes and 1 hour and 30 minutes are one and the samething.

ARRANGE THE DIGITS:

Arrange the digits 1,2,3,4,5,6,7,8,9 in order from left to right and use only + or _ signs so as to produce a result of 100?

Sol:

123-45-67+89

DIVISION OF 45:

Can you divide the number 45 into four parts such that when 2 is added to the first part, 2 is subtracted from the second part, 2 is multiplied

by the third part, and the fourth part is divided by 2.All the four results to be the same number.

Sol:

Let us take A,B,C,D are the four equal parts and their sum is equal to 45.

A+B+C+D = 45

Given that,

A+2 = B-2 = C*2 = D/2 ----------------(1)

=>A = B-4 ; C = (B-2)/2 ; D = 2(B-2)

=>B-4 + [(B-2)/2] + 2(B-2) = 45

=>B=12

:. A=8 , C=5 , D=20

And condition (1) is satisfied.

i .e; 8+2 = 12-2 = 5*2 = 20/2

SPECIAL NUMBER:

What is the special about the number 1729?

Sol:

This is popularly known as Ramanujan's number. This is the known number that is a sum of two cubes in two different ways.

i .e; (10*10*10) + (9*9*9) = 1729

( 12*12*12) + (1*1*1) = 1729

PRICE OF A BOTTLE:

A bottle and its cork together cost Rs 1.10, and the bottle costs Rs 1.00 more than its cork. What is the price of the bottle?

Sol:

Let us assume that,

B = Price of the bottle

C = Price of the cork

It is given that,

B + C = Rs 1.10 -------------------(1)

and B â€“ C = Rs 1.00 -------------------(2)

From the equations (1) and (2) it is clear that

B = Rs 1.05

C = Rs 0.05

A QUESTION OF DISTANCE:

It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen. I was driving from Banglore to Brindavan Gardens. It took me one hour and 30 minutes to complete the journey. After lunch I returned to banglore. I drove for 90 minutes. How do you

explain it?

Sol:

90 minutes = 1 hour 30 minutes.

Hence, the driving time there and back is absolutely same because 90 minutes and 1 hour 30 minutes are one and the same.

FOR THE CHARITIES:

One day when I was walking on the road in New Delhi, a group of boys approached me for donation for their poor boys' fund. I gave them a

Rupee more than half the money I had in my purse. I must have walked a few more yards when a group of women approached me for donation, for an orphanage. I gave them 2 Rupees more than half the money I had in my

purse. Then after a few yards I was approached by a religious group for a donation to the temple they were building. I gave them 3 Rupees more

than half of what I had in my purse.

At last I returned to my hotel room, I found that I had only one Rupee remaining in my purse.

How much money did I have in my purse when I started?

Sol:

Suppose that the money in his purse when he started = x -------------------(1)

For poor boys fund he gave x/2 + 1 Rs/- --------------------------------------(2)

i .e; 1 Rupee more than half the amount he had.

Now he left with [x - (x/2 + 1)] =(x-2)/2 Rs/-

For Orphanage he gave [(x/2 â€“ 1)/2] +2 = (x+6)/4 Rs/- ----------------(3)

Now he left with [(x-2)/2] â€“ [(x+6)/4] = (x-10)/4 Rs/-

For temple building he gave [(x-10)/4]/2 + 3 = (x+14)/8 Rs/- ---------(4)

Now he left with [(x-10)/4] â€“ [(x+14)/8] = (x-34)/8 Rs/-

Finally he had 1 Rupee in his purse.

i .e; Actual amount â€“ Expended amount = 1

:. from (1),(2),(3) and (4) we have

x-{ [ (x+2)/2 ] + [ (x+6)/4 ] + [ (x+14)/8 ] } = 1

=>x-34=8

=> x=42

:. The original amount in his purse at the beginning = Rs 42/-

A PAIR OF PALLINDROMES

Multiply 21978 by 4.Comment about the result?

Solution:

21978*4=87912.

If we clearly observe the two numbers 21978 and 87912, the resultant number ie; 87912 is the reverse number of the number 219780.

There fore these two numbers are a pair of paindromes.

A COMPUTING PROBLEM

Compute: [5-2/(4-5)]pow(2).

Solution:

[5-2/(4-5)]pow(2)

=[5-2/(-1)]pow(2)

=[5+2]pow(2)=49.

CONTINUE THE SERIES

1,3,6,10.Name the next three numbers in the series.

Solution:

The series is +2,+3,+4,-----------------

There fore next three numbers are:

10+5,10+5+6,10+5+6+7 = 15,21,28.

NAME FIVE TERMS OF ANOTHER SERIES

These are the numbers that are the first five terms of a series that add upto 150.Can you name five terms of another series without fractons

that add upto 153?[ex: 10,20,30,40,50. sum=150.]

Solution:

Each term in this series is a factorial, in other words,the product of a the numbers from 1 to that particular term considered.

The first five terms of the series are,there fore 1,2,6,24,120.

Their sum is 153 and are factorials of 1,2,3,4,5 respectively.

FIND OUT THE TIME

What does 1408 hours mean?

Solution:

1408 hours is actually 8 minutes past 2 PM.

This is the system of twenty-four-hour cock.Writing the hours and minutes this way is a sensible means of avoiding confusion between

AM and PM.

FIND OUT TTHE PATTERN

What do you think the pattern is? 6,24,60,120,210,336,â€¦â€¦â€¦â€¦â€¦â€¦

Solution:

The series is 1.2.3, 2.3.4, 3.4.5, 4.5.6, 5.6.7, 6.7.8,----------

The next numbers would be 7.8.9, 8.9.10,-------

(or) 504, 720,-----------------------

THE TRAIN AND THE CYCLIST

A railway track runs parallel to a road until a bend brings the road to a level crossing. A cyclist rides along to work along the road every day

at a constant speed of 12 miles per hour.

He normally meets a train that travels in the same direction at the crossing

One day he was late by 25 minutes and met the train 6 miles ahead of the level crossing. Can you figure out the speed of the train?

Solution:

Suppose that the train and the cyclist meet everyday at the crossing at 8:00A.M. i. e; starts at 7:00A.M

Since the cyclist is late by 25 minutes, he starts at 7:25A.M

As his speed is 12 miles per hour, he reaches the crossing at

7:25A.M + 1 Hour = 8:25A.M

By 8:30A.M the train is 6 miles ahead of the cyclist

The difference between their timings = 8:30A.M â€“ 8:25A.M = 5 Minutes

The difference between their distances = 6 Miles

Therefore,the train travels 6 miles in 5 minutes

In 1 minute it travels ---------------------?

= [(1 * 6) /5] * 60

= 72 Miles/hour

HEIGHT OF THE PALM TREE

A palm tree was 90 cm high, when it was planted. It grows by an equal number of cm each year, and at the end of the seventh year it was one ninth taller than at the end of the sixth year. Can you tell how tall

was the tree at the end of the twelfth year?

Solution:

Suppose that the tree grows x cm each year

Height of the tree at the end of the sixth year = (90 + 6x) cm

Growth in seventh year is,

X = 1/9(90 + 6x) cm

x = 10 + 2x/3

x = 30

Therefore the height of the tree at the end of the twelfth year=(90+12*30)=450cm

PROBLEM OF AGE

Recently I attended a cocktail party. There was a beautiful young lady, who seemed very vitty and intelligent. She was posed a question, â€œ how

old are you? â€.�

She answered , â€œ my age 3 years hence munltiply by 3 and from that subtracted

3 times my age 3 years ago will give you my exact age? How od is the lady?

Solution:

Let the age be x

Age after 3 years wi be (x + 3)

Age before 3 years = (x â€“ 3)

Hence 3(x + 3) â€“ 3(x â€“ 3) = x

x = 3x + 9 â€“ 3x + 9

x = 18

Therefore the age of the lady = 18 years

CONSECUTIVE NATURASL NUMBERS

There are two consecutive natural numbers whose product is equal to the product of three consecutive natural numbers,for example x(x+1) = y(y+1)

(y+2).What are the two numbers?

Solution:

14 * 15 = 5 * 6 * 7

SOME GLUTTON

A man sitting beside me at a hotel ate idlis one after the other by ordering plate by plate. He said to me after drinking some water the

last one I ate was my hundredth idli in last five days. Each day I ate 6 more than the previous day. Can you tell me how many he ate

yesterday?

Soluton:

First day the number of idlis he ate be x

Second day the count is (x+6)

Third day ------------------------- (x+12)

Fourth day ------------------------(x+18)

Fifth day ---------------------------(x+24)

Total is 5x + 6(1+2+3+4) =100

5x + 60 =100

x = 8

Day Idlis

1 8

2 14

3 20

4 26

5 32

So, on fourth day the number of idlis the man ate were 26.

Reasoning

Reasoning CLASSIFICATION TYPE

Read the following information carefully and answer the questionsthat follow:

1)

There are six cities A,B,C,D,E and FA is not a hill station.B and E are not historical places.D is not an industrial city.A and D are not historical craj_konkepudiities.A and B are not alike.

1.Which two cities are industrial centres?a)A,B b)E,F* c)C,D d)B,F e)A,D

2.Which two cities are historical places?a)A,C b)B,F c)C,F* d)B,E e)A,D

3.Which two cities are hill stations?a)A,B b)C,A c)B,D* d)A,F e)none

4.Which city is a hill station and an industrial centre but not ahistorical place?a)E* b)F c)A d)B e)C

5.Which two cities are neither historical places nor industrialcentres?a)A,B b)D,E c)F,C d)B,D* e)none

Solution:A B C D E F

Historical x x y x x y

Industrial y x y x y y

Hillstation x y y y y y

2)

Five friends Indu,Pinki,Sai,Srujan,Pavan travelled to five differentcities of Chennai,Calcutta,Delhi,Bangalore and Hyderabad by five differnet modes of transport of Bus,Train,Aeroplane,Car and Boat fromMumbai. The person who travelled to Delhi did not travel by Boat.Sai went to Banalore by Car and Pinki went to Calcutta by Aeroplane.Srujan travelled by Boat whereas Pavan travelled by Train.Mumbai is not connected by bBus to Delhi and Chennai.

1.Which of the following combinations of person and mode is not correct?a)Indu-Bus b)Pinki-Aeroplane c)Sai-Car d)Srujan-Boat e)Pavan-Aeroplane*

2.Which of the following combinations is true for Srujan?a)Delhi-Bus b)Chennai-Bus c)Chennai-Boat*d)Data inadequate

3.Which of the following combinations of place and mode is not correct?a)Delhi-Bus* b)Calcutta-Aeroplane c)Bangalore-Car d)Chennai-Boat

4.Person travelling to Delhi went by which of the following modes?a)Bus b)Train* c)Aeroplane d)Car e)Boat

5.Who among the following traj_konkepudiravelled to Delhi?a)Sai b)Srujan c)Pavan* d)Data inadequate e)none

Solution:Sai travels by Car.Pinki travels by Aeropraj_konkepudilane.Srujan travels by Boat.Pavan travels by Train.Indu travels by Bus.Sai goes to Bangalore.Pinki goes to Calcutta.Bus facility is not there for Delhi or Chennai.Indu goes to Hyderabad by Bus.

From given information it is clear that Srujan goes by Boat but notto Delhi.Hence Srujan goes to Chennai.Pavan goes to Delhi.

PLACE MODE

Indu Hyderabad Bus

Pinki Calcutta Aeroplane

Sai Bangalore Car

Srujan Chennai Boat

Pavan Delhi Train

3)

Four youngmen Thirbhuvan,Thrishanth,trinath,Trived are friendly

with four girls Indira,Madhuri,Swetha and Dimple.Indira and Swethaare friends.Trinath's girlfriend does not like Indira and Swetha.Madhuri does not care for Trinath.Trishanth's girlfriend is friendly with Indira.Indira does not like Thribhu.

1.Who is Thribhu's girlfriend?a)Indira b)Madhuri* c)Swetha d)Dimple

2.With whom is Indira friendly?a)Thribhuvan b)Thrishanth c)Trinath d)Trived*

3.who is Dimple's boyfriend?a)Trived b)Trinath* c)Thrishanth d)Thribhuvan

4.Who does not like Indira and Swetha?a)Dimple* b)Thribhuvan c)Trived d)Trinath

Solution:Given Indira and Swetha are friends.Thrishanth's girlfriend is friendly with Indira.Hence,Thrishanth's girlfriend is Swetha.Given,Trinath's girlfriend does not like Indira and Swetha.=>She might be Madhuri or Dimple.But Madhuri does not care for Trinath.=>Trinath's girlfriend is Dimple.Given,Indira does not like Thrinbhuvan.=>Thribhuvan's girlfriend is Madhuri.Clearly,Trived's girlfriend is Indira.

BOY GIRLThribhuvan MadhuriThrishanth SwethaTrinath DimpleTrived Indira

COMPARISION TYPE QUESTIONS

1)Clues will be given regarding comparisions among a ssset of personsor things with respest to one or more qualities.After analysing aproper ascending ,descending sequence is formed and then are supposed to answer.There are five friends-Sachin,Sourav,Rahul,Zaheer and Yuvi.Sachin is shorter than Sourav but taller than Yuvi.Rahul is the tallest.Zaheer is little shorter than Sourav and little taller than Sachin.

1.Who is the shortest?a)Yuvi* b)Sachin c)Zaheer d)Sourav e)none

2.If they stand inorder of their heights,who will be in the middle?a)Sourav b)Yuvi c)Sachin d)Zaheer* e)none

3.If they stand inorder of their increasing heights,who will be the second?a)Zaheer b)Sachin* c)Yuvi d)Sourav e)none

4.Who is the second tallest?a)Sachin b)Sourav* c)Zaheer d)Yuvi e)none

5.Who is taller than Zaheer but shorter than Rahul?a)Sourav* b)Yuvi c)Sachin d)datainadequate e)none

Solution:Sachin < Sourav Sourav > YuviRahul is tallestZaheer < Sourav Zaheer > Sachin

Yuvi < Sachin < Sourav < Rahul and Sachin < Zaheer < SouravYuvi < Sachin < Zaheer < Sourav < Rahul

2)

Among five boys Vineeth is taller than Manick,but not as tall asRavi,Jacob is taller than Dilip but shorter than Manick.Who is the tallest in their group?

a)Ravi* b)Manick c)Vineeth d)can't say e)none

Solution:Manick < Vineeth ; Vineeth < Ravi ; Dilip < Jacob ; Jacob < Mainck=>Dilip < Jacob < Manick < Vineeth < Ravi

3)Sudhanshu is as much older than Kokila as he is younger than Praveen.Nitin is as old as Kokila.Which of the following is wrong?a)Kokila is younger than Praveen.b)Nitin is younger than Praveen.c)Sudhanshu is older than Nitin.d)Praveen is not the oldest.*e)Kokila is younger than Sudhanshu.

Solution:=>Kokila < Sudhanshu; Praveen > Sudhanshu; Nitin = Kokila=>Nitin = Kokila < Sudhanshu < Praveen=>Praveen is the oldest. SELECTION BASED ON GIVEN CONDITIONS

1)

A few essential criteria for selection of group of items are given:From amongst five doctors A,B,C,D and E;four engineers G,H,K and L;six teachers M,N,O,P,Q and R some teams are to be selected.A,B,G,H,O,Pand Q are females and the rest are males.Where ever there is a male doctor ,no female teacher.Where ever there is a male engineer, no female doctor.There shall not be more than two male teachers in any team.

1.If the team consists of 2 doctors,3 female teachers and 2 engineers,the members of the team are:

a)A B O P Q G H* b)C D O P Q G H c)C D K L O P Qd)D E G H O P Q

2.If the team consists of 2 doctors,1 engineer and 4 teachers all thefollowing are possible excepta)A B G M N O P b)A B H M O P Q c)A B H M R P Q d)A B K N R P Q*

3.If the team consists of2 doctors,2 female teachers and two engineersall the following teams are possible except:a)A B G H O Q b)A B G H P Q c)A B K L P Q d)O P G H A B

4.If the team consists of 3 doctors,2 male engineers and 2 teachers,the members of the team could be:a)A B C K L M Rb)B C D K L N Rc)C D E K L M Nd)C D E K L P R

5.if the team consists of two doctors,two engineers and two teachers,all the following teams are possible except:a)A B G H O P b)A B G H M N c)C E K L N R d)C D K L O P

Explanation:Doctors A B C D E Engineers G H K L Teachers M N O P Q R

Given Females A B G H O P Q

Males C D E K L M N RMale doctor and female teachermale engineer and female doctorMore than two teachers/team combination are not allowed.

1Q) Explanation:

Doctors are A B C D EFemale teachers are O P QEngineers are G H K LMale doctor and no female teacher => A,B are Doctors.So the Team consists of A B O P Q G H.

2Q) Explanation:

Teachers are M N O P Q RFour are neededThree are male teachers.Female teachers are also to be selected.Hence no male doctors C D Eare selected.So Doctors => A BBoth doctors are Females =>no male Engineer to be selected.Ans:Team =>A B K N R P Q.

3Q) Explanation:

The doctors are A B C D E,Female Teachers are O P Q,Engineers are G H K L,

Since 2 Female Teachers are to be selected.So male doctors i.e;C D Ecannot be selected.So, 2 Doctors selected will be A,B.Both Doctors are Females.So, male Engineer K L cannot be selected.G H are chosen.=>A B K L P Q are selected.

4Q)Explanation:

Doctors A,B,C,D,EMale Engineers K,LTeachers M,N,O,P,Q,RMale Engineers =>no female Doctors=>no A,B3 Doctors to be selected are C,D,E who are all males.

FAMILY BASED PROBLEMS

1)There is a group of six persons A,B,C,D,E and F from a family.They are psychologist,manager,lawyer,jeweler,doctor and engineer.The doctor is the grandfather of F who is a psychologist.The manager D is married to A.C,the jeweler is married to the lawyer.B is the mother of F and E.There are two married couples in the family.

1.What is the profession of E?a)Doctor b)Jeweler c)Manager d)Psychologist e)None*

2.How is A related to E?a)Brother b)Uncle c)Father d)Grandfather*

3.How many male members are there in the family?a)One b)Three c)Four d)Data insufficient*

4.What is the profession of A?a)Doctor* b)Lawyer c)Jeweler d)Manager

5.Which of the following is one of the pairs of couples in the family?a)AB b)AC c)AD* d)canâ€™t say

Solution: F is a Psychologist. B is the mother of F and E. =>E is brother or sister of F. There are two married couples in the family. Since D is married to A,C the jeweler is married to lawyer B. Manager D is married to A means A is doctor,grandfather of E and F.So no one else is engineer till this point.=>E is engineer.

AD--->CB------>EF

2)Prashanth Arora has three children-Sangeeta,Vimal and Ashish.Ashishmarried Monika,the eldest daughter of Mr.andMrs.Roy.The Roys married

their youngest daughter to the eldest son of Mr.and Mrs.Sharma,and they had two children named Amit and Shashi.The Roys have two more children,Roshan and Vandana,both elder to Veena.Sameer and Ajay are sons of Ashish and Monika.Rashmi is the daughter of Amit.

Prashanth Arora-------> 1.Sangeeta 2.Vimal 3.Ashish(Monika)--------------------> 1.Sameer 2.Ajay

Mr & Mrs.Sharma----------------> 1.Amit(Veena)---------------->1.Rashmi 2.Shashi

Mr & Mrs.Roy---------------------> 1.Monika 2.Roshan 3.Vandana 4.Veena

1.What is the surname of Rashmi?a)Sharma* b)Roy c)Arora d)canâ€™t say e)none

2.How is Sameer related to father of Monika?a)Grandson* b)Son c)Cousin d)Son-in-law e)none

3.What is the surname of Sameer?a)Roy b)Sharma c)Arora* d)canâ€™t say e)none

4.How is Mrs.Roy related to Ashish?a)aunt b)Mother-in-law* c)mother d)sister-in-law

3)

In a family of six persons,there are three generations.Each person has seperate profession and also they like different colours.There are twocouples in the family.Rohan is a CA and his wife neither is a doctornor likes green colour. Engineer likes red colour and his wife is a teacher.Mohini is mother-in-law of Sunita and she likes orange colour.Vinod is grandfather of Tanmay,Tanmay is principal likes black colour.Nanu is granddaughter of Mohini and she likes blue colour.Nanuâ€™smother likes white colour.

1.Who is an Engineer?a)Nanu b)Mohini c)Sunita d)canâ€™t say e)none*

2.What is the profession of Sunita?a)engineer b)doctor c)teacher* d)canâ€™t say e)none

3.Which of the following is the correct pair?a)Mohini-Vinod and Rohan-Sunita*b)Vinod-Mohini and Rohan-Nanuc)Rohan-Sunita and Tanmay-Nanud)canâ€™t say e)none

4.How many ladies are there in the family?a)2 b)3* c)4 d)canâ€™t say e)none

5.Which colour is liked by CA?a)green* b)white c)white/green d)canâ€™t say e)none

Solution:(Engineer) Vinod- Mohini(Teacher)------------->(CA)Rohan-Sunita----- ----->Nanu&Tanmay(principal)

Vinod------------RedMohini-----------OrangeRohan-------------GreenSunita--------------WhiteNanu---------------BlackTanmay-----------Black

Mohini is mother-in-law of Sunita and grandmother of Nanu.Vinod is grandfather of Tanmay.So,Nanu and Tanmay represent third generation.Mohini and vinod form a couple and represent first generation.Rohan and Sunita form the other couple and represdent second generation.Rohan is a CA.Since,engineer is married,Vinod is engineer.Given Vinod loves red colour.Mohini is ateacher and likes Orange colour.Nanu likes blue colour.Tanmay is principal and likes black colour.Sunita Nanuâ€™s mother likes white colour.=>Rohan likes green colour.

54). A soldiar looses his way in a thick jungle at random walks

from his camp but mathematically in an interestingg fashion.

First he walks one mile east then half mile to north. Then 1/4

mile to west, then 1/8 mile to south and so on making a loop.

Finally hoe far he is from his camp and in which direction.

ans: in north and south directions

1/2 - 1/8 + 1/32 - 1/128 + 1/512 - and so on

= 1/2/((1-(-1/4))

similarly in east and west directions

1- 1/4 + 1/16 - 1/64 + 1/256 - and so on

= 1/(( 1- ( - 1/4))

add both the answers

55). there are six boxes containing 5 , 7 , 14 , 16 , 18 , 29

balls of either red or blue in colour. Some boxes contain only

red balls and others contain only blue . One sales man sold one

box out of them and then he says " I have the same number of red

balls left out as that of blue ". Which box is the one he solds

out ?

Ans : total no of balls = 89 and (89-29 /2 = 60/2 = 30

and also 14 + 16 = 5 + 7 + 18 = 30

56) Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 for 30 cows . How many cows can eat away the same in 96 days.? Ans : 20 g - grass at the beginning r - rate at which grass grows, per day y - rate at which cow eats grass, per day n - no of cows, to be found g + 24*r = 70 * 24 * y g + 60*r = 30 * 60 * y g + 96*r = n * 96 * y Solvin

57)Three friends divided some bullets equally. After all of them shot 4 bullets the total no.of remaining bullets is equal to that of one has after division. Find the original number divided.

Ans. x x x x-4 x-4 x-4 3x-12 = x x= 6 ans is 18 2 marks

g, n = 20.

58)A ship went on a voyage after 180 miles a plane statrted with 10 times speed that of the ship. Find the distance when they meet from starting point.

Ans. 180 + (x/10) = x x = 20 ans is 180+20=200miles. 2 marks

59) There N stations on a railroad. After adding x stations 46 additional tickets have to be printed. Find N and X.

Ans. let N(N-1) = t; (N+x)(N+x-1) = t+46; trail and error method x=2 and N=11 4 marks

(60). A + B + C +D = D + E + F + G = G + H + I =17.IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9. 8MARKSANS : A = 4 ,B = 2, C =6, D = 5, E = 3, F = 8,G = 1, H = 7, I = 9.

61)Brain Teaser No : 00338

Insert mathematical functions to convert the 3 numbers on the left side of the equation to equal 6. I filled in the 2's (using the addition function twice) for you to get you started.

1 1 1 = 6

2 + 2 + 2 = 6

3 3 3 = 6

4 4 4 = 6

5 5 5 = 6

6 6 6 = 6

7 7 7 = 6

8 8 8 = 6

9 9 9 = 6

Answer

(1 + 1 + 1)! = 6

2 + 2 + 2 = 6

(3 * 3) - 3 = 6

sqrt(4) + sqrt(4) + sqrt(4) = 6

5 + (5 / 5) = 6

6 + 6 - 6 = 6

7 - (7 / 7) = 6

8 - sqrt(sqrt(8+8)) = 6

(sqrt(9) * sqrt(9)) - sqrt(9) = 6


500 men are arranged in an array of 10 rows and 50 columns according to their heights.

Tallest among each row of all are asked to come out. And the shortest among them is A.

Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B.

Now who is taller A or B ?

Answer

No one is taller, both are same as A and B are the same person.

As it is mentioned that 500 men are arranged in an array of 10 rows and 50 columns according to their heights. Let's assume that position numbers represent their heights. Hence, the shortest among the 50, 100, 150, ... 450, 500 is person with height 50 i.e. A. Similarly the tallest among 1, 2, 3, 4, 5, ..... 48, 48, 50 is person with height 50 i.e. B

Now, both A and B are the person with height 50. Hence both are same.


A, B, C and D are related to each other.

One of the four is the opposite sex from each of the other three. D is A's brother or only daughter. A or B is C's only son. B or C is D's sister.

How are they related to each other?

Answer

A, B & D are males; C is female. B is C's only son. A & D are C's brothers. A(male) --- C(female) --- D(male)

|

|

B(male)Work out which relation can hold and discard the contradictory options.

From (2) and (4), D can not be a only daughter and have a sister (B or C). Hence, D is A's brother i.e. D is a Male.

From (4), let's say that B is D's sister i.e. B is Female. From (3), A is C's only son i.e. A is Male. But D is A's brother which means that A is not C's only son. Hence, our assumption was wrong.

Thus, C is D's sister i.e. C is Female. And B must be C's only son.

Now it is clear that D & B are Males and C is Female. A must be a Male as only one of them is of opposite sex from each of the other three. And he is C & D's brother.


There are 10 cups placed on a table such that 3 are face up and 7 are bottom up. A move is defined as inverting a pair (compulsorily) of cups.

What is the minimum number of moves required to make all the cups face the same way?

Answer

It is not at all possible. There is no way that one can do that with given Move.

A move is defined as inverting a pair of cups, compulsorily. Also, there are odd number of cups face up (3)and odd number of cups bottom up (7). Now whenever you make a move you have to

invert 2 cups compulsorily. Hence, always odd number of cups will be face up and bottom up, whatever move you make.


There is a Tibetan monastery filled with a number of monks. The number of monks is more than one, and not unreasonably large.

God visits the monks one night and tells them that he is planning on destroying the world unless they can solve a puzzle. He puts a red dot on the foreheads of a certain number of monks, X. X is greater than zero.

Then he takes away all means of communication between the monks, and all means by which the monks can see themselves. No mirrors, no hand signals, no nothing. A single monk can see all the other monks, and see if another monk has a dot on his head or no, but he cannot see himself, and he cannot communicate in any way with any other monk.

God then says that all of the monks with dots, and only those monks with dots, must jump off of the cliff next to the monastery and sacrifice themselves to save the world, at EXACTLY the same moment, on EXACTLY the same day.

The next morning, all of the monks from the monastery walk out to the edge of the cliff. They pause and look around, and then walk quickly back to the monastery.

The second morning, all of the monks from the monastery walk to the edge of the cliff. They pause, look, and walk back to the monastery.

The third morning, all of the monks walk to the edge of the cliff. And then all of the monks with dots, and only those monks with dots walk to the cliff edge and jump off at the exact same moment.

The question: how many monks jumped, and how did they organize the jump?

Answer

Total 3 monks jumped in, on the third day.

There are two important points. One is that there is atleast one monk with red dot. And the second is that all the monks are intelligent ;-)

On the day 1 If there is only one monk with red dot, he would see all monks without the red dot. So he would know that he is the one and jump in. But no one jumped.

On the day 2 Now, all the monks know that there are atleast 2 monks with the red dots. If there are exactly two monks with the red dot, they would see only one monk with the red dot. So they would know that they have to jump in. But no one jumped.

On the day 3 Now, all the monks know that there are atleast 3 monks with the red dots. If there are exactly three monks with the red dot, they would see only two monks with the red dot. So they would know that they have to jump in.

Thus total 3 monks jumped in on the third day.

P.S. If monks jumped in on Nth days, then there are total N monks with the red dots.


There is a cup with water and ice in it.

Will the water level change if the ice melts?

Answer

The water level will not change.

The ice displaces water equal to its mass and when ice melts, its mass remains the same.

However, if the ice is not floating in the water - either supercooled or not being allowed to float by pressure or not enough water to float - the level may change.


Here is the family tree of Mr. RAHUL

RAHUL

|

---------------------------------------------

| | | |

RATISH YASH OM TRILOK

| | ?

-------- ------------------

| | | | |

AMAR AMIT RAM HARSH ASHOK

| |

----- -------

| | | | |

How many children does Mr. TRILOK have?

Answer

TRILOK have 5 children.

Name of the person and number of his children are related by some pattern.

Assign each vowel following values. A=0 E=1 I=2 O=3 U=4

The number of children to any person is the sum of the values represented by vowels in his name. RATISH = 0(A) + 2(I) = 2 OM = 3(O) = 3 AMIT = 0(A) + 2(I) = 2 ASHOK = 0(A) + 3(O) = 3 TRILOK = 2(I) + 3(O) = 5

Hence, TRILOK have 5 children.


Which animal is next in this sequence: Gorilla, Hedgehog, Camel, Mouse, Fox, Bear, Mole, ?

Choose from: Porcupine, Badger, Cow, Giraffe, Squirrel, Rhinoceros

Do explain your answer.

Answer

The next animal in the series is Badger.

The pattern is - number of characters in the animal name is equal to number of characters in the ordinally respective calendar month name, starting with January.

First animal (January-7) : Gorilla Second animal (February-8) : Hedgehog Third animal (March-5) : Camel Fourth animal (April-5) : Mouse Fifth animal (May-3) : Fox Sixth animal (June-4) : Bear Seventh animal (July-4) : Mole

Now, the next animal must be 6 characters long (for August). Thus, the next animal in the series is Badger.

If you follow the pattern, then the next possible animals (from the given list) will be Porcupine, Giraffe, Squirrel, etc.


What is the last digit of 746? In other words, what will the remainder be, if 746 is divided by 10?

Don't try to solve this on calculator, you may get the wrong answer. Also, do explain your answer.

Answer

The last digit of 746 is 9.

The powers of any number have a repeating pattern for the last digit. It can be found easily without performing the entire multiplication of each power.

Let's consider the powers of 7.

7N Value Last Digit

70 1 1

71 7 7

72 49 9

73 343 3

74 2401 1

75 16807 1

76 117649 1

Note that there is a repeating pattern of four numbers (1, 7, 9, 3) for the powers of 7. Hence, the last digit of 744 will be 1, of 745 will be 7 and of 746 will be 9.

Also, there is no need to actually perform the entire multiplication. Start with 1, multiply it by 7, discard all digits except units and multiply again by 7 and so on.


Someone at a party introduces you to your mother's only sister's husband's only sister-in-law. She has no brothers.

What do you call this lady?

Answer

Mother or Mom


There is a man who is looking at a picture. He is the only son in the family and his dad has no siblings.

He says, "This person's father is my father's son".

Who is this person in the picture?

Answer

The person in the picture is his CHILD.

He is saying that "this person's father, is my father's son." Now, my father's son is himself as he is the only son in the family. And he is the father of the person in picture. Hence the person in picture is his son or daughter.


In Laloo's family, each son has the same number of sisters and brothers. Also, each daughter has twice the number of brothers than sisters.

How many sons and daughters does Laloo have?

Answer

4 sons and 3 daughters

Laloo must be having one more son then daughter, as each son has same number of sisters and brothers. Using this and little trial-and-error, we can get the result i.e. 4 sons and 3 daughters.

Each brother has 3 sisters and 3 brothers. Each sister has 2 sisters and 4 brothers.


How can 1,000,000,000 be written as a product of two factors neither of them containing zeros?

Answer

29 * 59 or

(2 * 5)9

Solution submitted by Craig : 512 * 1,953,125 = 1,000,000,000


How do you cut a square cake into 8 equal piece only using 3 straight cuts?

Note that you can not move the pieces as you cut the cake.

Answer

Cut cake into 4 quarters by 2 vertical cuts which are perpendicular to each other. Now slice the cake horizontially which makes total 8 pieces.


There are 4 army men. They have been captured by a rebel group and have been held at ransom. An army intelligent officer orders them to be burried deep in dirt up to their

necks. The format of their burrial are as shown in the figure.

Conditions

They each have hats on their heads. either black(b) or white (w) look at diagram above. There are total 2 white hats and 2 black hats.

They only look in front of them not behind. They are not allowed to communicate by talking.

Between army man 1 and 2, there is a wall. Captive man 4 can see the colour of hats on 2 and 3 3 can only see 2's hat 2 can only see a wall and 1 can see a wall too, but is on the other side

The officer speaks up, "If one of you can correctly tell me the colour of your hat, you will all go scott free back to your contries. If you are wrong, you will all be killed.

How can one of them be certain about the hat they are wearing and not risk the lives of their fellow souldiers by taking a 50/50 guess!

Answer

Either soldier 3 or soldier 4 can save the life as soldier 1 and soldier 2 can not see colour of any hat, even not their own.. In our case soldier 3 will tell the colour of his hat.

Soldier 4 can see the hat on soldier 2 and soldier 3. If both are white, then he can be sure about colour of his hat which will be black and vice-versa. But if one of them is white and one is black, then soldier 4 can not say anything as he can have either of them. So he will keep mum.

If soldier 4 won't say anyhing for a while, then soldier 3 will know that soldier 4 is not in position to tell the colour of hat on his hat. It means that colour of soldier 3's hat is opposite of colour of soldier 2's hat. So soldier 3 can tell correctly the colour of hat on his head which is Black.

Here, we are assuming that all the soldiers are intelligent enough. Also, this solution will work for any combination of 2 Black hats and 2 White hats.


Given a rectangular cake with a rectangular piece removed (any Size or Orientation).

How would you cut the remainder of the cake into two equal halves with one straight cut of a knife?

Answer

If you cut a rectangular thing along the center (horizontally, vertically or at any angle), you will get two halves.

Find the centers of both - the original cake and the removed piece. Now, cut the reminder along the line connecting these two centers.

This is true because this line cut both - the original cake and the removed piece - in half, thus the remainder into two halves.


When Socrates was imprisoned for being a disturbing influence, he was held in high esteem by his guards. All four of them hoped that something would occur that would facilitate his escape. One evening, the guard who was on duty intentionally left the cell door open so that Socrates could leave for distant parts.

Socrates did not attempt to escape, as it was his philosophy that if you accept society's rules, you must also accept it's punishments. However, the open door was considered by the authorities to be a serious matter. It was not clear which guard was on that evening. The four guards make the following statements in their defense:

Aaron: A) I did not leave the door open. B) Clement was the one who did it.

Bob: A) I was not the one who was on duty that evening. B) Aaron was on duty.

Clement: A) Bob was the one on duty that evening. B) I hoped Socrates would escape.

David: A) I did not leave the door open. B) I was not surprised that Socrates did not attempt to escape.

Considering that, in total, three statements are true, and five statements are false, which guard is guilty?

Answer

David is the guilty.

Note that "All four of them hoped that something would occur that would facilitate his escape". It makes Clement's statement B True and David's statement B False.

Now consider each of them as a guilty, one at a time.

Aaron Bob Clement David TrueStmts A B A B A B A B

If Aaron is guilty False False True True False True True False 4

If Bob is guilty True False False False True True True False 4

If Clement is guilty True True True False False True True False 5

If David is guilty True False True False False True False False 3

Since in total, three statements are true and five statements are false. It is clear from the above table that David is the guity.


A cube is made of a white material, but the exterior is painted black.

If the cube is cut into 125 smaller cubes of exactly the same size, how many of the cubes will have atleast 2 of their sides painted black?

Answer

44

36 of the cubes have EXACTLY 2 of their sides painted black, but because a cube with 3 of its sides painted black has 2 of its sides painted black, you must also include the corner cubes.

This was a trick question, but hopefully the title of the puzzle tipped you off to this.


There are N secret agents each know a different piece of secret information. They can telephone each other and exchange all the information they know. After the telephone call, they both know anything that either of them knew before the call.

What are the minimum number of telephone calls needed so that all of the them know everything?

Answer

(2N - 3) telephone calls, for N = 2,3 (2N - 4) telephone calls, for N > 3

Divide the N secret agents into two groups. If N is odd, one group will contain one extra agent.

Consider first group: agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. Similarly in second group, agent 1 will call up agent 2, agent 2 will call up agent 3 and so on. After (N - 2) calls, two agents in each the group will know anything that anyone knew in his group, say they are Y1 & Y2 from group 1 and Z1 & Z2 from group 2.

Now, Y1 will call up Z1 and Y2 will call up Z2. Hence, in next two calls total of 4 agents will know everything.

Now (N - 4) telephone calls are reqiured for remaining (N - 4) secret agents.

Total telephone calls require are = (N - 2) + 2 + (N - 4) = 2N - 4

Let\'s take an example. Say there are 4 secret agents W, X, Y & Z. Divide them into two groups of 2 each i.e. (W, X) and (Y, Z). Here, 4 telephone calls are required.

1. W will call up X. 2. Y will call up Z. 3. W, who knows WX will call up Y, who knows YZ. 4. X, who knows WX will call up Z, who knows YZ.

Take an another example. Say there are 5 secret agents J, K, L, M & N. Divide them into two groups i.e. (J, K) and (L, M, N). Here, 6 telephone calls are required.

1. J will call up K. 2. L will call up M. 3. M will call up N. Now M and N know LMN. 4. J, who knows JK will call up M, who knows LMN. 5. K, who knows JK will call up N, who knows LMN.

6. L will call up to anyone of four.


One side of the bottom layer of a triangular pyramid has 12 balls. How many are there in the whole pyramid?

Note that the pyramid is equilateral and solid.

Answer

There are total 364 balls.

As there are 12 balls along one side, it means that there are 12 layers of balls. The top most layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55, 66 and 78 balls in the remaining layers.

Hence, the total number of balls are = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 = 364 balls


A woman took a certain number of eggs to the market and sold some of them.

The next day, through the industry of her hens, the number left over had been doubled, and she sold the same number as the previous day.

On the third day the new remainder was tripled, and she sold the same number as before.

On the fourth day the remainder was quadrupled, and her sales the same as before.

On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stock.

What is the smallest number of eggs she could have taken to market the first day, and how many did she sell daily? Note that the answer is not zero.

Answer

She took 103 eggs to market on the first day and sold 60 eggs everyday.

Let's assume that she had N eggs on the first day and she sold X eggs everyday. Putting down the given information in the table as follow.

Days Eggs at the start of the day Eggs Sold Eggs Remaining

Day 1 N X N-X

Day 2 2N-2X X 2N-3X

Day 3 6N-9X X 6N-10X

Day 4 24N-40X X 24N-41X

Day 5 120N-205X X 120N-206X

It is given that she disposed of her entire stock on the fifth day. But from the table above, the number of eggs remaining are (120N-206X). Hence, 120N - 206X = 0 120N = 206X 60N = 103X

The smallest value of N and X must be 103 and 60 respectively. Hence, she took 103 eggs to market on the first day and sold 60 eggs everyday.


Mr. Khemani have a triangular, Royal Blue colored dining table in his house. The three sides of the table are 13, 17 and 43 inches long respectively. He wants to change the color of the table to Olive Green. According to the painter, 50ml of paint per square inch of the table is required for that.

How much paint is required to change the color of the table to Olive Green?

Answer

A trick question.

A triangular table with sides 13, 17 and 43 inches does not exist. It is impossible to have such a triangle.

Summation of any two sides of the triangle should be greater than the third side. It is true for any type of triangle. The dimension of the table is 13, 17 and 43, where 13 + 17 < 43. Hence such table does not exist and no need to paint it.


If Sean is Louis' brother, If Marie is Bruno's sister, And if Paul is Christopher's brother,

Then who is Hamilton's sister - Mona, Sue or Linda? And Why so?

Answer

Sue is Hamilton's sister.

Each pair of names contains all the five vowels (a, e, i, o, u) once.

1. Sean, Louis 2. Marie, Bruno 3. Paul, Christopher

Now Hamilton has 3 vowels - a, i and o. Hence, his sister's name should have remaining 2 vowels i.e. e and u. Thus Sue is his sister.


Can you decode the following Cryptogram?

R K A P B R G S G X R R K G J N X

R K G X G X R K A Z A X R P T

U I I Y P F I E X. R K A B A X X G S G X R

T A U F X R K G X G X R F D A.

Answer

There are two statements in the given cryptogram, both are starting with RKA. It must be "THE". Also, there are words like RKGX and GX, which must be "THIS" and "IS" respectively. The rest is simple substitution and application of language-knowledge to form the possible words. R K A P B R G S G X R R K G J N X

T H E O P T I M I S T T H I N K S

R K G X G X R K A Z A X R P T

T H I S I S T H E B E S T O F

U I I Y P F I E X. R K A B A X X G S G X R

A L L W O R L D S. T H E P E S S I M I S T

T A U F X R K G X G X R F D A.

F E A R S T H I S I S T R U E.


Mark ate half of a pizza on Monday. He ate half of what was left on Tuesday and so on. He followed this pattern for one week.

How much of the pizza would he have eaten during the week?

Answer

Mark would have ate 127/128 (99.22%) of the pizza during the week.

Mark ate half the pizza on Monday. On Tuesday, he would have ate half of the remaining pizza i.e. 1/4 of the original pizza. Similarly, he would have ate 1/8 of the original pizza on Wednesday and so on for the seven days.

Total pizza Mark ate during the week is = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = 127/128 = 99.22% of the original pizza


To move a Safe, two cylindrical steel bars 7 inches in diameter are used as rollers.

How far will the safe have moved forward when the rollers have made one revolution?

Answer

The safe must have moved 22 inches forward.

If the rollers make one revolution, the safe will move the distance equal to the circumference of the roller. Hence, the distance covered by the safe is = PI * Diameter (or 2 * PI * Radius) = PI * 7 = 3.14159265 * 7 = 21.99115 = 22 inches approx.


The average scores of the class for exam are as follow: Average score of the boys = 90 Average score of the girls = 81 Average score of the class = 84

Find whether Class contains more Boys or Girls?

Answer

Assume that there are B boys and G girls in the Class.

Hence from the given data :

90*B + 81*G = 84*(B+G)

90B + 81G = 84B + 84G

6B = 3G

2B = G

Hence number of Girls in the Class are twice the number of Boys.


Find the values of each of the alphabets.

N O O N

S O O N

+ M O O N

----------

J U N E

Answer

Using trial and error. There are 2 solutions to it and may be more.

2 4 4 2

1 4 4 2

+ 5 4 4 2

----------

9 3 2 6

4 1 1 4

5 1 1 4

+ 0 1 1 4

----------

9 3 4 2


What are the next two numbers in this series?

1, 4, 1, 5, 9, 2, _, _

Answer

Next two numbers are 6 and 5.

The patten is the each digit in the value of PI after the decimal point. The value of mathematical constant PI is 3.1415926536.


A positive integer that, when added to 1000 gives a sum which is greater than when multiplied by 1000.

Find the positive integer.

Answer

The positive integer is 1.

Sum of 1 and 1000 = 1 + 1000 = 1001 Multiplication of 1 and 1000 = 1 * 1000 = 1000

Thus, sum of 1 and 1000 is greater than the multiplication of 1 and 1000.


Which of the following numbers is the odd one out, and why? 1, 2, 3, 5, 9, 13, 21

Note that 2 is not the odd one.

Answer

The odd number is 9.

It is a Fibonacci Series - a series in which next number is summation of previous 2 numbers.

The first two numbers are 1 and 2. Third number is = 1 + 2 = 3 Fourth number is 3 + 5 = 8 Fifth number is 5 + 8 = 13 Sixth number is 8 + 13 = 21

Hence 9 is the odd number It should 8.


Replace the letters with the correct numbers.

T W O

X T W O

---------

T H R E E

Answer

T=1, W=3, O=8, H=9, R=2, E=4 1 3 8

x 1 3 8

------------

1 9 0 4 4You can reduce the number of trials. T must be 1 as there is multiplication of T with T in hundred's position. Also, O can not be 0 or 1. Now, you have to find three digit number whose square satisfies above conditions and square of that has same last two digits. Hence, it must be between 102 and 139.


Mrs. Watsherface had a garage sale. A custmer named Gina bought an old lamp and a rug. She paid a total of $5.25 for everything. The rug cost 25 cents more than the lamp.

How much did each cost?

Answer

The lamp cost $ 2.50 and the rug cost $ 2.75

A simple one.

Assume that the lamp cost $ L. Hence the rug must have cost $ (L + 0.25) Also, total cost is $ 5.25, Hence the equation : L + L + 0.25 = 5.25 2 * L = 5 L = 2.50

Hence, the lamp cost $ 2.50 and the rug cost $ 2.75


A group of friends went on a holiday to a hill station. It rained for 13 days. But when it rained in the morning, the afternoon was lovely. And when it rained in the afternoon, the day was preceded by clear morning.

Altogether there were 11 very nice mornings and 12 very nice afternoons. How many days did their holiday last?

Answer

The holiday last for 18 days.

Let's assume the number of days as follows: Rain in the morning and lovely afternoon = X days Clear morning and rain in the afternoon = Y days No rain in the morning and in the afternoon = Z days

Number of days with rain = X + Y = 13 days Number of days with clear mornings = Y + Z = 11 days Number of days with clear afternoons = X + Z = 12 days

Solving above 3 equations, we get X = 7, Y = 6 and Z = 5

Hence, total number of days on holiday = 18 days


If you added together the number of 2's in each of the following sets of numbers, which set would contain the most 2's: 1-333, 334-666, or 667-999?

Answer

1-333

The reason why is because 200-299 each begins with a 2!


Five students - Akash, Chintan, Jignesh, Mukund and Venky - appeared for an exam. There were total five questions - two multiple choice (a, b or c) and three true/false questions. They answered five questions each and answered as follow.

I II III IV V

--------------------------------------------------

Chintan c b True True False

Akash c c True True True

Jignesh a c False True True

Mukund b a True True False

Venky b b True False True

--------------------------------------------------Also, no two students got the same number of correct answers.

Can you tell which are the correct answers? What are their individual score?

Answer

The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).

As no two students got the same number of correct answers, the total number of correct answers must be either 15 (1+2+3+4+5) or 10 (0+1+2+3+4).

Let's find out the maximum number of correct answers possible from the answers given by them. For Question I = 2 (b or c) For Question II = 2 (b or c) For Question III = 4 (True) For Question IV = 4 (True) For Question V = 3 (True)

Thus, the maximum number of correct answers possible are 15 (2+2+4+4+3) which means that Akash would have given all correct answers as only he answered True for questions III, IV and V. But then Chintan and Jignesh would have exactly 3 correct answers. And also, Mukund and Venky would have 2 correct answers. So no one got all five correct. One can also arrive at this conclusion by trial-and-error, but that would be bit lengthy.

Now, it is clear that total number of correct answers are 10 (0+1+2+3+4). Questions III and IV both can not be False. If so, total number of correct answers would not be 10. So the student who got all wrong can not be Chintan, Akash and Mukund.

If Venky got all wrong, then Chintan, Jignesh and Mukund each would have atleast 2 correct answers. It means that Akash would have to be the student with only one correct answer and the correct answers for questions I and II would be a and a respectively. But then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus, Jignesh is the student with all wrong answers. The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).


Two people enter a race in whick you run to a point and back. Person A runs 20 mph to and from the point. Person B runs to the point going 10 mph and 30 mph going back.

Who came in first?

Answer

Person A came in first.

Let's assume that the distance between start and the point is D miles.

Total time taken by Person A to finish = (D/20) + (D/20) = D/10 = 0.1D

Total time taken by Person B to finish = (D/10) + (D/30) = 2D/15 = 0.1333D

Thus, Person A is the Winner.

Alternatively (if you don't like mathematics ;)), analyse the situation as follow:

Note that initially speed of Person A (20 mph) was twice the speed of Person B (10 mph). Hence, when Person A (20 mph forward) reached the point, Person B (10 mph forward) was halfway. When Person A (20 mph back) finished, Person B (still 10 mph forward) reached the point.

Thus, Person A wins the race and by that time Person B covers only half the distance, no matter how far the point is!!!


Consider a game of Tower of Hanoi (like the one that you can play on BrainVista).

If the tower has 2 discs, the least possible moves with which you can move the entire tower to another peg is 3.

If the tower has 3 discs, the least possible moves with which you can move the entire tower to another peg is 7.

What is the least possible moves with which you can move the entire tower to another peg if the tower has N discs?

Answer

There are number of ways to find the answer.

To move the largest disc (at level N) from one tower to the other, it requires 2(N-1) moves. Thus, to move N discs from one tower to the other, the number of moves required is = 2(N-1) + 2(N-2) + 2(N-3) + ..... + 22 + 21 + 20 = 2N - 1

For N discs, the number of moves is one more than two times the number of moves for N-1 discs. Thus, the recursive function is F(1) = 1 F(N) = 2*[F(N-1)] + 1 where N is the total number of discs

Also, one can arrive at the answer by finding the number of moves for smaller number of discs and then derive the pattern. For 1 disc, number of moves = 1 For 2 discs, number of moves = 3 For 3 discs, number of moves = 7 For 4 discs, number of moves = 15 For 5 discs, number of moves = 31

Thus, the pattern is 2N - 1


On evey Sunday Amar, Akbar and Anthony lunch together at Preetam-Da-Dhaba where they order lassi based on following facts.

1. Unless neither Amar nor Akbar have lassi, Anthony must have it. 2. If Amar does not have lassi, either Akbar or Anthony or both have it. 3. Anthony has lassi only if either Amar or Akbar or both have it. 4. Akbar and Anthony never have lassi together.

Who order(s) lassi?

Answer

Amar and Anthony both have lassi whereas Akbar never does.

Fact (2) can be alternatively stated that "either Amar or Akbar or Anthony must have lassi".

From Fact (3), it can be infered that either Amar or Akbar must have lassi.

Now, from Fact (1), it is apparent that Anthony too must have lassi. But according to Fact (4), Akbar cannot have lassi when Anthony does.

Thus Amar and Anthony both have lassi whereas Akbar never does.


A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?

Answer

Let the base be X.

Therefore (X*X + X*0 + 0) = (2*X +4) + (3*X + 2) X*X = 5*X + 6 X*X - 5*X -6 = 0 (X-6)(X+1) = 0

Therefore base is 6


What two numbers comes next in this series?: 1, 2, 3, 6, 7, 14, 15, 30, 31, ? ?

Answer

Next two numbers are 62 and 63.

The pattern is double the number, add one, double the number, add one...

Starting with 1, for next number double the number = 2 For next number add one = 3 Again, double the number = 6 Next add one = 7 and so on ...

Now at 31, for next number double the number = 62 Add one for next number = 63


What are the next two number in the series? 1, 2, 4, 7, 28, 33, 198, ?, ?

Answer

The next two numbers are 205 and 1640.

The pattern is : 1 + 1 * 2 + 3 * 4 + 5 * 6 + 7 * 8 + 9 * 10 or to the current number add and multiply alternatively by the number's position to get the next number.

First number = 1 Second number = 1 + 1 = 2 Third number = 2 * 2 = 4 Fourth number = 4 + 3 = 7 Fifth number = 7 * 4 = 28 Sixth number = 28 + 5 = 33 Seventh number = 33 * 6 = 198 Eight number = 198 + 7 = 205 Ninth number = 205 * 8 = 1640 Tenth number = 1640 + 9 = 1649


There is a two digit number, the second digit of which is 4 less than its first digit. Also, the number is divisible by the sum of its digits and if you do so, the quotient would be 7.

Find the number.

Answer

The number is 84.

Let's assume that the first digit is N. Hence, the second digit is (N-4) and the number is = 10N + (N-4) = 11N - 4

Now, it is given that the number is divisible by the sum of its digit and the quotient would be 7. (11N - 4) / (N + N - 4) = 7 (11N - 4) / (2N - 4) = 7 (11N - 4) = 7 * (2N - 4) 11N - 4 = 14N - 28 3N = 24

N = 8

Thus, the first digit is 8, the second digit is 4 and the required number is 84.


A farmer built a fence around his 25 cows in a square region. He built it in such a way that one can see 5 poles from either of the four sides.

What are the minimum number of poles the farmer must have used?

Answer

16 poles X X X X X

X X

X X

X X

X X X X X

One pole at each corner and three poles along each side so that one can always see 5 poles from either of the four sides. The corner pole is shared by two sides and hence reducing the number of poles to 16.


A man is at a river with a 9 gallon bucket and a 4 gallon bucket. He needs exactly 6 gallons of water.

How can he use both buckets to get exactly 6 gallons of water?

Note that he cannot estimate by dumping some of the water out of the 9 gallon bucket or the 4 gallon bucket.

Answer

For the sack of explanation, let's identify 4 gallon bucket as Bucket P and 9 gallon bucket as Bucket Q.

Operation4 gallon bucket

(Bucket P)9 gallon bucket

(Bucket Q)

Initially 0 0

Fill the bucket Q with 9 gallon water 0 9

Pour 4 gallon water from bucket Q to bucket P 4 5

Empty bucket P 0 5


Empty bucket P 0 1


Fill the bucket Q with 9 gallon water 1 9


9 gallon bucket contains 6 gallon of water, as required.


Using the numbers 0 to 9 once each, what combination will add up to exactly 100?

Note that you can use ONLY ADDITION.

Answer

It is impossible to get 100 using digits 0-9 once each and only addition.

All digits from 0 to 9 add up to 45. Now, take any two digits (and make a two-digit number) and add it to the remaining digits. The result will always increase by a multiple of 9. e.g Let's take 1 and 2 i.e. 12 and add all remaining digits to it, the total is 54 i.e. 45 + 9.

Thus, whatever you do the answer will always be 45 + 9X, where X is an integer. In other words, we are trying to solve 45 + 9X =100 for integer value of X.

Thus, it is impossible to get 100 using digits 0-9 once each and only addition


A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives.

He paid five rupees per day to each man, four ruppes to each woman and one rupee to each child. He gave out 200 rupees every evening.

How many men, women and children were working with the constructor?

Answer

16 men, 12 women and 72 children were working with the constructor.

Let's assume that there were X men, Y women and Z children working with the constructor. Hence,

X + Y + Z = 100 5X + 4Y + Z = 200

Eliminating X and Y in turn from these equations, we get X = 3Z - 200 Y = 300 - 4Z

As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get

if Y = X, 300 - 4Z = 3Z - 200 7Z = 500 Z = 500/7 i.e. 71.428

if Y = X/2, 300 - 4Z = (3Z - 200)/2 600 - 8Z = 3Z - 200 11Z = 800 Z = 800/11 i.e. 72.727

But Z must be an integer, hence Z=72. Also, X=16 and Y=12

There were 16 men, 12 women and 72 children working with the constructor.


If a man holds an Apple in his left hand, and an Orange in his right, What time does the 10:00 train come at?

Answer

10:00 O'Clock

The train is scheduled for 10:00 and it will come at 10:00 clock unless it is late - no matter what the guy holds in his hands.


There were an electrician and a plumber waiting in line for admission to the Home Show. One of them was the father of the other's son.

How could this be possible?

Answer

The electrician and the plumber are Husband-Wife.

Hence, one is the father and the other is the mother


Put the appropriate plus or minus signs between the numbers below, in the correct places, so that the value of the expression on the left will equal the value on the right:

0 1 2 3 4 5 6 7 8 9 = 1

Answer

A simple one. There are many solutions to it.

0 + 1 - 2 - 3 - 4 + 5 - 6 - 7 + 8 + 9 = 1

0 + 1 + 2 + 3 + 4 + 5 - 6 - 7 + 8 - 9 = 1

0 - 1 - 2 + 3 - 4 + 5 - 6 + 7 + 8 - 9 = 1

0 + 1 - 2 - 3 - 4 - 5 + 6 + 7 - 8 + 9 = 1

0 + 1 + 2 + 3 + 4 – 5 + 6 + 7 - 8 - 9 = 1


I bought a car with a peculiar 5 digit numbered licence plate which on reversing could still be read. On reversing value is increased by 78633.

Whats the original number if all digits are different?

Answer

Only 0 1 6 8 and 9 can be read upside down. So on rearranging these digits we get the answer as 10968.


Shivangi is the 11th girl from either end of the raw of girls.

How many girls are there in a row?

Answer

There are total 21 girls in a row.

Shivangi is the 11th girl from either end of the raw of girls. It means that there are 10 girls each, on both sides of her. Hence, there are total 21 girls.


Three friends check into a hotel for the night and the clerk tells them the bill is Rs. 30, payable in advance. So, they each pay the clerk RS. 10 and go to their room.

A few minutes later, the clerk realizes he has made an error and overcharged the trio by Rs 5. He asks the hotel-boy to return Rs. 5 to the 3 friends who had just checked in. The hotel-boy sees this as an opportunity to make Rs. 2 as he reasons that the three friends would have a tough time dividing Rs. 5 evenly among them; so he decides to tell them that the clerk made a mistake of only Rs. 3, giving a rupee back to each of the friends. He pockets the leftover Rs. 2 and goes home for the day!

Now, each of the three friends gets a rupee back, thus they each paid Rs. 9 for the room which is a total of Rs. 27 for the night. We know the hotel-boy pocketed Rs. 2 and adding that to the Rs. 27, you get Rs. 29, not Rs. 30 which was originally spent.

Where did the other rupee go???

Answer

The facts in this riddle are clear: There is an initial Rs. 30 charge. It should have been Rs. 25, so Rs.5 must be returned and accounted for. Rs. 3 is given to the 3 friends, Rs. 2 is kept by the hotel-boy - there you have the Rs. 5.

The trick to this riddle is that the addition and subtraction are done at the wrong times to misdirect your thinking - and quite successfully for most. Each of the 3 friends did indeed pay Rs. 9, not Rs. 10, and as far as the friends are concerned, they paid Rs. 27 for the night. But we know that the clerk will tell us that they were charged only Rs. 25 and when you add the Rs. 3 returned with the Rs. 2 kept by the hotel-boy, you come up with Rs. 30


A man is on a search for Atlantis and comes upon an island where all the inhabitants know whether Atlantis is still around or not.

However, all of the inhabitants are either Fairies or Trolls and they all use a spell to appear humanoid so you cannot tell which is which. And the Faries always tell the truth and the Trolls always lie, but there is a slight complication, some of the Fairies have gone insane and always lie and some of the Trolls have also gone insane and always tell the truth.

So here is your task: you must ask the first inhabitant that you come to ONE question and from that ONE question you must determine wether Atlantis is still around or not.

What is the question that you must ask?

Answer

There are 2 answers to it:

Answer I"Is the statement that you are reliable equivalent to the statement that Atlantis is still around?"

Answer II"Do you believe that the Statement that you are a Fairy is equivalent to the statement that Atlantis is still around?"


If a butcher's clerk is 5 ft 10 inches. What does he weigh?

Answer

The butcher's clerk weighs meat.

The height is given only to misguide you so that you think of clerk's weight


There is a river and 3 people. Two people weigh 50 lbs and one of them weighs 100 lbs. The only way you can cross the river is by using a boat.

How can they all cross the river if the boat only can hold 100 lbs?

Answer

1. Both 50 pound persons will cross the river. 2. One of the 50 pound person will come back. 3. Now, 100 pound person will cross the river alone. 4. 50 pound person will come back who is at the other end of the river.

5. Again both 50 pound persons will cross the river together.


A hole that is 3 feet deep, and 6 inches in diameter, has how much dirt in it..?

Answer

A tricky one.

It's a hole. So there is no dirt in it.


Shahrukh speaks truth only in the morning and lies in the afternoon, whereas Salman speaks truth only in the afternoon and lies in the morning.

A says that B is Shahrukh.

Is it morning or afternoon and who is A - Shahrukh or Salman?

Answer

It is Afternoon and A can be Salman or Shahrukh. If A is Salman, he is speaking truth. If A is Shahrukh, he is lying.

Want to confirm it? Consider following 4 possible answers and check for its truthness individually.

1. It is Morning and A is Shahrukh 2. It is Morning and A is Salman 3. It is Afternoon and A is Shahrukh

4. It is Afternoon and A is Salman


A bus driver was going down a street. He was on the wrong side. He went at a stop sign and didn't signal at a turn. He went under the minimum speed and yet he didn't get in trouble for his actions.

WHY not?

Answer

He was Walking. He is a Bus driver. But it does not say that he was driving.


Tom and Jerry and the boys in the bar were exchanging old war stories.

Tom offered one about how his grandfather led a battalion against a German division during World War I. Through brilliant maneuvers, he defeated them and captured valuable territory. After the battle he was awarded a medal that was inscribed with:

"For Bravery, Daring and Leadership - World War I.

From the Men of Battalion 8."Jerry looked at Tom and said, "You really don't expect anyone to believe that yarn, do you?"

What's wrong with the story?

Answer

World War I wasn't called "World War I" until World War II.


A murderer is condemned to death. He has to choose between three rooms. The first is full of raging fires, the second is full of assassins with loaded guns, and the third is full of lions that haven't eaten in 3 years.

Which room is the safest for him?

Answer

The third room - full of lions - is the safest.

The third room is full of lions that HAVEN'T EATEN IN 3 YEARS. It is obvious that they won't survive without eating for 3 years, they must be dead.


Amit, Bhavin, Himanshu and Rakesh are sitting around a table.

The Electonics Engineer is sitting to the left of the Mechanical Engineer. Amit is sitting opposite to Computer Engineer. Himanshu likes to play Computer Games. Bhavin is sitting to the right of the Chemical Engineer.

Can you figure out everyone's profession?

Answer

Amit is the Mechanical Engineer. Bhavin is the Computer Engineer. Himanshu and Rakesh are either Chemical Engineer or Elecronics Engineer.

Amit and Bhavin are sitting opposite to each other. Whereas Chemical Engineer and Elecronics Engineer are sitting opposite to each other.

We cannot find out who is Chemical Engineer and Elecronics Engineer as data provided is not sufficient.


Sita has six pairs of black gloves and six pairs of brown gloves in her drawer.

In complete darkness, how many gloves must she take from the drawer in order to be sure to get a matching pair (i.e. left hand and right hand golves of the same color)? Think carefully!!

Answer

13

She could possibly take out 6 black left hand gloves and then 6 brown left hand gloves, the next one would have to be either the right hand or left hand match


In your sock drawer, you have a ratio of 5 pairs of blue socks, 4 pairs of brown socks, and 6 pairs of black socks.

In complete darkness, how many socks would you need to pull out to get a matching pair of the same color?

Answer

4 If you don't agree, try it yourself!


You have a bucket of jelly beans. Some are red, some are blue, and some green. With your eyes closed, pick out 2 of a like color.

How many do you have to grab to be sure you have 2 of the same?

Answer

If you select 4 Jelly beans you are guarenteed that you will have 2 that are the same color.


There are 70 employees working with BrainVista of which 30 are females. Also,

30 employees are married 24 employees are above 25 years of age 19 married employees are above 25 years, of which 7 are males 12 males are above 25 years of age 15 males are married.

How many unmarried females are there and how many of them are above 25?

Answer

15 unmarried females & none are above 25 years of age.

Simply put all given information into the table structure and you will get the answer.

Married Unmarried

Below 25 Above 25 Below 25 Above 25

Female 3 12 15 0

Male 8 7 20 5


There is a one floor house, and everything in it is blue, green or purple. The people are purple, the carpeting is green and the walls are blue.

What colour do you suppose the stairs are?

Answer

It is a one floor house and one floor house do not have stairs. Hence, there is no question of colour of stairs as there are no stairs. 0;-)


Mr. Black, Mr. White and Mr. Grey were chatting in the Yahoo conference. They were

wearing a black suit, a white suit and a grey suit, not necessarily in the same order.

Mr. Grey sent message, "We all are wearing suit that are of the same color as our names but none of us is wearing a suit that is the same color as his name."

On that a person wearing the white suit replied, "What difference does that make?"

Can you tell what color suit each of the three persons had on?

Answer

Mr. Grey is wearing Black suit. Mr. White is wearing Grey suit. Mr. Black is wearing White suit.

Mr. Grey must not be wearing grey suit as that is the same colour as his name. Also, he was not wearing white suit as the person wearing white suit responded to his comment. So Mr Grey must be wearing a black suit.

Similarly, Mr. White must be wearing either black suit or grey suit. But Mr. Grey is wearing a black suit. Hence, Mr. White must be wearing a grey suit.

And, Mr. Black must be wearing white suit.


On one side of the river there is a man, a wolf, a goat, and some cabbage. The man needs to take each of them across to the other side of the river. He has a boat, but in it he can only fit one of them as well as himself.

If he takes the wolf across first, the goat will eat the cabbage. If he takes the cabbage across, the wolf will eat the goat.

Write the correct order in which he can take them across.

Answer

There are two answers to the teaser, similar but slightly different. The key is never keep Wolf-Goat and Goat-Cabbage alone.

Initially, the man, the wolf, the goat and the cabbage is on one side of the river i.e. (MWGC, 0)

Answer : 1 1. Take the goat across leaving the wolf and the cabbage. (WC, MG) 2. Come back. (WCM, G) 3. Take the cabbage across leaving the wolf alone. (W, MGC) 4. Take the goat back. (MWG, C)

5. Take the wolf across. (G, MWC) 6. Come back. (MG, WC) 7. Take the goat across. (0, MWGC)

Answer : 2 1. Take the goat across leaving the wolf and the cabbage. (WC, MG) 2. Come back. (WCM, G) 3. Take the wolf across leaving the wolf alone. (C, MWG) 4. Take the goat back. (MGC, W) 5. Take the cabbage across. (G, MWC) 6. Come back. (MG, WC)

7. Take the goat across. (0, MWGC)


What is the next number in the given series?

10^3, 10^9, 10^27, 10^2, 0, 4, 8, 3, ?

Note that 10^3 means 10 raised to the power of 3 i.e. 1000.

Answer

A tought one !!!

The next number in the series is 5.

The pattern is - the series of numbers containing first letter of the English alphabet starting from "A".

10^3 = one thousAnd

10^9 = one Billion

10^27 = one oCtillion

10^2 = one hunDreD

0 = zEro

4 = Four

8 = eiGht

3 = tHree

5 = fIve


In another galaxy, a certain nation of creatures speak a language known as Algramara. In this language, "harvec tood zevac" means "my tooth hurts", "lavec lina zevac" means "my delicious food", & "zevac corma lina" means "eating my food".

What does "corma harvec zevac tood" mean?

Answer

There are 3 sentances: 1. "harvec tood zevac" means "my tooth hurts," 2. "lavec lina zevac" means "my delicious food," 3. "zevac corma lina" means "eating my food."

From (1), (2) & (3) "zevac" means my From (2) & (3) "lina" means food So From (3) "corma" means eating and From (2) "lavec" means delicious

Now, two unknown words are there: "harvec" and "tood" which means tooth or hurts. So there are two options: 1. Eating hurts my tooth 2. Eating tooth my hurts

Only first option makes sence. So "corma harvec zevac tood" means "Eating hurts my tooth"


There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in colour. Some boxes contain only red balls and others contain only blue.

One sales man sold one box out of them and then he says, "I have the same number of red balls left out as that of blue."

Which box is the one he solds out?

Answer

Total no of balls = 5 + 7 + 14 + 16 + 18 + 29 = 89

Total number of balls are odd. Also, same number of red balls and blue balls are left out after selling one box. So it is obvious that the box with odd number of balls in it is sold out i.e. 5, 7 or 29.

Now using trial and error method, (89-29) /2 = 60/2 = 30 and

14 + 16 = 5 + 7 + 18 = 30

So box with 29 balls is sold out


Sam and Mala have a conversation.

Sam says I am certainly not over 40 Mala says I am 38 and you are atleast 5 years older than me Now Sam says you are atleast 39

All the statements by the two are false. How old are they really?

Answer

Sam is 41 and Mala is 37.

Let's invert the teaser and read it like this : Sam says I am certainly over 40 Mala says I am not 38 and you are atmost 4 years older than me Now Sam says you are atmost 38

From first statement it is clear that Sam is over 40. Also, from next 2 statements it is clear that Mala is less then 38. Hence the possibilities are : Sam = 41, 42, 43, 44, 45, ...... Mala = 37, 36, 35, 34, 33, ......

It also says that the difference between their age is maximum 4 years. Hence, there is only one possible pair i.e. 41 and 37, all other combination have differences more then 4.

Hence the answer - Sam is 41 and Mala is 37.


How many squares are there in a 5 inch by 5 inch square grid? Note that the grid is made up of one inch by one inch squares.

Answer

There are 55 squares in a 5 by 5 grid.

There are 25 squares of one grid. There are 16 squares of four grids i.e. 2 by 2 There are 9 squares of nine grids i.e. 3 by 3 There are 4 squares of sixteen grids i.e. 4 by 4 There is 1 square of twenty-five girds i.e. 5 by 5

Hence, there are total 25 + 16 + 9 + 4 + 1 = 55 squares.

You must have noticed one thing that total number squares possible of each size is always a perfact square i.e. 25, 16, 9, 4, 1

For a grid of N by N, the possible number of squares are = N2 + (N - 1)2 + (N - 2)2 + (N - 3)2 + ......... + 32 + 22 + 12

For 1 by 1 grid, total squares = 12 = 1 For 2 by 2 grid, total squares = 22 + 12 = 5 For 3 by 3 grid, total squares = 32 + 22 + 12 = 14 For 4 by 4 grid, total squares = 42 + 32 + 22 + 12 = 30 For 5 by 5 grid, total squares = 52 + 42 + 32 + 22 + 12 = 55


You are given an arbitrarily long string of parentheses i.e. a string comprising of '(' and ')'.

Give an efficient way of finding whether a given string contains a balanced parentheses or not. Note that the string can be of any length.

A string with balanced parentheses is a string containing same number of left parentheses '(' and right parentheses ')'. Also, they must be in correct order i.e. ( ( ) ( ) ) is balanced whereas ( ) ) ( ( ) is not.

Answer

Start a counter with 0 and read the given string from left to right. Add 1 for each left parentheses and subtract 1 for each right parentheses.

If count is negative at any time, the given string is not balanced.

If the final count is non-zero, the given string is not balanced.

If the final count is zero, the given string is balanced.


Four couples are going to the movie. Each row holds eight seats. Betty and Jim don't want to sit next to Alice and Tom. Alice and Tom don't want to sit next to Gertrude and Bill. On the otherhand, Sally and Bob don't want to sit next to Betty and Jim.

How can the couples arrange themselves in a row so that they all sit where they would like?

Answer

From the given data, it can be inferred that: (Sally & Bob) NOT (Betty & Jim) NOT (Alice & Tom) NOT (Gertrude & Bill)

(A) NOT (B) means A and B can not seat next to each other.

Now, it is obvious that (Betty & Jim) and (Alice & Tom) will occupy the corner seats as both of them can have only one neighbour. Therefore, (Gertrude & Bill) will seat next to (Betty & Jim) (Sally & Bob) will seat next to (Gertrude & Bill) (Alice & Tom) will seat next to (Sally & Bob)

Thus, there are two possible arrangements - a mirror images of each other.

1. (Betty & Jim) - (Gertrude & Bill) - (Sally & Bob) - (Alice & Tom) 2. (Alice & Tom) - (Sally & Bob) - (Gertrude & Bill) - (Betty & Jim)


There are 4 mugs placed upturned on the table. Each mug have the same number of marbles and a statement about the number of marbles in it. The statements are: Two or Three, One or Four, Three or One, One or Two.

Only one of the statement is correct. How many marbles are there under each mug?

Answer

A simple one.

As it is given that only one of the four statement is correct, the correct number can not appear in more than one statement. If it appears in more than one statement, then more than one statement will be correct.

Hence, there are 4 marbles under each mug.


Do you know the missing number in the given series?

10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000

Answer

The missing number is 31.

The pattern is - Sixteen in the base n for n = 16, 15, ..., 2.

16 in the base 16 = 10 16 in the base 15 = 11

16 in the base 14 = 12 16 in the base 13 = 13 16 in the base 12 = 14 16 in the base 11 = 15 16 in the base 10 = 16 16 in the base 9 = 17 16 in the base 8 = 20 16 in the base 7 = 22 16 in the base 6 = 24 16 in the base 5 = 31 16 in the base 4 = 100 16 in the base 3 = 121 16 in the base 2 = 10000

Thus, the complete series is - 10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31, 100, 121, 10000


There are 10 boxes containing 10 balls each. 9 boxes contain 10 balls of 10 kg each and one box contains 10 balls of 9 kg each. Tool is available for proper weighing. How can you find out the box containing 9 kg balls?

You are allowed to weigh only once. You can remove balls from the boxes. All balls are of same size and color.

Answer

1. Mark the boxes with numbers 1, 2, 3, 4, ... upto 10 2. Take 1 ball from box 1, take 2 balls from box 2, take 3 balls from box 3, take 4 balls

from box 4 and so on 3. Put all of them on the scale at once and take the measurement. 4. Now, subtract the measurement from 550 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10 + 6*10 +

7*10 + 8*10 + 9*10 + 10*10)

5. The result will give you the box number which has a ball of 9 Kg


There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd coin?

Answer

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

1. Take 8 coins and weigh 4 against 4. o If both are not equal, goto step 2

o If both are equal, goto step 3

2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.

o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3. If both are equal, L4 is the odd coin and is lighter. If L2 is light, L2 is the odd coin and is lighter. If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

If both are equal, there is some error. If H1 is heavy, H1 is the odd coin and is heavier. If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

If both are equal, L1 is the odd coin and is lighter. If H3 is heavy, H3 is the odd coin and is heavier. If H4 is heavy, H4 is the odd coin and is heavier.

3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.

o If both are equal, there is some error. o If X is heavy, X is the odd coin and is heavier.

o If X is light, X is the odd coin and is lighter.


Below is a Quiz written by Einstein in the lst century. He said 98% of the people in the world cannot solve the quiz. Are you among the other 2%? FACTS

1. There are 5 houses in 5 different colors. 2. In each house lives a person with a different nationality. 3. These 5 owners drink a certain beverage, smoke a certain brand of cigar and keep a

certain pet.

4. No owners have the same pet, smoke the same brand of cigar or drink the same drink.

HINTS 1. The Brit lives in a red house. 2. The Swede keeps dogs as pets. 3. The Dane drinks tea. 4. The green house is on the immediate left of the white house. 5. The green house owner drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the yellow house smokes Dunhill. 8. The man living in the house right in the center drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes blend lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The owner who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the Blue House. 15. The man who smokes blend has a neighbor who drinks water.

THE QUESTION IS....WHO KEEPS FISH?

There is no trick to this - it needs deductive reasoning and definitely a pen and paper.

Answer

Nationality Beverage Cigar Pet House Color

------------------------------------------------------

Norwegian Water Dunhill Cat Yellow

Dane Tea Blend Horses Blue

Brit Milk Pall Mall Bird Red

German Coffee Prince Fish Green

Swede Beer Blue Master Dog WhiteTherefore the answer is the German.

I wonder if U could work it out by realizing that Einstien was a german and kept fish but that would probably be cheating :)

There is one more possible answer, if we remove "immediate" from the Hint 4 i.e. read it as "The green house is on the left of the white house, not necessarily on the immediate left"Nationality Beverage Cigar Pet House Color

------------------------------------------------------

Norwegian Coffee Blend Fish Green

German Water Prince Cats Blue

Swede Milk Dunhill Dogs Yellow

Brit Beer Blue Master Horses Red

Dane Tea Pall Mall Birds WhiteThus, either German or Norwegian keeps the Fish, if the green house is not necessarily on the immediate left of the white house.

Thanks to Katie Crowe and friends for the second answer !!!


A professor went to a Museum of Natural History. She saw the following five exhibits :

1. A prehistoric arrowhead made of copper. 2. The fossil skeleton of a dinosaur no bigger than a chicken. 3. An ancient Roman coin marked 120 B.C. 4. A red diamond ring. 5. An ancient Egyptian cat mummy.

The professor knew one of these is fake. Do you?

Answer

An ancient Roman coin marked 120 B.C. is fake.

B.C. stands for "Before Christ". The ancient Roman people did not know that they were before Christ. or what B.C. stands for. Infact nobody would have knew that they were 120 years before Christ !!!


Two women, X and Y, were selling Oranges in the market. X were selling 3 oranges for a Rupee and Y were selling 2 oranges for a Rupee.

One day each of them had 30 oranges unsold. They put together the two lots of oranges and decided to sell 5 oranges for Rs. 2. According to their calculation, 3 oranges for Rs. 1 and 2 oranges for Rs. 1 was exactly the same as 5 oranges for Rs. 2

Now, they were expecting to get Rs. 25 for the oranges, as they would have got, if sold separately. But to their surprise they got only Rs. 24 for the entire lot of the 60 oranges.

Where did the one rupee go?

Answer

Woman X was selling 3 oranges for a Rupee. Hence, the average price of one orange is (1/3)

= Rs. 0.3333

Woman Y was selling 2 oranges for a Rupee. Hence, the average price of one orange is (1/2) = Rs. 0.5

Now they put together 60 oranges. Hence, the average price of one orange is (0.3333+0.5)/2 = Rs. 0.4167

But, they were selling 5 oranges for Rs. 2. Hence, the average price of one orange is 2/5 = Rs. 0.4

The differnce is (0.4167 - 0.4) = 0.0167 There are 60 oranges. Hence (0.0167 * 60) = 1

So here is that one rupee. Thus, one rupee didn't go anywhere, they did their math wrong.


A blindfolded man is asked to sit in the front of a carrom board. The holes of the board are shut with lids in random order, i.e. any number of all the four holes can be shut or open.

Now the man is supposed to touch any two holes at a time and can do the following.

Open the closed hole. Close the open hole. Let the hole be as it is.

After he has done it, the carrom board is rotated and again brought to some position. The man is again not aware of what are the holes which are open or closed.

How many minimum number of turns does the blindfolded man require to either open all the holes or close all the holes?

Note that whenever all the holes are either open or close, there will be an alarm so that the blindfolded man will know that he has won.Submitted by : Vikrant Ramteke

Answer

The blindfolded man requires 5 turns.

1. Open two adjacent holes.

2. Open two diagonal holes. Now atleast 3 holes are open. If 4th hole is also open, then you are done. If not, the 4th hole is close.

3. Check two diagonal holes. o If one is close, open it and all the holes are open. o If both are close, open any one hole. Now, two holes are open and two are

close. The diagonal holes are in the opposite status i.e. in both the diagonals,

one hole is open and one is close.

4. Check any two adjacent holes. o If both are open, close both of them. Now, all holes are close. o If both are close, open both of them. Now, all holes are open. o If one is open and one is close, invert them i.e. close the open hole and open

the close hole. Now, the diagonal holes are in the same status i.e. two holes in one diagonal are open and in other are close.

5. Check any two diagonal holes. o If both are open, close both of them. Now, all holes are close.

o If both are close, open both of them. Now, all holes are open.


What are the next two numbers in the series? 34, 58, 56, 60, 42, 52, 65, ?, ?

Answer


The pattern is : the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX, SEVEN, EIGHT and NINE when A=1, B=2, C=3, D=4 and so on.

First number = 15(O) + 14(N) + 5(E) = 34 Second Number = 20(T) + 23(W) + 15(O) = 58 Third Number = 20(T) + 8(H) + 18(R) + 5(E) + 5(E) = 56 Fourth Number = 6(F) + 15(O) + 21(U) + 18(R) = 60 Fifth Number = 6(F) + 9(I) + 22(V) + 5(E) = 42 Sixth Number = 19(S) + 9(I) + 24(X) = 52 Seventh Number = 19(S) + 5(E) + 22(V) + 5(E) + 14(N) = 65 Eight Number = 5(E) + 9(I) + 7(G) + 8(H) + 20(T) = 49 Ninth Number = 14(N) + 9(I) + 14(N) + 5(E) = 42

146. There are 20 poles with a constant distance between each pole A car takes 24 second to reach the 12th pole. How much will it take to reach the last pole.

Ans: 41.45 seconds (2 marks) Let the distance between two poles = x Hence 11x:24::19x:?

147. Ann, Boobie, Cathy and Dave are at their monthly business meeting.

Their occupations are author, biologist, chemist and doctor, but not necessarily in that order. Dave just told the biologist that Cathy was on her way with doughnuts. Ann is sitting across from the doctor and next to the chemist. The doctor was thinking that Boobie was a goofy name for parent's to choose,but didn't say anything. What is each person's occupation?

Ans: Since Dave spoke to the biologist and Ann sat next to the chemist and across the doctor, Cathy must be the author and Ann the biologist. The doctor didn't speak, but David did, so Bobbie is the doctor and Dave the chemist.


There is a girl running from some robbers. She holds two solid gold balls each weighing 20 pounds. She comes to an old rope bridge and a sign that says max weight 130 pounds.

Knowing her weight was 100 pounds, she walked across the bridge with both gold balls, without the bridge breaking. How did she do that?

Answer

She juggled the balls so that one ball was always in the air when she walked across the bridge.


A man throws an object a certain distance; it stops momentarily and returns. It is not a boomerang and does not touch anything. How?

Answer

The man throws the object vertically upward, perpendicular to the ground.

When you throw an object vertically up, it stops for a moment and then returns back.


Everyday in his business a merchant had to weigh amounts from 1 kg to 121 kgs, to the nearest kg. What are the minimum number of different weights required and how heavy should they be?

Answer

The minimum number is 5 and they should weigh 1, 3, 9, 27 and 81 kgs.


There was a man who had just hung himself. When you search the room where it happend all you can see is the young man hanging and a puddle of water below him. There are no objects or furniture that could have helped him to commit suicide.

How did he kill himself?

Answer

The man stood on a block of ice. When it melted, he hung himself or he jumped off and hung himself.


A woman had two sons who were born on the same hour of the same day of the same month of the same year. But they were not twins.

How could this be so?

Answer

This is a lateral thinking puzzle.

They were two of a set of triplets (or quadruplets, etc.).

This puzzle stumps many people. They try outlandish solutions involving test-tube babies or surrogate mothers. Why does the brain search for complex solutions when there is a much simpler one available?


A class of 100 students. 24 of them are girls and 32 are not. Which base am I using?

Answer

Let the base be X.

Therefore (X*X + X*0 + 0) = (2*X +4) + (3*X + 2)

X*X = 5*X + 6 X*X - 5*X -6 = 0 (X-6)(X+1) = 0

Therefore base is 6


Suppose I declare : "I am now lying."

Am I lying or telling the truth?

Answer

It's a Paradox.

If you say that he is lying then his declaration becomes true i.e. he is telling truth; and If you say that he is telling truth then his declaration becomes false i.e. he is lying.


A woman goes into a hardware store to buy something for her house. When asked the price, the clerk replies, "the price of one is twelve cents, the price of forty-four is twenty-four cents, and the price a hundred and forty-four is thirty-six cents.

What does the woman want to buy?

Answer

House numbers.

Price of one i.e. 1 = 12 cents (as only number 1) Price of forty-four i.e. 44 = 24 cents (as there are two numbers) Price of a hundred and forty-four i.e. 144 = 36 cents (as there are three numbers)


A man is wearing black. Black shoes, socks, trousers, jumper, gloves and balaclava. He is walking down a black street with all the street lamps off. A black car is coming towards him with its light off but somehow manages to stop in time.

How did the driver see the man?

Answer


It was day time.


A farmer needs 8 gallons of water. He has only three unmared buckets, two 6 gallon and one 11 gallon bucket.

How can he collect 8 gallons of water using three unmarked buckets? Provide solution with minimal water wastage.

Answer

Here is the solution with 10 gallon water wastage.

OPERATIONS 6 6 11

Fill 6 gallon bucket with water 6 0 0

Empty 6 gallon bucket into 11 gallon bucket 0 0 6

Fill 6 gallon bucket with water 6 0 6

Fill 11 gallon bucket to full using filled 6 gallon bucket. This will leave 1 gallon water in 6 gallon bucket

1 0 11

Empty 11 gallon bucket into second 6 gallon bucket. 1 6 5

Empty 11 gallon bucket - wastage of 5 gallon water 1 6 0

Empty second 6 gallon bucket into 11 gallon bucket 1 0 6

Fill seccond 6 gallon bucket with water 1 6 6

Fill 11 gallon bucket to full using filled second 6 gallon bucket. This will leave 1 gallon water in second 6 gallon bucket

1 1 11

Fill first 6 gallon bucket with 1 gallon water which is in second 6 gallon bucket 2 0 11

Empty 11 gallon bucket into second 6 gallon bucket. 2 6 5

Empty 11 gallon bucket - wastage of 5 gallon water 2 6 0

Fill 11 gallon bucket with water in both the 6 gallon buckets 0 0 11


An electric train is going south but the wind is blowing westward.

Which direction will the smoke from the train be blowing?

Answer

It is an Electric Train. And Electric trains do not produce any smoke. Therefore there is no smoke to be blown


Ankit and Tejas divided a bag of Apples between them.

Tejas said, "It's not fair! You have 3 times as many Apples I have." Ankit said, "OK, I will give you one Apple for each year of your age." Tejas replied, "Still not fair. Now, you have twice as many Apples as I have." "Dear, that's fair enough as I am twice older than you.", said Ankit.

Ankit went to Kitchen to drink water. While Ankit was in Kitchen, Tejas took apples from Ankit's pile equal to Ankit's age.

Who have more apples now?

Answer

At the end, Ankit and Tejas, both have the same number of apples.

Let's assume that initially Tejas got N apples and his age is T years. Hence, initially Ankit got 3N apples and his age is 2T years.

Operation Ankit's Apples Tejas's Apples

Initially 3N N

Ankit gave T apples to Tejas(equals age of Tejas)

3N – T N + T

Tejas took 2T apples from Ankit's pile(equals age of Ankit)

3N - 3T N + 3T

It is given that after Ankit gave T apples to Tejas, Ankit had twice as many apples as Tejas had. 3N - T = 2*(N + T) 3N - T = 2N + 2T N = 3T

From the table, at the end Ankit have (3N - 3T) apples and Tejas have (N + 3T) apples. Substituting N = 3T, we get Ankit's apples = 3N - 3T = 9T - 3T = 6T Tejas's apples = N + 3T = 3T + 3T = 6T

Thus, at the end Ankit and Tejas, both have the same number of apples.


In a small town, there are three temples in a row and a well in front of each temple. A pilgrim came to the town with certain number of flowers.

Before entering the first temple, he washed all the flowers he had with the water of well. To his surprise, flowers doubled. He offered few flowers to the God in the first temple and moved to the second temple. Here also, before entering the temple he washed the remaining flowers with the water of well. And again his flowers doubled. He offered few flowers to the God in second temple and moved to the third temple. Here also, his flowers doubled after washing them with water. He offered few flowers to the God in third temple.

There were no flowers left when pilgrim came out of third temple and he offered same number of flowers to the God in all three temples.

What is the minimum number of flowers the pilgrim had initially? How many flower did he offer to each God?

Answer

The pilgrim had 7 flowers, initially and he offered 8 flowers to each God.

Assume that the pilgrim had X flowers initially and he offered Y flowers to each God.

From the above figure, there are (8X - 7Y) flowers when the pilgrim came out of the third temple. But it is given that there were no flowers left when he came out of third temple. It means that (8X - 7Y) = 0

8X = 7Y

The minimum values of X and Y are 7 and 8 respectively to satisfy above equation. Hence, the pilgrim had 7 flowers and he offered 8 flowers to each God.

In general, the pilgrim had 7N flowers initially and he offered 8N flowers to each God, where N = 1, 2, 3, 4, .....


There were two men standing on a street. The one says to the other, "I have 3 daughters, the product of their ages is 36. What is the age of the OLDEST daughter?"

The second guy says, "I need more information." So, the first guy says, "The sum of their ages is equal to the address of the house across the street."

The second guy looks at the address and says, "I still need more information." So, the first guy says, "My oldest daughter wears a red dress."

Answer

The answer is 9 years.

First you need to find all the possible sets of three numbers that when multiplied equals 36:

1 1 36 1 2 18 1 3 12 1 4 9 1 6 6 2 2 9 2 3 6 3 3 4

Then you add the numbers together to find the sum 1 1 36 = 38 1 2 18 = 21 1 3 12 = 16 1 4 9 = 14 1 6 6 = 13 2 2 9 = 13 2 3 6 = 11 3 3 4 = 10

Even though we don't know the address the guy knows it. For him to need more information that means that at least two of the sets of numbers has the same sum. Two of them do, 1 6 6 and 2 2 9.

When the first guy said that his OLDEST daugher wears a red dress that meant that there had to be the oldest. So 1 6 6 can't possibly be the answer. So the possible possiblity is 2 2 9 and the OLDEST daughter is 9 years old.

Therefore, the answer is 9.


You have someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day.

If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

Answer

Make two breaks such that you get three segments with 1 piece, 2 pieces and 4 pieces and follow as below:

Day 1 : Give a single segment to the worker. Day 2 : Ask the worker to return the segment you gave him on day 1 and give him the segment with Two connected pieces. Day 3 : GIve the worker the 1 piece segment you have. Day 4 : Ask the worker to return all pieces he has (one segment and 2 segment pieces) and give him the segment with 4 pieces on it. Day 5 : Give him the the segment with 1 piece. Day 6 : Ask him to return the 1 piece segment and give him the segment with two pieces. Day 7 : Give him the 1 piece segment you have.


You are in a room with 2 doors leading out. Behind 1 door is a coffer overflowing with jewels & gold, along with an exit. Behind the other door is an enormous, hungry lion that will pounce on anyone opening the door. You do not know which door leads to the treasure & exit, & which door leads to the lion.

In the room you are in, are 2 individuals. The first is a knight, who always tells the truth, & a knave, who always lies. Both of these individuals know what is behind each door. You do not know which individual is the knight, or which one is the knave.

You may ask 1 of the individuals exactly 1 question. What should you ask in order to be certain that you will open the door with the coffer behind it, instead of the hungry lion?

Answer

There are 2 possible answers for this puzzle:

Answer 1: You ask one of the individuals what the other one would say if you asked him or her which door you should open to get to the coffer. In this case, you would open the other door.

Answer 2: You ask one of the individuals what the other one would say if you asked him or her which door is holding back the hungry lion. In this case, you would open this door.


Last Saturday Milan went for the late night show and came late. In the morning family members asked him which movie did he see. He gave different answers to everyone.

He told to his father that he had gone to see MONEY. According to his mom, he saw either JOHNY or BABLU. His elder brother came to know that he saw BHABI. To his sister, he told ROBOT. And his grandpa heard that he saw BUNNY.

Thus, Milan gave six movie names, all five letter words. But he saw some other movie with five letter word. Moreover, each of the six movie names mentioned above has exactly two letters common with the movie he saw. (with the same positions)

Can you tell which movie did Milan see?

Answer

Milan saw BOBBY.

The six movie names are - MONEY, JOHNY, BABLU, BHABI, ROBOT and BUNNY.

Compare MONEY and JOHNY. They have O common at the second place and Y common at the fifth place. Also, they can't have two different letters each, common with the required movie as the letters in remaining three places are all different. Thus, the required movie must have either O at the second place or Y at the fifth place or both.

Similarly, comparing JOHNY and BUNNY - the required movie must have either N at the fourth place or Y at the fifth place or both. Also, comparing MONEY and BUNNY - the required movie must have either N at the third place or Y at the fifth place or both.

From the above 3 deduction, either Y is at fifth place or O is at the second place and N is at the third & fourth place. The later combination is not possible as BABLU, BHABI & ROBOT will need at least 3 other letters which makes the required movie 6 letter long. Hence, the required movie must have Y at the fifth place.

Now Y is not there in BABLU and BHABI at the fifth place and they have only B common at the first place. Hence, B must be the first letter.

As B is at the first place and Y is at the fifth place and every movie has exactly 2 letters common with the required movie. From BUNNY, the required movie do not have U at the second place and N at the third and fourth place. Now looking at JOHNY and MONEY, they must have O common at the second place.

Using the same kind of arguments for BABLU, BHABI and ROBOT, we can conclude that Milan saw BOBBY.


You are locked inside a room with 6 doors - A, B, C, D, E, F. Out of which 3 are Entrances only and 3 are Exits only.

One person came in through door F and two minutes later second person came in through door A. He said, "You will be set free, if you pass through all 6 doors, each door once only and in correct order. Also, door A must be followed by door B or E, door B by C or E, door C by D or F, door D by A or F, door E by B or D and door F by C or D."

After saying that they both left through door B and unlocked all doors. In which order must you pass through the doors?

Answer

The correct order is CFDABE

It is given that one person came in through door F and second person came in through door A. It means that door A and door F are Entrances. Also, they both left through door B. Hence, door B is Exit.

As Exit and Entrance should alter each other and we know two Entrances, let's assume that the third Entrance is W. Thus, there are 6 possibilities with "_" indicating Exit.

(1) _W_A_F (2) _W_F_A (3) _F_W_A (4) _F_A_W (5) _A_W_F (6) _A_F_W

As door A must be followed by door B or E and none of them lead to the door F, (1) and (6) are not possible.

Also, door D must be the Exit as only door D leads to the door A and door A is the Entrance.

(2) _W_FDA (3) _F_WDA (4) _FDA_W (5) DA_W_F

Only door D and door C lead to the door F. But door D is used. Hence, door C must be the Exit and precede door F. Also, the third Exit is B and the W must be door E.

(2) BECFDA (3) CFBEDA (4) CFDABE (5) DACEBF

But only door B leads to the door C and both are Exits. Hence, (2) and (5) are not possible. Also, door F does not lead to door B - discard (3). Hence, the possible order is (4) i.e. CFDABE.


A person travels on a cycle from home to church on a straight road with wind against him. He took 4 hours to reach there.

On the way back to the home, he took 3 hours to reach as wind was in the same direction.

If there is no wind, how much time does he take to travel from home to church?

Answer

Let distance between home and church is D.

A person took 4 hours to reach church. So speed while travelling towards church is D/4.

Similarly, he took 3 hours to reach home. So speed while coming back is D/3.

There is a speed difference of 7*D/12, which is the wind helping person in 1 direction, & slowing him in the other direction. Average the 2 speeds, & you have the speed that person can travel in no wind, which is 7*D/24.

Hence, person will take D / (7*D/24) hours to travel distance D which is 24/7 hours.

Answer is 3 hours 25 minutes 42 seconds


Consider a number 235, where last digit is the sum of first two digits i.e. 2 + 3 = 5.

How many such 3-digit numbers are there?

Answer

There are 45 different 3-digit numbers.

The last digit can not be 0.

If the last digit is 1, the only possible number is 101. (Note that 011 is not a 3-digit number)

If the last digit is 2, the possible numbers are 202 and 112.

If the last digit is 3, the possible numbers are 303, 213 and 123.

If the last digit is 4, the possible numbers are 404, 314, 224 and 134.

If the last digit is 5, the possible numbers are 505, 415, 325, 235 and 145.

Note the pattern here - If the last digit is 1, there is only one number. If the last digit is 2, there

are two numbers. If the last digit is 3, there are three numbers. If the last digit is 4, there are four numbers. If the last digit is 5, there are five numbers. And so on.....

Thus, total numbers are 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Altogether then, there are 45 different 3-digit numbers, where last digit is the sum of first two digits.


What are the next three numbers in the given series?

0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 ? ? ?

Answer

The next three numbers are 4, 2 and 2

The pattern is - the number of factors in prime factorization of positive integers. Note that all prime number positions are 1 in the given series.

Number Prime Factorization Factors in Prime Factorization

1 1 0

2 2 1

3 3 1

4 2*2 2

5 5 1

6 2*3 2

7 7 1

8 2*2*2 3

9 3*3 2

10 2*5 2

Thus, the next few numbers in the given series are: 4 2 2 3 3 1 3 1 5 2 2 2 4 1 2 2 4 1 3 1 3 3 2 1 5 2 3 2 3 1 4 2 4 2 2 1 ...

Note that 1 is neither a Prime number nor a Composite number.


Two planes take off at the same exact moment. They are flying across the Atlantic. One leaves New York and is flying to Paris at 500 miles per hour. The other leaves Paris and is flying to New York at only 450 miles per hour ( because of a strong head wind ).

Which one will be closer to Paris when they meet?

Answer

They will both be the same distance from Paris when they meet!!!


There is a river and 3 people. Two people weigh 50 lbs and one of them weighs 100 lbs. The only way you can cross the river is by using a boat.

How can they all cross the river if the boat only can hold 100 lbs?


How many possible combinations are there in a 3x3x3 rubics cube?

In other words, if you wanted to solve the rubics cube by trying different combinations, how many might it take you (worst case senerio)?

How many for a 4x4x4 cube?

Answer

There are 4.3252 * 10^19 possible combinations for 3x3x3 Rubics and 7.4012 * 10^45 possible combinations for 4x4x4 Rubics.

Let's consider 3x3x3 Rubics first.

There are 8 corner cubes, which can be arranged in 8! ways. Each of these 8 cubes can be turned in 3 different directions, so there are 3^8 orientations altogether. But if you get all but one of the corner cube into chosen positions and orientations, only one of 3 orientations of the final corner cube is possible. Thus, total ways corner cubes can be placed = (8!) * (3^8)/8 = (8!) * (3^7)

Similarly, 12 edge cubes can be arranged in 12! ways. Each of these 12 cubes can be turned in 2 different directions, so there are 2^12 orientations altogether. But if you get all but one of the edge cube into chosen positions and orientations, only one of 2 orientations of the final edge cube is possible. Thus, total ways edge cubes can be placed = (12!) * (2^12)/2 = (12!) * (2^11)

Here, we have essentially pulled the cubes apart and stuck cubes back in place wherever we please. In reality, we can only move cubes around by turning the faces of the cubes. It turns out that you can't turn the faces in such a way as to switch the positions of two cubes while returning all the others to their original positions. Thus if you get all but two cubes in place, there is only one attainable choice for them (not 2!). Hence, we must divide by 2.

Total different possible combinations are = [(8!) * (3^7)] * [(12!) * (2^11)] / 2 = (8!) * (3^7) * (12!) * (2^10) = 4.3252 * 10^19

Similarly, for 4x4x4 Rubics total different possible combinations are = [(8!) * (3^7)] * [(24!)] * [(24!) / (4!^6)] / 24 = 7.4011968 * 10^45

Note that there are 24 edge cubes, which you can not turn in 2 orientations (hence no 2^24 / 2). Also, there are 4 center cubes per face i.e. (24!) / (4!^6). You can switch 2 cubes without affecting the rest of the combination as 4*4*4 has even dimensions (hence no division by 2). But pattern on one side is rotated in 4 directions over 6 faces, hence divide by 24.


A rich old Arab has three sons. When he died, he willed his 17 camels to the sons, to be divided as follows:

First Son to get 1/2 of the camels Second Son to get 1/3rd of the camels Third Son to get 1/9th of the camels.

The sons are sitting there trying to figure out how this can possibly be done, when a very old wise man goes riding by. They stop him and ask him to help them solve their problem. Without hesitation he divides the camels properly and continues riding on his way.

How did he do it?

Answer

The old man temporarily added his camel to the 17, making a total of 18 camels.

First son got 1/2 of it = 9

Second son got 1/3 of it = 6

Third son got 1/9 of it = 2

For a total of 17. He then takes his camel back and rides away......


Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way :

The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared upon the scene. Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer.

Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and the youngest 4. One camel remained : this was his, which he mounted and rode away.

Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it possible?

Answer

They took their percentages from 40 and not from 39, so they got more than their share.

The oldest son got 1/2 of 40 = 20 which is 0.5 more The second son got 1/4 of 40 = 10 which is 0.25 more The third son got 1/8 of 40 = 5 which is 0.125 more The youngest son got 1/10 of 40 = 4 which is 0.1 more

And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)

All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.


Because cigars cannot be entirely smoked, a Bobo who collects cigar butts can make a cigar to smoke out of every 3 butts that he finds.

Today, he has collected 27 cigar butts. How many cigars will he be able to smoke?

Answer

13 not 12

He makes 9 originals from the 27 butts he found, and after he smokes them he has 9 butts left for another 3 cigars. And then he has 3 butts for another cigar.

So 9+3+1=13


You have 13 balls which all look identical. All the balls are the same weight except for one. Using only a balance scale, can find the odd one out with only 3 weighings?

Is it possible to always tell if the odd one out is heavier or lighter than the other balls?

Answer

It is always possible to find odd ball in 3 weighings and in most of the cases it is possible to tell whether the odd ball is heavier or lighter. Only in one case, it is not possible to tell the odd ball is whether heavier or lighter.

1. Take 8 balls and weigh 4 against 4. o If both are not equal, goto step 2 o If both are equal, goto step 3

2. One of these 8 balls is the odd one. Name the balls on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the balls on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one ball from the remaining 5 balls in intial weighing.

o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3. If both are equal, L4 is the odd ball and is lighter. If L2 is light, L2 is the odd ball and is lighter. If L3 is light, L3 is the odd ball and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

If both are equal, there is some error. If H1 is heavy, H1 is the odd ball and is heavier. If H2 is heavy, H2 is the odd ball and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

If both are equal, L1 is the odd ball and is lighter. If H3 is heavy, H3 is the odd ball and is heavier. If H4 is heavy, H4 is the odd ball and is heavier.

3. One of the remaining 5 balls is the odd one. Name the balls as C1, C2, C3, C4, C5.

Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3 are any three balls from the first weighing of 8 balls.

o If both are equal, one of remaining 2 balls is the odd i.e. either C4 or C5. Weigh C4 with X1

If both are equal, C5 is the odd ball. But you can not tell whether it is heavier or lighter.

If C4 is heavy, C4 is the odd ball and is heavier. If C4 is light, C4 is the odd ball and is lighter.

o If (C1, C2, C3) is heavier side, one of C1, C2, C3 is the odd ball and is heavier. Weigh C1 and C2.

If both are equal, C3 is the odd ball and is heavier. If C1 is heavy, C1 is the odd ball and is heavier. If C2 is heavy, C2 is the odd ball and is heavier.

o If (C1, C2, C3) is lighter side, one of C1, C2, C3 is the odd ball and is lighter. Weigh C1 and C2.

If both are equal, C3 is the odd ball and is heavier. If C1 is light, C1 is the odd ball and is lighter.

If C2 is light, C2 is the odd ball and is lighter.


There are four people in a room (not including you). Exactly two of these four always tell the truth. The other two always lie.

You have to figure out who is who IN ONLY 2 QUESTIONS. Your questions have to be YES or NO questions and can only be answered by one person. (If you ask the same question to two different people then that counts as two questions). Keep in mind that all four know each other's characteristics whether they lie or not.

What questions would you ask to figure out who is who? Remember that you can ask only 2 questions.

Answer

0-represents a Liar 1-represents a Truth Teller A B C D

-------

0 0 1 1 A and B are Liars

0 1 0 1 A and C are Liars

0 1 1 0 A and D are Liars

1 0 0 1 B and C are Liars

1 0 1 0 B and D are Liars

1 1 0 0 C and D are Liars

As you can see there are 6 possible cases. If you give only yes or no questions and get only yes or no responses the best you can do is eliminate half of the cases for every question asked. That means after asking the first question the best (worst case) scenario you can end up with is to eliminate all but three cases. That means that the next question addresses three cases and can at best eliminate only 1!

That means that you cannot solve this in 2 questions "giving only yes or no questions and get ONLY yes or no responses". The trick is to ask a question that has the potential of being answered with: "yes", "no", or no response at all.

First Question Ask A: is B a Truth teller AND is C a Liar. (Asking A: does B=1 AND does C=0)

if answer to first Question is YES: then do {PART 1} if answer to first Question is NO: then skip to {PART 2}

---------------------{PART 1}--------------------- we know it is one of the following scenarios: A B C D

-------

0 0 1 1

1 1 0 0

0 1 1 0

Second Question: Ask C: Let's assign a value of 0 for every Liar and a value of 1 for every Truth teller in the room. Let's also suppose that B has a secret number in his head that ONLY HE KNOWS. All we know about this number is that it is greater than 0, less than 2, and is not an integer. Would B say that the sum of A and B is greater than the secret number he is thinking of? (Asking C: would B say (A+B) > n; where 0>n>2 and n is not an integer) If C answers YES:

A B C D

-------

0 0 1 1 A and B are the Liars

If A and B are Liars then their sum (0+0) will 0. We know the number B is thinking of is greater than 0. Therefore B will lie and say YES. C will tell us the truth of what B would say and so he says YES. If C answers NO:

A B C D

-------

1 1 0 0 C and D are the Liars

If A and B are Truth tellers then their sum (1+1) will 2. We know the number B is thinking of is less than 2. Therefore B will tell the truth and say YES. HOWEVER, C will LIE about what B will say and his response will be NO. IF C DOESN'T answer:

A B C D

-------

0 1 1 0 A and D are the Liars

If A is a liar and B is a Truth teller then their sum (0+1) will 1. We know the number B is thinking of is less than 2 and greater than 0 but ONLY B KNOWS for sure what its value is. Therefore C does not know for sure what B will say (even though he knows it will be the truth). Therefore C doesn't answer because if he did he could run the risk of telling a lie which, of course, he never does. [ALL DONE]

---------------------{PART 2}--------------------- We know it is one of the following scenarios: A B C D

-------

1 0 0 1

1 0 1 0

0 1 0 1

Second Question: Ask C: Let's assign a value of 0 for every Liar and a value of 1 for every Truth teller in the room. Let's also suppose that B has a secret number in his head that ONLY HE KNOWS. All we know about this number is that it is greater than 0, less than 2, and is not an integer. Would B say that the sum of D and B is greater than the secret number he is thinking of? (Asking C: would B say (D+B) > n; where 0>n>2 and n is not an integer) If C answers YES:

A B C D

-------

1 0 1 0 B and D are the Liars

If D and B are Liars then their sum (0+0) will 0. We know the number B is thinking of is greater than 0. Therefore B will lie and say YES. C will tell us the truth of what B would say and so he says YES. If C answers NO:

A B C D

-------

0 1 0 1 A and C are the Liars

If D and B are Truth tellers then their sum (1+1) will 2. We know the number B is thinking of is less than 2. Therefore B will tell the truth and say YES. HOWEVER, C will LIE about what B will say and his response will be NO. IF C DOESN'T answer:

A B C D

-------

1 0 0 1 B and C are the Liars

If B is a liar and D is a Truth teller then their sum (0+1) will 1. We know the number B is thinking of is less than 2 and greater than 0 but ONLY B KNOWS for sure what its value is. Therefore C does not know for sure what B will say (even though he knows it will be a lie). Therefore C doesn't answer because if he did he could run the risk of telling the truth which, of course, he never does. [ALL DONE]

Thanks to Ryan Hutcherson !!!

Here is one more valid answer (and much more simpler) from Carmel.

Lets call our people A, B, C & D

T = always tells the truth L = always lies

"Same" means both liars or both truthtellers. "Different" means one is a liar and one tells the truth.Ask D "IF A AND B ARE DIFFERENT, is A a truth-teller?"

If D answers yes,

Ask A "Are B and C the same?"

If A answers yes

A=T, B=L, C=L, D=T

If A answers no

A=L, B=T, C=T, D=L

If D answers no,

Ask A "Are B and D the same?"

If A answers yes

A=T, B=L, C=T, D=L

If A answers no

A=L, B=T, C=L, D=T

If D doesn't answer, then it means that

A and B are the same.

Ask A "Are C & D the same?"

If A answers yes,

A=T, B=T, C=L, D=L

If A answers no,

A=L, B=L, C=T, D=T

Explanation: there are six different outcomes, you only get 2 questions. The questions can only be answered yes or no. There is no stipulation that anyone MUST answer a question.


A man is on a search for Atlantis and comes upon an island where all the inhabitants know whether Atlantis is still around or not.

However, all of the inhabitants are either Fairies or Trolls and they all use a spell to appear humanoid so you cannot tell which is which. And the Faries always tell the truth and the Trolls always lie, but there is a slight complication, some of the Fairies have gone insane and always lie and some of the Trolls have also gone insane and always tell the truth.

So here is your task: you must ask the first inhabitant that you come to ONE question and from that ONE question you must determine wether Atlantis is still around or not.

What is the question that you must ask?

Answer

There are 2 answers to it:

Answer I"Is the statement that you are reliable equivalent to the statement that Atlantis is still around?"

Answer II"Do you believe that the Statement that you are a Fairy is equivalent to the statement that Atlantis is still around?"


Yesterday in a party, I asked Mr. Shah his birthday. With a mischievous glint in his eyes he replied. "The day before yesterday I was 83 years old and next year I will be 86."

Can you figure out what is the Date of Birth of Mr. Shah? Assume that the current year is 2000.

Answer

Mr. Shah's date of birth is 31 December, 1915

Today is 1 January, 2000. The day before yesterday was 30 December, 1999 and Mr. Shah was 83 on that day. Today i.e. 1 January, 2000 - he is 84. On 31 December 2000, he will be 85 and next year i.e. 31 December, 2001 - he will be 86. Hence, the date of birth is 31 December, 1915.

Many people do think of Leap year and date of birth as 29th February as 2000 is the Leap year and there is difference of 3 years in Mr. Shah's age. But that is not the answer.


A man and his son are in a car accident. The father dies on the scene, but the child is

rushed to the hospital. When he arrives the surgeon says, "I can't operate on this boy, he is my son!"

How can this be?

Answer


The surgeon was his mother.

Also, the doctor could be the boy's biological father, and the man in the car could be the step or legal father.


A man bought a parrot at a pet-shop. The owner of the shop guaranteed the parrot could repeat any word it hears. The man bought the parrot.

It never spoke a word. But, the owner of the store spoke the absolute truth. It could repeat any word it heard.

Why didn't it talk?

Answer

The parrot was deaf.

The pet-shop owner guaranteed that the parrot could repeat any word that it hears, but if he couldn't hear, he couldn't repeat.

The other possible answer is the man was mute. As he could not speak, there was nothing to hear and hence the parrot never spoke a word.


In 1990, a person is 15 years old. In 1995, that same person is 10 years old.

How is this possible?

Answer

The time is BC (Before Christ) i.e. years are counted backwards.

The years are counted backwards in BC. In 1995 BC, the person would be 10 years old. 5 years later the time would have been 1990 BC and the person would be 15 years old.


A bus driver was going down a street. He was on the wrong side. He went at a stop sign and didn't signal at a turn. He went under the minimum speed and yet he didn't get in trouble for his actions.

WHY not?

Answer

He was Walking. He is a Bus driver. But it does not say that he was driving.


One day Kerry celebrated her birthday. Two days later her older twin brother, Terry, celebrated his birthday.

How? Note that they both celebrated their birthday on their actual birthdays.

Answer


At the time she went into labor, the mother of the twins was traveling by boat. The older twin, Terry, was born first early on March 1st. The boat then crossed a time zone and Kerry, the younger twin, was born on February the 28th. Therefore, the younger twin celebrates her birthday two days before her older brother.

There is one more answer submitted by Beano. She obviously had an older set of twin brothers and their birthday was 2 days after hers, they just didn't mention that it was also her other twin brothers birthday.

Also, according to Robbie TWINS DONT HAVE TO HAVE THE SAME BIRTHDAY! SOME TWINS ARE BORN 300 DAYS APART !!!


When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you will put to the sword, and if you tell the truth you will be hanged."

The soldier could make only one statement. He made that statement and went free. What did he say?

Answer

The soldier said, "You will put me to the sword."

The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!


There are 4 mathematicians - Brahma, Sachin, Prashant and Nakul - having lunch in a hotel. Suddenly, Brahma thinks of 2 integer numbers greater than 1 and says, "The sum of the numbers is..." and he whispers the sum to Sachin. Then he says, "The product of the numbers is..." and he whispers the product to Prashant. After that following conversation takes place :

Sachin : Prashant, I don't think that we know the numbers. Prashant : Aha!, now I know the numbers. Sachin : Oh, now I also know the numbers. Nakul : Now, I also know the numbers.

What are the numbers? Explain your answer.

Answer

The numbers are 4 and 13.

As Sachin is initially confident that they (i.e. he and Prashant) don't know the numbers, we can conclude that - 1) The sum must not be expressible as sum of two primes, otherwise Sachin could not have been sure in advance that Prashant did not know the numbers. 2) The product cannot be less than 12, otherwise there would only be one choice and Prashant would have figured that out also.

Such possible sum are - 11, 17, 23, 27, 29, 35, 37, 41, 47, 51, 53, 57, 59, 65, 67, 71, 77, 79, 83, 87, 89, 93, 95, 97, 101, 107, 113, 117, 119, 121, 123, 125, 127, 131, 135, 137, 143, 145, 147, 149, 155, 157, 161, 163, 167, 171, 173, 177, 179, 185, 187, 189, 191, 197, ....

Let's examine them one by one.

If the sum of two numbers is 11, Sachin will think that the numbers would be (2,9), (3,8), (4,7) or (5,6).

Sachin : "As 11 is not expressible as sum of two primes, Prashant can't know the numbers."

Here, the product would be 18(2*9), 24(3*8), 28(4*7) or 30(5*6). In all the cases except for product 30, Prashant would know the numbers.

- if product of two numbers is 18: Prashant : "Since the product is 18, the sum could be either 11(2,9) or 9(3,6). But if the sum was 9, Sachin would have deduced that I might know the numbers as (2,7) is the possible prime numbers pair. Hence, the numbers must be 2 and 9." (OR in otherwords, 9 is not in the Possible Sum List)

- if product of two numbers is 24: Prashant : "Since the product is 24, the sum could be either 14(2,12), 11(3,8) or 10(4,6). But 14 and 10 are not in the Possible Sum List. Hence, the numbers must be 3 and 8."

- if product of two numbers is 28: Prashant : "Since the product is 28, the sum could be either 16(2,14) or 11(4,7). But 16 is not in the Possible Sum List. Hence, the numbers must be 4 and 7."

- if product of two numbers is 30: Prashant : "Since the product is 30, the sum could be either 17(2,15), 13(3,10) or 11(5,6). But 13 is not in the Possible Sum List. Hence, the numbers must be either (2,15) or (5,6)." Here, Prashant won't be sure of the numbers.

Hence, Prashant will be sure of the numbers if product is either 18, 24 or 28.

Sachin : "Since Prashant knows the numbers, they must be either (3,8), (4,7) or (5,6)." But he won't be sure. Hence, the sum is not 11.

Summerising data for sum 11:

Possible Sum PRODUCT Possible Sum

2+9 18 2+9=11 (possible)3+6=9

3+8 24 2+12=143+8=11 (possible)4+6=10

4+7 28 2+12=143+8=11 (possible)4+6=10

5+6 30 2+15=17 (possible)3+10=135+6=11 (possible)

Following the same procedure for 17:

Possible Sum PRODUCT Possible Sum

2+15 30 2+15=17 (possible)3+10= 135+6=11 (possible)

3+14 42 2+21=23 (possible)3+14=17 (possible)6+7=13

4+13 52 2+26=284+13=17 (possible)

5+12 60 2+30=323+20=23 (possible)4+15=195+12=17 (possible)6+10=16



8+9 72 2+36=383+24=27 (possible)4+18=226+12=188+9=17 (possible)

Here, Prashant will be sure of the numbers if the product is 52.

Sachin : "Since Prashant knows the numbers, they must be (4,13)."

For all other numbers in the Possible Sum List, Prashant might be sure of the numbers but Sachin won't.

Here is the step by step explaination:

Sachin : "As the sum is 17, two numbers can be either (2,15), (3,14), (4,13), (5,12), (6,11), (7,10) or (8,9). Also, as none of them is a prime numbers pair, Prashant won't be knowing numbers either."

Prashant : "Since Sachin is sure that both of us don't know the numbers, the sum must be one of the Possible Sum List. Further, as the product is 52, two numbers can be either (2,26) or (4,13). But if they were (2,26), Sachin would not have been sure in advance that I don't know the numbers as 28 (2+26) is not in the Possible Sum List. Hence, two numbers are 4 and 13."

Sachin : "As Prashant now knows both the numbers, out of all possible products - 30(2,15),

42(3,14), 52(4,13), 60(5,12), 66(6,11), 70(7,10), 72(8,9) - there is one product for which list of all possible sum contains ONLY ONE sum from the Possible Sum List. And also, no such two lists exist. [see table above for 17] Hence, two numbers are 4 and 13."

Nakul figured out both the numbers just as we did by observing the conversation between Sachin and Prashant.

It is interesting to note that there are no other such two numbers. We checked all the possible sums till 500 !!!


Replace the letters with the correct numbers.

T W O

X T W O

---------

T H R E E

Answer

T=1, W=3, O=8, H=9, R=2, E=4 1 3 8

x 1 3 8

------------

1 9 0 4 4You can reduce the number of trials. T must be 1 as there is multiplication of T with T in hundred's position. Also, O can not be 0 or 1. Now, you have to find three digit number whose square satisfies above conditions and square of that has same last two digits. Hence, it must be between 102 and 139.


Decipher this sentence.

H S E N O W S J S U D, Z Q' T

E R K V Z K W Z T J; H S E N O W S J

Q A S, Z Q' T W D T D K W H P

Answer

Start with ZQ'T it must be "it's". That leaves ERKVZKWZTJ ending with "is_". The last letter must be "m" (as "t" or "h" is not possible after few try). It leaves 4-letter word OWSJ ending with "m". Try some common words and "from" will fit. Similarly, try some trial-n-error for the remaining words.

H S E N O W S J S U D, Z Q' T

c o p y f r o m o n e, i t' s

E R K V Z K W Z T J; H S E N O W S J

p l a f i a r i s m; c o p y f r o m

Q A S, Z Q' T W D T D K W H P

t w o, i t' s r e s e a r c h


Substitute digits for the letters to make the following addition problem true.

W H O S E

T E E T H

A R E

+ A S

-------------------

S W O R D SNote that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter H, no other letter can be 3 and all other H in the puzzle must be 3.

Answer

It is obvious that S=1 and T=9.

Also, (H + E) should be greater than 10 and hence, (E + H + E) must 20. Thus, there are 3 possible values for (E, H) pair: (6, 8) or (7, 6) or (8, 4). Use trial-n-error and everything will fit-in.

W H O S E 2 8 5 1 6

T E E T H 9 6 6 9 8

A R E 4 7 6

+ A S + 4 1

------------------- -------------------

S W O R D S 1 2 5 7 3 1


Decipher this sentence.

B R W Q H L F K W H J K Q I B W K

Q I C E D W Z B G W K K M I K E

Z B G Q H S K Z B G J K Z K W

B U U Z B G J D B H F W.

Answer

Start with ZBG and ZBGJ. It should be either "the/then" or "you/your" combination as they appear more.

B R W Q H L F K W H J K Q I B W K

o b s t a c l e s a r e t h o s e

Q I C E D W Z B G W K K M I K E

t h i n g s y o u s e e w h e n

Z B G Q H S K Z B G J K Z K W

y o u t a k e y o u r e y e s

B U U Z B G J D B H F W.

o f f y o u r g o a l s.

d189)Brain Teaser No : 00020

Multiply by 7 the number of 8's immediately followed by 3, but not by 5, in the number below:

381654783298514285838385737983256941837408326

Answer

This is simple one.

Simply count the occurrence of 83 There are 6 (six) 8's followed by 3 Now multiply it by 7

Hence the answer is 42.


Next number in the series is 1, 2, 4, 13, 31, 112, ?

Answer

224

A tough one. But analyse the series carefully. No number has digits more than 4. So try to convert them to decimal numbers from base 5 numbers. (just try that) So it turns out to be:

1, 2, 4, 8, 16 , 32 , ?

Got it !!! Next number should be 64. But all numbers in actual series are base 5 numbers So convert 64 to base 5 and that is 224.


If 6 x 4 = 12 8 x 6 = 32 11 x 8 = 66 then 10 x 10 = ??

Answer

80

The pattern is multiply the first number by second after reducing second number by 2. One can view is as follow:

6 x (4 - 2) = 12 8 x (6 - 2) = 32 11 x (8 - 2) = 66 10 x (10 - 2) = 80


There are 3 ants at 3 corners of a triangle, they randomly start moving towards another corner.

What is the probability that they don't collide?

Answer

Let's mark the corners of the triangle as A,B,C. There are total 8 ways in which ants can move. 1. A->B, B->C, C->A 2. A->B, B->C, C->B 3. A->B, B->A, C->A 4. A->B, B->A, C->B 5. A->C, C->B, B->A 6. A->C, C->B, B->C 7. A->C, C->A, B->A 8. A->C, C->A, B->C

Out of which, there are only two cases under which the ants won't collide : A->B, B->C, C->A A->C, C->B, B->A

Therefore, probability of ants not colliding : 2/8 = 1/4


How do you cut a square cake into 8 equal piece only using 3 straight cuts?

Note that you can not move the pieces as you cut the cake.

Answer

Cut cake into 4 quarters by 2 vertical cuts which are perpendicular to each other. Now slice the cake horizontially which makes total 8 pieces.


Which number in the series does not fit in the given series: 1 4 3 16 6 36 7 64 9 100

Answer

This is a series with odd positions containing position number whereas even positions containing square of the position.i.e. even position numbers are 4 16 36 64 100 and odd position numbers are 1 3 5 7 9

Hence, 6 does not fit in the series. It should be 5.


What is the four-digit number in which the first digit is 1/3 of the second, the third is the sum of the first and second, and the last is three times the second?

Answer

The 4 digit number is 1349.

It is given that the first digit is 1/3 of the second. There are 3 such possibilities. 1. 1 and 3 2. 2 and 6 3. 3 and 9

Now, the third digit is the sum of the first and second digits. 1. 1 + 3 = 4 2. 2 + 6 = 8 3. 3 + 9 = 12

It is clear that option 3 is not possible. So we are left with only two options. Also, the last digit is three times the second, which rules out the second option. Hence, the answer is 1349.


A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he need?

Answer

The sign-maker will need 192 zeroes.

Divide 1000 building numbers into groups of 100 each as follow: (1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.

For group numbers 2 to 9, he will require 20 zeroes each. And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are = 11 + 8*20 + 21 = 11 + 160 + 21 = 192


What are the next 2 numbers in the pattern?

2, 7, -1, 10, -4, 13, _, _

Answer

The next two number are -7 and 16

The pattern is : To get odd position number, subtract 3 from the previous odd position number. Similarly, to get even position number, add 3 to the previous even position number.

The odd position numbers are : 2, -1, -4 The next odd position numbers are : -7, -10, -13 and so on.

The even position numbers are : 7, 10, 13 The next even position numbers are : 16, 19, 22 and so on.

Hence the series is 2, 7, -1, 10, -4, 13, -7, 16, -10, 19 ....


If it is given that: 25 - 2 = 3 100 x 2 = 20 36 / 3 = 2

What is 144 - 3 = ?

Answer

There are 3 possible answers to it.

Answer 1 : 9 Simply replace the first number by its square root. (25) 5 - 2 = 3 (100) 10 x 2 = 20 (36) 6 / 3 = 2 (144) 12 - 3 = 9

Answer 2 : 11 Drop the digit in the tens position from the first number. (2) 5 - 2 = 3

1 (0) 0 x 2 = 20 (3) 6 / 3 = 2 1 (4) 4 - 3 = 11

You will get the same answer on removing left and right digit alternatively from the first number i.e remove left digit from first (2), right digit from second (0), left digit from third (3) and right digit from forth (4). (2) 5 - 2 = 3 10 (0) x 2 = 20 (3) 6 / 3 = 2 14 (4) - 3 = 11

Answer 3 : 14 Drop left and right digit alternatively from the actual answer. 25 - 2 = (2) 3 (drop left digit i.e. 2) 100 * 2 = 20 (0) (drop right digit i.e. 0) 36 / 3 = (1) 2 (drop left digit i.e. 1) 144 - 3 = 14 (1) (drop right digit i.e. 1)


Let's say that you were a bus driver. At the 1st stop 10 people got on the bus, 2nd stop 3 people got on the bus and 2 got off, 3rd stop 23 people got on and 9 got off, 4th stop 53 people got on, 5th stop 30 people got off another 2 people got off and then got back on, and 4 got on at this bus stop. Now at the last stop before the bus came to its final destination 29 peolpe got on and 23 people got on.

What was the bus drivers name?

Answer

Your Name.

In the beginning of the riddle, it says "Let's say that you were a bus driver." So you are the Bus Driver and your name is the answer to this riddle.


You have 14 apples. Your Friend Marge takes away 3 and gives you 2. You drop 7 but pick up 4. Bret takes 4 and gives 5. You take one from Marge and give it to Bret in exchange for 3 more. You give those 3 to Marge and she gives you an apple and an orange. Frank comes and takes the apple Marge gave you and gives you a pear. You give the pear to Bret in exchange for an apple. Frank then takes an apple from Marge, gives it to Bret for an orange, gives you the orange for an apple.

How many pears do you have?

Answer

None

Frank gave you a pear in exchange of the apple which Marge gave you. And you gave that pear to Bret in exchange for an apple. All the others exchanges involved apples and/or organges.


A champion weight lifter has a brother that's a doctor, but the same doctor doesn't have a brother. How are they related then?

Answer

The champion weight lifter is a sister of the doctor i.e. they are Brother and Sister

The champion weight lifter is a Woman. Her brother is a doctor. The doctor doesn't have a brother, but a sister.


If today is Wednesday, what is one day before the day, after the day, three days after the day before yesterday?

Answer

Thursday

Start backwards. Today is Wednesday. The day before yesterday is Monday. Three days after Monday is Thursday. The day after Thursday is Friday The day before Friday is Thursday.

Also, note that the first two conditions cancel each other out as one day before the day, one day after the day is the same day. Hence, it can be reduced to "three days after the day before yesterday".


How many months have 29 days in year 2000?

Which month(s)?

Answer

All 12 months have 29 days in year 2000.


You have four 9's and you may use any of the (+, -, /, *) as many times as you like. I want to see a mathematical expression which uses the four 9's to = 100

How many such expressions can you make?

Answer

There are 5 such expressions.

99 + (9/9) = 100

(99/.99) = 100

(9/.9) * (9/.9) = 100

((9*9) + 9)/.9 = 100

(99-9)/.9 = 100


A girl has a certain number of pets. All but two are dogs, all but two are cats and all but two are goats.

How many pets does this girl have?

Answer

The answer is 3 i.e. 1 dog, 1 cat and 1 goat

It says "all but two are dogs", which means that 2 are not dog. Similarly, 2 are not cat and 2 are not goat. Thus solution is there are 3 pets out of which one is dog, one is cat and one is goat.

Also, there is one more aspect to it. The girl might have only 2 pets and none of them is dog or cat or goat.


A boy goes into a shop to buy some sweets. He uses a £1 coin and buys 40p worth of

sweets. The shop assistant gives him two coins for his change.

One of them was not a 50p so how could the boy have gotten his exact change?

Answer

One of them was 10p and the other one was 50p.

The boy got two coins, 50p and 10p. One of them was not a 50p. But the other one was.


Find all sets of consecutive integers that add up to 1000.

Answer

There are total 8 such series: 1. Sum of 2000 numbers starting from -999 i.e. summation of numbers from -999 to

1000. (-999) + (-998) + (-997) + ..... + (-1) + 0 + 1 + 2 + ..... + 997 + 998 + 999 + 1000 = 1000

2. Sum of 400 numbers starting from -197 i.e. summation of numbers from -197 to 202. (-197) + (-196) + (-195) + ..... + (-1) + 0 + 1 + 2 + ..... + 199 + 200 + 201 + 202 = 1000

3. Sum of 125 numbers starting from -54 i.e. summation of numbers from -54 to 70. (-54) + (-53) + (-52) + ..... + (-1) + 0 + 1 + 2 + ..... + 68 + 69 + 70 = 1000

4. Sum of 80 numbers starting from -27 i.e. summation of numbers from -27 to 52. (-27) + (-26) + (-25) + ..... + (-1) + 0 + 1 + 2 + ..... + 50 + 51 + 52 = 1000

5. Sum of 25 numbers starting from 28 i.e. summation of numbers from 28 to 52. 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 = 1000

6. Sum of 16 numbers starting from 55 i.e. summation of numbers from 55 to 70. 55 + 56 + 57 + 58 + 59 +60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 = 1000

7. Sum of 5 numbers starting from 198 i.e. summation of numbers from 198 to 202. 198 + 199 + 200 +201 + 202 = 1000

8. Sum of 1 number starting from 1000. 1000 = 1000


There are 10 statements written on a piece of paper:

1. At least one of statements 9 and 10 is true. 2. This either is the first true or the first false statement. 3. There are three consecutive statements, which are false. 4. The difference between the numbers of the last true and the first true statement divides

the number, that is to be found. 5. The sum of the numbers of the true statements is the number, that is to be found. 6. This is not the last true statement.

7. The number of each true statement divides the number, that is to be found. 8. The number that is to be found is the percentage of true statements. 9. The number of divisors of the number, that is to be found, (apart from 1 and itself) is

greater than the sum of the numbers of the true statements. 10. There are no three consecutive true statements.

Find the minimal possible number?

Answer

The numebr is 420.

If statement 6 is false, it creates a paradox. Hence, Statement 6 must be true.

Consider Statement 2: If it is true, it must be the first true statement. Otherwise, it creates a paradox. If it is false, it must be the second false statement. Otherwise, it creates a paradox.

In both the cases, Statement 1 is false.

As Statement 1 is false, Statement 9 and Statement 10 both are false i.e. there are three consecutive true statements.

1 2 3 4 5 6 7 8 9 10

False - - - - True - - False False

Let\'s assume that Statement 3 is false i.e. there are no three consecutive false statements. It means that Statement 2 and Statement 8 must be true, else there will be three consecutive false statements.

1 2 3 4 5 6 7 8 9 10

False True False - - True - True False False

Also, atleast two of Statements 4, 5 and 7 must be true as there are three consecutive true statements.

According to Statement 8, the number that is to be found is the percentage of true statements. Hence, number is either 50 or 60. Now if Statement 7 is true, then the number of each true statement divides the number, that is to be found. But 7 and 8 do not divide either 50 or 60. Hence, Statement 7 is false which means that Statement 4 and 5 are true. But Statement 5 contradicts the Statement 8. Hence, our assumption that Statement 3 is false is wrong and Statement 3 is true i.e. there are 3 consecutive false statements which means that Statement 8 is false as there is no other possibilities of 3 consecutive false statements.

Also, Statement 7 is true as Statement 6 is not the last true statement.

1 2 3 4 5 6 7 8 9 10

False - True - - True True False False False

According to Statement 7, the number of each true statement divides the number, that is to be found. And according to Statement 5, the sum of the numbers of the true statements is the number, that is to be found. For all possible combinations Statement 5 is false.

There 3 consecutive true statements. Hence, Statement 2 and Statement 4 are true.

1 2 3 4 5 6 7 8 9 10

False True True True False True True False False False

Now, the conditions for the number to be found are: 1. The numebr is divisible by 5 (Statement 4) 2. The numebr is divisible by 2, 3, 4, 6, 7 (Statement 7) 3. The number of divisors of the number, that is to be found, (apart from 1 and itself) is

not greater than the sum of the numbers of the true statements. (Statement 9)

The minimum possible number is 420.

The divisors of 420, apart from 1 and itself are 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210. There are total of 22 divisors. Also, the sum of the numbers of the true statements is 22 (2+3+4+6+7=22), which satisfies the third condition.


Mr. X wants to send Diamond Necklace to his fiancee Y who lives in some other country. Mr. X puts the Diamond Necklace inside a box to parcel. Note that:

1. The parcel must be locked. 2. The box is made up of such a material that nobody can break it. 3. The box has a handle large enough to put a lock on it. 4. Mr. X can not send a key with the parcel. 5. Combination locks and telephones are not available.

How will Mr. X send Diamond Necklace to his fiancee Y?

Answer

Mr. X and his fiancee Y will follow the following steps. 1. Mr. X will put Diamond Necklace inside a box and will lock it. 2. He will send the parcel to his fiancee Y, without the key. 3. Y will put her lock on it and will send box back to Mr. X. 4. Mr. X will remove his lock and will send the box back to Y.

5. Y will remove her lock and will receive the Necklace.


Find the remainder of (2^n)/n

Answer

There is no regular pattern as such, the remainder just depends on the value of n. If n is the power of 2 (i.e 1, 2, 4, 8, ...), the ramainder is 0. If n is the prime number and not the power of 2 (i.e. 3, 5, 7, 11, 13, ...), the ramainder

is 2.

For remaining cases like, even numbers other than the power of 2 and odd numbers other than the prime numbers, there is no regular pattern. If you get such pattern, do let all know !!!


Tic-Tac-Toe is being played. One 'X' has been placed in one of the corners. No 'O' has been placed yet.

Where does the player that is playing 'O' has to put his first 'O' so that 'X' doesn't win?

Assume that both players are very intelligent. Explain your answer.

Answer

"O" should be placed in the center.

Let's number the positions as: 1 | 2 | 3

---------

4 | 5 | 6

---------

7 | 8 | 9It is given that "X" is placed in one of the corner position. Let's assume that its at position 1.

Now, let's take each position one by one. If "O" is placed in position 2, "X" can always win by choosing position 4, 5 or 7. If "O" is placed in position 3, "X" can always win by choosing position 4, 7 or 9. If "O" is placed in position 4, "X" can always win by choosing position 2, 3 or 5. If "O" is placed in position 6, "X" can always win by choosing position 3, 5 or 7. If "O" is placed in position 7, "X" can always win by choosing position 2, 3 or 9. If "O" is placed in position 8, "X" can always win by choosing position 3, 5 or 7. If "O" is placed in position 9, "X" can always win by choosing position 3, or 7.

If "O" is placed in position 5 i.e. center position, "X" can't win unless "O" does something foolish ;))

Hence, "O" should be placed in the center.


What is the remainder left after dividing 1! + 2! + 3! + … + 100! By 7?

Think carefully !!!

Answer

A tricky one.

7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left after dividing 7! + 8! + 9! + ... + 100! is 0.

The only part to be consider is = 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873

The remainder left after dividing 873 by 7 is 5

Hence, the remainder is 5.


If you started a business in which you earned Rs.1 on the first day, Rs.3 on the second day, Rs.5 on the third day, Rs.7 on the fourth day, & so on.

How much would you have earned with this business after 50 years (assuming there are exactly 365 days in every year)?

Answer

Rs.333,062,500

To begin with, you want to know the total number of days: 365 x 50 = 18250.

By experimentation, the following formula can be discovered, & used to determine the amount earned for any particular day: 1 + 2(x-1), with x being the number of the day. Take half of the 18250 days, & pair them up with the other half in the following way: day 1 with day 18250, day 2 with day 18249, & so on, & you will see that if you add these pairs together, they always equal Rs.36500.

Multiply this number by the total number of pairs (9125), & you have the amount you would have earned in 50 years.

Math gurus may use series formula to solve it.(series: 1,3,5,7,9,11.....upto 18250 terms)


Complete the series : 5, 20, 24, 6, 2, 8, ?

Answer

12

Note the sequence of operations: 5 * 4 = 20 20 + 4 = 24 24 / 4 = 6 6 - 4 = 2 2 * 4 = 8

There is arithmatic operation on the current number to get next number in the series. The sequence of operations is *, +, / and - by 4.

So after multiplying by 4, next operation is addition of 4. So the next number is 8 + 4 = 12

215)Which of the following numbers is the odd one out, and why? 1, 2, 3, 5, 9, 13, 21

Note that 2 is not the odd one.

Answer

The odd number is 9.

It is a Fibonacci Series - a series in which next number is summation of previous 2 numbers.

The first two numbers are 1 and 2. Third number is = 1 + 2 = 3 Fourth number is 3 + 5 = 8 Fifth number is 5 + 8 = 13 Sixth number is 8 + 13 = 21

Hence 9 is the odd number It should 8.


Find next two numbers in the series : 100, 365, 24, 60, ?, ?

Answer


The pattern is breaking down the time from Century to miliseconds. There are 100 years in a century. There are 365 days in a year.

There are 24 hours in a day. There are 60 minutes in an hour. There are 60 seconds in a minute. There are 1000 miliseconds in a second.

Hence, the sequence is : 100, 365, 24, 60, 60, 1000


Pinto says, "The horse is not Black." Sandy says, "The horse is either Brown or Grey." Andy says, "The horse is Brown."

At least one is telling truth and at least one is lying.

Can you tell the color of the horse?

Answer

The color of the horse can be any color other than Black and Brown.

If the color of the horse is Black - all are lying.

If the color of the horse is Brown - all are telling truth.

Thus, the horse is neither Black nor Brown.

If the color of the horse is Grey - Pinto and Sandy are telling truth whereas Andy is lying.

If the color of the horse is other than Black, Brown and Grey - Pinto is telling truth whereas Sandy and Andy are lying.

You must have noticed that for the given conditions, Pinto is always telling truth whereas Andy is always lying.


One side of the bottom layer of a triangular pyramid has 12 balls. How many are there in the whole pyramid?

Note that the pyramid is equilateral and solid.

Answer

There are total 364 balls.

As there are 12 balls along one side, it means that there are 12 layers of balls. The top most

layer has 1 ball. The second layer has 3 (1+2) balls. The third layer has 6 (1+2+3) balls. The fourth layer has 10 (1+2+3+4) balls. The fifth layer has 15 (1+2+3+4+5) balls. Similarly, there are 21, 28, 36, 45, 55, 66 and 78 balls in the remaining layers.

Hence, the total number of balls are = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 = 364 balls


A man received a cheque. The rupees has been transposed for paise and vice versa. After spending 5 rupees 42 paise, he discovered that he now had exactly six times the value of the correct cheque amount.

What amount should he have received?

Answer

He received 6 rupees and 44 paise.

Let's assume that he received a cheque of X rupees and Y paise i.e. (100X + Y)

The amount received by him = 100Y + X

After spending 5 rupees 42 paise, the remaining amount is (100Y + X - 542) which is 6 times the original amount. Thus, (100Y + X - 542) = 6*(100X + Y) 100Y + X - 542 = 600X + 6Y 94Y = 599X + 542

Using trial-n-error, we get X=6 and Y=44

Hence, he should have received 6 rupees and 44 paise.


Each of the five characters in the word BRAIN has a different value between 0 and 9. Using the given grid, can you find out the value of each character?

B R A I N 31

B B R B A 31

N I A B B 32

N I B A I 30

I R A A A 23

37 29 25 27 29The numbers on the extreme right represent the sum of the values represented by the characters

in that row. Also, the numbers on the last raw represent the sum of the values represented by the characters in that column. e.g. B + R + A + I + N = 31 (from first row)

Answer

B=7, R=6, A=4, I=5 and N=9

Make total 10 equations - 5 for rows and 5 for columns - and sovle them.

From Row3 and Row4, N + I + A + B + B = N + I + B + A + I + 2 B = I + 2

From Row1 and Row3, B + R + A + I + N = N + I + A + B + B - 1 R = B - 1

From Column2, R + B + I + I + R = 29 B + 2R + 2I = 29 B + 2(B - 1) + 2I = 29 3B + 2I = 31 3(I + 2) + 2I = 31 5I = 25 I = 5

Hence, B=7 and R=6

From Row2, B + B + R + B + A = 31 3B + R + A = 31 3(7) + 6 + A = 31 A = 4

From Row1, B + R + A + I + N = 31 7 + 6 + 4 + 5 + N = 31 N = 9

Thus, B=7, R=6, A=4, I=5 and N=9


Subhash is 14 inches taller than Jatin. The difference between Subhash and Sanjeev is two inches less than between Sanjeev and Jatin. Subhash at 6'6" is the tallest.

How tall are Sanjeev and Jatin?

Answer

Sanjeev is 6' and Jatin is 5'4"

It is given that Subhash at 6'6" is the tallest and also he is 14 inches taller than Jatin. It means that Jatin is 5'4".

Now as the difference between Subhash and Sanjeev is two inches less than between Sanjeev and Jatin, Sanjeev's height is more than Jatin's. And that is 6'.

Thus, Sanjeev is 6' and Jatin is 5'4".


Consider the sum: ABC + DEF + GHI = JJJ

If different letters represent different digits, and there are no leading zeros, what does J represent?

Answer

The value of J must be 9.

Since there are no leading zeros, J must be 7, 8, or 9. (JJJ = ABC + DEF + GHI = 14? + 25? + 36? = 7??)

Now, the remainder left after dividing any number by 9 is the same as the remainder left after dividing the sum of the digits of that number by 9. Also, note that 0 + 1 + ... + 9 has a remainder of 0 after dividing by 9 and JJJ has a remainder of 0, 3, or 6.

The number 9 is the only number from 7, 8 and 9 that leaves a remainder of 0, 3, or 6 if you remove it from the sum 0 + 1 + ... + 9. Hence, it follows that J must be 9.


In the village called TALAJA, only three TV channels are available - Moon Plus, Mony and Mee TV.

Out of 4000 TV viewers in the village, 1500 watch Moon TV, 2000 watch Mony and 2500 watch Mee TV.

Amongst these, 500 viewers watch Moon Plus and Mony, 800 watch Moon Plus and Mee TV, and 1000 watch Mony and Mee TV.

How many viewers watch all three channels?

Answer

300 viewers watch all three channels.

Let's assume that total X viewers watch all three channels.

total viewers who watch only Moon Plus and Mony = 500-X total viewers who watch only Moon Plus and Mee TV = 800-X total viewers who watch only Mony and Mee TV = 1000-X

total viewers who watch only Moon Plus = 1500 - (500-X) - (800-X) - X = 200 + X

total viewers who watch only Mony = 2000 - (500-X) - (1000-X) - X = 500 + X

total viewers who watch only Mee TV = 2500 - (1000-X) - (800-X) - X = 700 + X

We know that total viewers are 4000. Summing up all 7 values, X + (500 - X) + (800 - X) + (1000 - X) + (200 + X) + (500 + X) + (700 + X) = 4000 X + 3700 = 4000 X = 300

Hence, total 300 viewers watch all three channels.


Replace each letter by a digit. Each letter must be represented by the same digit and no beginning letter of a word can be 0.

O N E

O N E

O N E

+ O N E

-------

T E N

Answer

Use trial and error. 0 =1, N = 8 ,E = 2, T = 7

1 8 2

1 8 2

1 8 2

+ 1 8 2

------

7 2 8


Insert mathematical functions to convert the 3 numbers on the left side of the equation to equal 6. I filled in the 2's (using the addition function twice) for you to get you started.

1 1 1 = 6

2 + 2 + 2 = 6

3 3 3 = 6

4 4 4 = 6

5 5 5 = 6

6 6 6 = 6

7 7 7 = 6

8 8 8 = 6

9 9 9 = 6

Answer

(1 + 1 + 1)! = 6

2 + 2 + 2 = 6

(3 * 3) - 3 = 6

sqrt(4) + sqrt(4) + sqrt(4) = 6

5 + (5 / 5) = 6

6 + 6 - 6 = 6

7 - (7 / 7) = 6

8 - sqrt(sqrt(8+8)) = 6

(sqrt(9) * sqrt(9)) - sqrt(9) = 6


Five horses ran in the race.

There were no ties. Sikandar did not come first. Star was neither first nor last. Mughal Glory came in one place after Sikandar. Zozo was not second. Rangila was two place below Zozo.

In what order did the horses finish?

Answer

It's simple.

Let's find the possible places horses can finish. Possibilities are: Sikandar - 2,3,4 (not 5th as Mughal Glory came one place after him) Star - 2,3,4 Mughal Glory - 3,4,5 Zozo - 1,3 (not 4th & 5th as Rangila is two place after him) Rangila - 3,5

So the result is: 1 Zozo 2 Star 3 Rangila 4 Sikandar 5 Mughal Glory


Assume that you have just heard of a scandal and you are the first one to know. You pass it on to four person in a matter of 30 minutes. Each of these four in turn passes it to four other persons in the next 30 minutes and so on.

How long it will take for everybody in the World to get to know the scandal?

Assume that nobody hears it more than once and the population of the World is approximately 5.6 billions.

Answer

Everybody in the World will get to know the scandal in 8 hours.

You came to know of a scandal and you passed it on to 4 persons in 30 minutes. So total (1+4) 5 persons would know about it in 30 minutes.

By the end of one hour, 16 more persons would know about it. So total of (1+4+16) 21 persons

would know about it in one hour.

Similarly, the other (1+4+16+64) persons would have know about it in one and a half hours. (1+4+16+64+256) persons would have know about it in two hours and so on...

It can be deduced that the terms of the above series are the power of 4 i.e. 4^0, 4^1, 4^2, 4^3 and so on upto (2N+1) terms. Also, the last term would be 4^2N where N is the number of hours.

Sum of the above mentioned series = [4^(2N+1)-1]/3

The sum of the series must be 5.6 billions. Hence, equating the sum of the series with 5.6 billions, we get N=8 hours.

Scandals travel FAST !!!


Six cabins numbered 1-6 consecutively, are arranged in a row and are separated by thin dividers. These cabins must be assigned to six staff members based on following facts.

1. Miss Shalaka's work requires her to speak on the phone frequently throughout the day. 2. Miss Shudha prefers cabin number 5 as 5 is her lucky number. 3. Mr. Shaan and Mr. Sharma often talk to each other during their work and prefers to have

adjacent cabins. 4. Mr. Sinha, Mr. Shaan and Mr. Solanki all smoke. Miss Shudha is allergic to smoke and

must have non-smokers adjacent to her. 5. Mr. Solanki needs silence during work.

Can you tell the cabin numbers of each of them?

Answer

The cabins from left to right (1-6) are of Mr. Solanki, Mr. Sinha, Mr. Shaan, Mr. Sharma, Miss Shudha and Miss Shalaka.

From (2), cabin number 5 is assigned to Miss Shudha.

As Miss Shudha is allergic to smoke and Mr. Sinha, Mr. Shaan & Mr. Solanki all smoke, they must be in cabin numbers 1, 2 and 3 not necessarily in the same order. Also, Miss Shalaka and Mr. Sharma must be in cabin 4 and 6.

From (3), Mr. Shaan must be in cabin 3 and Mr. Sharma must be in cabin 4. Thus, Miss Shalaka is in cabin 6.

As Mr. Solanki needs silence during work and Mr. Shaan is in cabin 3 who often talks to Mr. Sharma during work, Mr. Solanki must be in cabin 1. Hence, Mr. Sinha is in cabin 2.

Thus, the cabins numbers are 1# Mr. Solanki, 2# Mr. Sinha,

3# Mr. Shaan, 4# Mr. Sharma, 5# Miss Shudha, 6# Miss Shalaka


In the middle of the confounded desert, there is the lost city of "Ash". To reach it, I will have to travel overland by foot from the coast. On a trek like this, each person can only carry enough rations for five days and the farthest we can travel in one day is 30 miles. Also, the city is 120 miles from the starting point.

What I am trying to figure out is the fewest number of persons, including myself, that I will need in our Group so that I can reach the city, stay overnight, and then return to the coast without running out of supplies.

How many persons (including myself) will I need to accomplish this mission?

Answer

Total 4 persons (including you) required.

It is given that each person can only carry enough rations for five days. And there are 4 persons. Hence, total of 20 days rations is available.

1. First Day : 4 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 15 days.

2. Second Day : The remaining three people use up 3 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 10 days.

3. Third Day : The remaining two people use up 2 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 5 days.

4. Fourth Day : The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 4 days of rations.

Thus, total 4 persons, including you are required.


Substitute digits for the letters to make the following relation true.

N E V E R

L E A V E

+ M E

-----------------

A L O N ENote that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter M, no other letter can be 3 and all other M in the puzzle must be 3.

Answer

A tough one!!!

Since R + E + E = 10 + E, it is clear that R + E = 10 and neither R nor E is equal to 0 or 5. This is the only entry point to

solve it. Now use trial-n-error method.

N E V E R 2 1 4 1 9

L E A V E 3 1 5 4 1

+ M E + 6 1

----------------- -----------------

A L O N E 5 3 0 2 1


For a TV talk show on Bollywood, the producer must choose a group of two Directors and two Musicians. At least one of them must be an Actor and at least one a Singer.

1. Available Directors are Mahesh Bhatt, Karan Johar, Subhash Ghai, Aditya Chopra and Ashutosh.

2. Available Musicians are A R Rehman, Annu Malik, Sandeep Chowta and Aadesh Srivastava.

3. Shubash Ghai, A R Rehman and Annu Malik are Actors. 4. Aditya Chopra and Aadesh Srivastava are Singers. 5. A R Rehman will not seat in the same room with Subhash Ghai, and will take part only if

Mahesh Bhatt is there. 6. Aditya Chopra refuses to take part with Annu Malik. 7. Ashutosh refuses to take part with Aadesh Srivastava.

How many acceptable groups can the producer put together?

Answer

The producer can put 9 acceptable groups together.

Your basic solution step is to make a table of the possible persons and qualifications on the Musicians side, with the possible combinations on the Directors side.

Note that A R Rehman insists on Mahesh Bhatt being present, but the reverse is not true. Also, remember that there must be at least one actor and at least one singer. The valid combinations are:

Musician Director

----------------------------------------------------------

(A R Rehman, Sandeep Chowta) (Mahesh Bhatt, Aditya Chopra)

(A R Rehman, (Mahesh Bhatt, Karan Johar)

Aadesh Srivastava) (Mahesh Bhatt, Aditya Chopra)

(Annu Malik, (Mahesh Bhatt, Karan Johar)

Aadesh Srivastava) (Mahesh Bhatt, Subhash Ghai)

(Karan Johar, Subhash Ghai)

(Sandeep Chowta, (Mahesh Bhatt, Subhash Ghai)

Aadesh Srivastava) (Karan Johar, Subhash Ghai)

(Aditya Chopra, Subhash Ghai)Thus, there are total 9 acceptable groups.


Mr. Subramaniam rents a private car for Andheri-Colaba-Andheri trip. It costs him Rs. 300 everyday.

One day the car driver informed Mr. Subramaniam that there were two students from Bandra who wished to go from Bandra to Colaba and back to Bandra. Bandra is halfway between Andheri and Colaba. Mr. Subramaniam asked the driver to let the students travel with him.

On the first day when they came, Mr. Subramaniam said, "If you tell me the mathematically correct price you should pay individually for your portion of the trip, I will let you travel for free."

How much should the individual student pay for their journey?

Answer

The individual student should pay Rs. 50 for their journey.

Note that 3 persons are travelling between Bandra and Colaba.

The entire trip costs Rs. 300 to Mr. Subramanian. Hence, half of the trip costs Rs. 150.

For Andheri-Bandra-Andheri, only one person i.e. Mr. Subramaniam is travelling. Hence, he would pay Rs. 150.

For Bandra-Colaba-Bandra, three persons i.e Mr. Subramaniam and two students, are travelling. Hence, each student would pay Rs. 50.


Mr. Wagle goes to work by a bus. One day he falls asleep when the bus still has twice as far to go as it has already gone.

Halfway through the trip he wakes up as the bus bounces over some bad potholes. When he finally falls asleep again, the bus still has half the distance to go that it has already travelled. Fortunately, Mr. Wagle wakes up at the end of his trip.

What portion of the total trip did Mr. Wagle sleep?

Answer

Mr. wagle slept through half his trip.

Let's draw a timeline. Picture the bus route on a line showen below:

---------------- ________ -------- ________________

Start 1/3 1/2 2/3 End

----- shows time for which Mr. Wagle was not sleeping

_____ shows time for which Mr. Wagle was sleeping

When Mr. Wagle fell asleep the first time, the bus sill had twice as far to go as it had already gone, that marks the first third of his trip.

He wake up halfway through the trip i.e slept from 1/3 mark to the 1/2 mark. He fell sleep again when the bus still had half the distance to go that it had already traveled i.e 2/3 mark.

Adding up, all sleeping times, = (1/2 - 1/3) + (1 - 2/3) = 1/6 + 1/3 = 1/2

Hence, Mr. wagle slept through half his trip.


Substitute digits for the letters to make the following addition problem true.

I

A G R E E

+ I T S

-------------------

T O U G HNote that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3.

Answer

I=7, A=2, G=9, R=4, E=5, T=3, S=6, O=0, G=9, H=8

It is obvious that T=A+1. Also, G=9, O=0 and R+I>10. Hence, T>1

There must be a carry from the units. Hence E+T=8. So (E, T) can be (6, 2), (5, 3), (3, 5), (2, 6) or (1, 7).

Now, use trial-n-error and solve it.

I 7

A G R E E 2 9 4 5 5

+ I T S + 7 3 6

------------------- -------------------

T O U G H 3 0 1 9 8


In the town called Alibaug, the following facts are true:

No two inhabitants have exactly the same number of hairs. No inhabitants has exactly 2025 hairs. There are more inhabitants than there are hairs on the head of any one inhabitants.

What is the largest possible number of the inhabitants of Alibaug?

Answer

2025

It is given that no inhabitants have exactly 2025 hairs. Hence there are 2025 inhabitants with 0 to 2024 hairs in the head.

Suppose there are more than 2025 inhabitants. But these will violate the condition that "There are more inhabitants than there are hairs on the head of any one inhabitants." As for any number more than 2025, there will be same number of inhabitants as the maximum number of hairs on the head of any inhabitant.


All of the students at a college are majoring in psychology, business, or both. 73% of the students are psychology majors, & 62% are business majors.

If there are 200 students, how many of them are majoring in both psychology & business?

Answer

70 students are majoring in both, psychology & business

If 73% of the students are psychology majors, we know that 27% are not psychology majors. By the same reasoning, 38% are not business majors, because 62% of the students do major in business. So: 27 + 38 = 65

65% of the students are not majoring in both psychology & business, so 35% are double majors, a total of 70 students.


If A+B=C, D-C=A and E-B=C, then what does D+F stands for? Provide your answer in letter terms as well as in number terms.

Answer

J or 10

A simple one.

Assume that each character represents the number equivalent to the position in the alphabet i.e. A = 1, B = 2, C = 3, D = 4 and so on. Now let's check our assumption.

A + B = C i.e. 1 + 2 = 3 D - C = A i.e. 4 - 3 = 1 E - B = C i.e. 5 - 2 = 3

Thus, our assumption was Correct. Hence, D + F = J i.e. 4 + 6 = 10


There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a perfectly straight floor i.e. the wall and the floor make a perfect right angle.

Can a perfect sphere of diameter 7 cms. pass through the space between the big sphere, the wall and the floor? Support your answer with valid arguments. Don't submit just "Yes" or "No".

Answer

For the sake of simplicity, consider two-dimension i.e view sphere as a two dimensional circle with diameter 40 cms.

From Figure I, (40 cms diameter sphere) OC2 = OD2 + CD2 OC2 = 202 + 202 OC = 28.28427 cms

Also, X is the closest point to origin O on the sphere. CX = 20 cms (radius) OX = OC - CX OX = 28.28427 - 20 OX = 8.28427 cms

From Figure II, (7 cms diameter sphere) OP2 = OQ2 + PQ2 OP2 = (3.5)2 + (3.5)2 OP = 4.94974 cms

Also, Y is the farthest point to origin O on the sphere. PY = 3.5 cms (radius) OY = OP + PY OY = 4.94974 + 3.5 OY = 8.44974 cms

Now, as OY > OX i.e. smaller sphere requires more space than the space available. Hence,

smaller sphere of 7 cms diameter can not pass through the space between the big sphere, the wall and the floor.

The puzzle can be solved by another method. Draw a line tangent to the big sphere at the point X such that X is the closest point to the origin O on sphere. The tanget will cut X and Y axes at A and B respectively such that OA=OB. [See Fig III] From above, OX=8.28427 cms.

From the right angle triangle OAB, we can deduct that OA = OB = 11.71572 cms AB = 16.56854 cms

Now, the diameter of the inscribed circle of right angle triangle is given by d = a + b - c where a <= b < c

The maximum possible diameter of the circle which can pass through the space between the big sphere, the wall and the floor is = OA + OB - AB = 11.71572 + 11.71572 - 16.56854 = 6.86291 cms

Hence, the sphere with 7 cms diameter can not pass through the space between the big sphere, the wall and the floor.


Five students - Akash, Chintan, Jignesh, Mukund and Venky - appeared for an exam. There were total five questions - two multiple choice (a, b or c) and three true/false questions. They answered five questions each and answered as follow.

I II III IV V

--------------------------------------------------

Chintan c b True True False

Akash c c True True True

Jignesh a c False True True

Mukund b a True True False

Venky b b True False True

--------------------------------------------------Also, no two students got the same number of correct answers.

Can you tell which are the correct answers? What are their individual score?

Answer

The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0),

Akash (1), Chintan (2), Venky (3) and Mukund (4).

As no two students got the same number of correct answers, the total number of correct answers must be either 15 (1+2+3+4+5) or 10 (0+1+2+3+4).

Let's find out the maximum number of correct answers possible from the answers given by them. For Question I = 2 (b or c) For Question II = 2 (b or c) For Question III = 4 (True) For Question IV = 4 (True) For Question V = 3 (True)

Thus, the maximum number of correct answers possible are 15 (2+2+4+4+3) which means that Akash would have given all correct answers as only he answered True for questions III, IV and V. But then Chintan and Jignesh would have exactly 3 correct answers. And also, Mukund and Venky would have 2 correct answers. So no one got all five correct. One can also arrive at this conclusion by trial-and-error, but that would be bit lengthy.

Now, it is clear that total number of correct answers are 10 (0+1+2+3+4). Questions III and IV both can not be False. If so, total number of correct answers would not be 10. So the student who got all wrong can not be Chintan, Akash and Mukund.

If Venky got all wrong, then Chintan, Jignesh and Mukund each would have atleast 2 correct answers. It means that Akash would have to be the student with only one correct answer and the correct answers for questions I and II would be a and a respectively. But then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus, Jignesh is the student with all wrong answers. The correct answers are b, a, True, False and False. Also, the scores are Jignesh (0), Akash (1), Chintan (2), Venky (3) and Mukund (4).


Adam, Burzin, Clark and Edmund each live in an apartment. Their apartments are arranged in a row numbered 1 to 4 from left to right. Also, one of them is the landlord.

1. If Clark's apartment is not next to Burzin's apartment, then the landlord is Adam and lives in apartment 1.

2. If Adam's apartment is right of Clark's apartment, then the landlord is Edmund and lives in apartment 4.

3. If Burzin's apartment is not next to Edmund's apartment, then the landlord is Clark and lives in apartment 3.

4. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin and lives in apartment 2.

Who is the landlord?

Answer

Clark is the landlord.

Assume each statement true, one at a time and see that no other statement is contradicted.

Let's assume that Statement (1) is true. Then, Adam is the landlord and lives in apartment 1. Also, other three's apartments will be on the right of his apartment - which contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Adam is not the landlord.

Let's assume that Statement (2) is true. Then, Edmund is the landlord and lives in apartment 4. Also, other three's apartments will be on the left of his apartment - which again contradicts Statement (4) i.e. If Edmund's apartment is right of Adam's apartment, then the landlord is Burzin. Thus, Edmund is not the landlord either.

Let's assume that Statement (3) is true. Then, Clark is the landlord and lives in apartment 3. It satisfies all the statements for (1) Adam - (2) Edmund - (3) Clark - (4) Burzin

Hence, Clark is the landlord.

Similarly, you can assume Statement (4) true and find out that it also contradicts.


The average scores of the class for exam are as follow: Average score of the boys = 90 Average score of the girls = 81 Average score of the class = 84

Find whether Class contains more Boys or Girls?

Answer

Assume that there are B boys and G girls in the Class.

Hence from the given data :

90*B + 81*G = 84*(B+G)

90B + 81G = 84B + 84G

6B = 3G

2B = G

Hence number of Girls in the Class are twice the number of Boys.


What are the next three numbers in the given series?

1 2 3 2 1 2 3 4 2 1 2 3 4 3 2 ? ? ?

Answer

The next three numbers are 3, 4 and 5.

The pattern is - the number of letters in the Roman numeral representation of the numbers i.e. number of letters in I, II, III, IV, V, VI, VII, VIII, IX, X, XI, XII, XIII, XIV, XV, .....

Hence, the next numbers in the given series are 3(XVI), 4(XVII), 5(XVIII), 3(XIX), 2(XX), 3(XXI), 4(XXII), 5(XXIII), 4(XXIV), 3(XXV), etc...


Three convicts are brought into the warden's office. He says he can parole on of them and to decide which one he will parole he takes out 5 hats (3 red and 2 white). He stands behind them and places a hat on each one of their heads and puts the other two remaining hats in a drawer.

He tells the prisioners they can look at the others hats and if they can tell which hat they have on they will be the one who is paroled.

The first man looks at the other two and says, "I don't know."

The second man looks at the others hats and says, "I don't know."

The third man who is blind says, "Even though I have not the gift of sight I can tell by what the others have said that the color of my hat is..."

What color is the blind mans hat and how does he know?

Answer

The color of blind man's hat is Red.

It is sure that the first man saw either both Red hats or one White hat and one Red hat. There are 6 such possibilities: 1) R R R

2) R R W

3) R W R

4) W R R

5) W R W

6) W W RIn all above possibilities, the first man won't be sure of the color of his hat.

Now, the second man knows that the first man saw either both Red hats or one White hat and

one Red hat. And, he also knows that its one of the above 6 possibilities. (like we know ;)) But he says, "I don't know". That means that (2) and (5) are not the possibilities as in either case he would be sure of the color of his hat (Red) by just looking at the third man's color of hat (White).

Now, the blind man knows that there are just 4 possibilities - (1), (3), (4), (6) - and in all, the color of his hat is Red.


Professors Ahmad and Joshi are extremely strange persons.

Prof. Ahmad lies on Mondays, Tuesdays and Wednesdays, but tells true on other days of the week.

Prof. Joshi lies on Thursdays, Fridays and Saturdays, but tells true on other days of the week.

They made the following statements: Prof. Ahmad : "Yesterday was one of my lying days." Prof. Joshi : "Yesterday was one of my lying days too." What day of the week was it?

Both Professors looked very alike and one day they said to a visitor to their department : First Prof: "I'm Ahmed." Second Prof: "I'm Joshi." Who was who? What day of the week was it?

On another occasion, both Professors made the following statements: First Prof : 1. "I lie on Saturdays." 2. "I lie on Sundays." Second Prof. : "I will lie tomorrow." What day of the week was it?

Answer

Mon Prof. AhmadLies

Prof. Joshitells truth

Tue

Lies Tells truth

Wed Lies Tells truth

Thu Tells truth Lies

Fri Tells truth Lies

Sat Tells truth Lies

Sun Tells truth Tells truth

Teaser 1 : Assume that Prof. Ahmad is telling truth => today is Thursday Assume that Prof. Ahmad is lying => today is Monday Similarly, Assume Prof. Joshi is telling truth => today is Sunday Assume that Prof. Joshi is lying => today is Thrusday. Hence, today is Thrusday, Prof. Ahmad is telling truth and Prof. Joshi is lying.

Teaser 2 : Assume that First Prof. is telling truth => Thursday, Friday, Saturday or Sunday Assume that First Prof. is lying => Thursday, Friday or Saturday Similarly, Assume Second Prof. is telling truth => Monday, Tuesday, Wednesday or Sunday Assume that Second Prof. is lying => Monday, Tuesday, Wednesday The only possibility is Sunday and both are telling truth.

Teaser 3 : A simple one. First Prof. says - "I lie on Sunday" which is false as both the Prof. tell truth on sunday. It means the first statement made by the First Prof. is also false. It means the First Prof. tells truth on Saturday. Hence First Prof. is Prof. Ahmad and he is lying. It means that today is either Monday, Tuesday or Wednesday.

It is clear that Second Prof. is Prof. Joshi. Assume that he is telling truth => today is Wednesday Assume that he is lying => today is Saturday.

Hence, today is Wednesday !!!


A man is going to an Antique Car auction. All purchases must be paid for in cash. He goes to the bank and draws out $25,000.

Since the man does not want to be seen carrying that much money, he places it in 15 evelopes numbered 1 through 15. Each envelope contains the least number of bills possible of any available US currency (i.e. no two tens in place of a twenty).

At the auction he makes a successful bid of $8322 for a car. He hands the auctioneer envelopes number(s) 2, 8, and 14. After opening the envelopes the auctioneer finds exactly the right amount.

How many ones did the auctioneer find in the envelopes?

Answer

Each envelope contains the money equal to the 2 raised to the envelope number minus 1. The sentence "Each envelope contains the least number of bills possible of any available US currency" is only to misguide you. This is always possible for any amount !!!

One more thing to notice here is that the man must have placed money in envelopes in such a way that if he bids for any amount less than $25000, he should be able to pick them in terms of envelopes.

First envelope contains, 20 = $1 Second envelope contains, 21 = $2 Third envelope contains, 22 = $4 Fourth envelope contains, 23 = $8 and so on...

Hence the amount in envelopes are $1, $2, $4, $8, $16, $32, $64, $128, $256, $512, $1024, $2048, $4096, $8192, $8617

Last envelope (No. 15) contains only $8617 as total amount is only $25000.

Now as he bids for $8322 and gives envelope number 2, 8 and 14 which contains $2, $128 and $8192 respectively.

Envelope No 2 conrains one $2 bill Envelope No 8 conrains one $100 bill, one $20 bill, one $5 bill, one $2 bill and one $1 bill Envelope No 14 conrains eighty-one $100 bill, one $50 bill, four $10 bill and one $2 bill

Hence the auctioneer will find one $1 bill in the envelopes


Veeru says to Jay, "Can you figure out how many Eggs I have in my bucket?" He gives 3 clues to Jay: If the number of Eggs I have

1. is a multiple of 5, it is a number between 1 and 19 2. is not a multiple of 8, it is a number between 20 and 29 3. is not a multiple of 10, it is a number between 30 and 39

How many Eggs does Veeru have in his bucket?

Answer

32 eggs

Let's apply all 3 condition separately and put all possible numbers together.

First condition says that if multiple of 5, then the number is between 1 and 19. Hence, the possible numbers are (5, 10, 15, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39)

Second condition says that if not a multiple of 8, then the number is between 20 and 29. Hence, the possible numbers are (8, 16, 20, 21, 22, 23, 25, 26, 27, 28, 29, 32)

Third condition says that if not a multiple of 10, then the number is between 30 and 39. Hence,

the possible numbers are (10, 20, 31, 32, 33, 34, 35, 36, 37, 38, 39)

Only number 32 is there in all 3 result sets. That means that only number 32 satisfies all three conditions. Hence, Veeru have 32 eggs in his bucket.


Five friends with surname Batliwala, Pocketwala, Talawala, Chunawala and Natakwala have their first name and middle name as follow.

1. Four of them have a first and middle name of Paresh. 2. Three of them have a first and middle name of Kamlesh. 3. Two of them have a first and middle name of Naresh. 4. One of them have a first and middle name of Elesh. 5. Pocketwala and Talawala, either both are named Kamlesh or neither is named Kamlesh. 6. Either Batliwala and Pocketwala both are named Naresh or Talawala and Chunawala

both are named Naresh. 7. Chunawala and Natakwala are not both named Paresh.

Who is named Elesh?

Answer

Pocketwala is named Elesh.

From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named Paresh.

From (6) and (5), if Pocketwala or Talawala both are named Kamlesh, then either of them will have three names i.e. Paresh, Kamlesh and Naresh. Hence, Pocketwala and Talawala both are not named Kamlesh. It means that Batliwala, Chunawala and Natakwala are named Kamlesh.

Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala is named Elesh.


Jim lies a lot. He tells the truth on only one day in a week.

One day he said: "I lie on Mondays and Tuesdays." The next day he said: "Today is either Sunday, Saturday or Thursday." The next day he said: "I lie on Fridays and Wednesdays."

On which day of the week does Jim tell the truth?

Answer

Jim tells the truth on Tuesday.

As Jim tells truth only on one day in a week, his statement on day 1 and day 3 both can not be false. Otherwise he tells truth on more than one days in a week. Also, all three statements are mad on three consecutive days, statement made on day 1 and day 3 both can not be true. Thus, either the statement made on day 1 or day 3 is true and other is false. Also, the statement made

on day 2 must be false i.e. day 1 is not Saturday, Friday or Wednesday.

Let's assume that the statement 1 is true. Then from the statement 3, day 1 must be either Friday or Wednesday. But it is already deduced that day 1 is not Saturday, Friday or Wednesday.

Hence, the statement made on day 1 is false and the last statement is true. then from the statement 1, day 3 must be either Monday or Tuesday. But it is already deduced that day 1 can not be Saturday i.e. day 3 can't be Monday. Hence, Jim tells the truth on Tuesday.


A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine, reverses itself immediately, and flies back to the other engine, repeating the process each time.

The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?

Answer

Initially, the trains were 240 km apart.

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.

It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of 60 km/h. So, each train traveled 120 km before they met.

Hence, the trains were 240 km apart initially.


You are stuck in a room with 2 doors. One door is the way out and the other one leads to Hell. In this room there are two computers, one always lies and one always tells the truth.

How do you get out safely? Once you open a door, you must go through it and can never turn back. You can ask only one question to any one of the computer.

Answer

Ask one of the computers "Would the other computer say that this door leads to Hell?" If he answers yes enter door, if he answers no use other door.

If you ask the computer who lies, it will tell you YES, if the door is the way out and NO, if the door leads to Hell. Similarly for the other computer who always tells truth.

Alternatively, you can ask one computer, "Which door would the other machine tell me is the

safe way out?" and whatever the answer, you take the other door


Dr. DoLittle always goes walking to the clinic and takes the same time while going and while coming back. One day he noticed something.

When he left the home, the hour hand and the minute hand were exactly opposite to each other and when he reached the clinic, they were together.

Similarly, when he left the clinic, the hour hand and the minute hand were together and when he reached the home, they were exactly opposite to each other.

How much time does Dr. DoLittle take to reach home from the clinic? Give the minimal possible answer.

Answer

32 minutes 43.6 seconds

In twelve hours, the minute hand and the hour hand are together for 11 times. It means that after every 12/11 hours, both the hands are together.

Similarly in twelve hours, the minute hand and the hour hand are exactly opposite to each other for 11 times. It means that after every 12/11 hours, both the hands are opposite.

Now, let's take an example. We know that at 12 both the hands are together and at 6 both the hands are exactly opposite to each other.

After 6, both the hands are in opposition at [6+(12/11)] hours, [6+2*(12/11)] hours, [6+3*(12/11)] hours and so on. The sixth such time is [6+6*(12/11)] hours which is the first time after 12. Thus after 12, both the hands are opposite to each other at 12:32:43.6

Hence, Dr. DoLittle takes 32 minutes and 43.6 seconds to reach home from the clinic.


Two trains starting at same time, one from Bangalore to Mysore and other in opposite direction arrive at their destination 1hr and 4hrs respectively after passing each other.

How much faster is one train from other?

Answer

The speed of Bangalore-Mysore train is TWICE the speed of Mysore-Bangalore train.

Let the distance between Bangalore and Mysore is D kms. Also, let speed of the train from Bangalore to Mysore is X km/hr and speed of the tain from Mysore to Bangalore is Y km/hr.

Now, assume that both the trains met each other at T kms from the Bangalore (point P in figure) Time taken by Bangalore-Mysore train to reach P = Time taken by Mysore-Bangalore train to reach P ( T / X ) = ( D - T ) / Y -----equ(I)

Also, Bangalore-Mysore train and Mysore-Bangalore train arrive destination 1 hr and 4 hrs respectively after passing each other. It means that Bangalore-Mysore train travels (D - T) kms in 1 hr at X km/hr and Mysore-Bangalore train travels T kms in 4 hrs at Y km/hr. Hence, ( D - T ) = X and T = 4 * Y

Substituting these values in equation I, we get ( 4 * Y ) / X = X / Y 4 * Y * Y = X * X 2 * Y = X

Hence, the speed of Bangalore-Mysore train is TWICE the speed of Mysore-Bangalore train.


Sita says to Gita, "I have 3 sons. They are all less than 10 years in age and all are of different age. The product of the ages of the two younger brothers is equal to the age of the eldest brother. Also, the sum of their ages is a Prime number."

How old is the eldest son?

Answer

The eldest son is 6 years old.

First list out all combinations of three numbers such that all of them are less than 10 and product of first two is third number. There are only 3 combination.

2 2 4 2 3 6 2 4 8

3 3 9

Also, note that age of the younger brother can not be 1 year. As in that case other two will be twins and it is mentioned in teaser that "the eldest brother" i.e. there is only one brother with the eldest age.

Now check fo the second condition i.e. the sum of their ages is a Prime number.

2 + 2 + 4 = 8 2 + 3 + 6 = 11 2 + 4 + 8 = 14 3 + 3 + 9 = 15

11 is the Prime number. Hence, the required combination is 2, 3 and 6. The eldest son is 6 years old.


In the following multiplication, certain digits have been replaced with asterisks (*). Replace all the asterisks such that the problem holds the result.

* * 7

X 3 * *

----------

* 0 * 3

* 1 *

* 5 *

-------------

* 7 * * 3

Answer

A simple one. 1 1 7

X 3 1 9

----------

1 0 5 3

1 1 7

3 5 1

-------------

3 7 3 2 3


Four friends - Arjan, Bhuvan, Guran and Lakha were comparing the number of sheep that they owned.

It was found that Guran had ten more sheep than Lakha.

If Arjan gave one-third to Bhuvan, and Bhuvan gave a quarter of what he then held to Guran, who then passed on a fifth of his holding to Lakha, they would all have an equal number of sheep.

How many sheep did each of them possess? Give the minimal possible answer.

Answer

Arjan, Bhuvan, Guran and Lakha had 90, 50, 55 and 45 sheep respectively.

Assume that Arjan, Bhuvan, Guran and Lakha had A, B, G and L sheep respectively. As it is given that at the end each would have an equal number of sheep, comparing the final numbers from the above table.

Arjan's sheep = Bhuvan's sheep 2A/3 = A/4 + 3B/4 8A = 3A + 9B 5A = 9B

Arjan's sheep = Guran's sheep 2A/3 = A/15 + B/5 + 4G/5 2A/3 = A/15 + A/9 + 4G/5 (as B=5A/9)

30A = 3A + 5A + 36G 22A = 36G 11A = 18G

Arjan's sheep = Lakha's sheep 2A/3 = A/60 + B/20 + G/5 + L 2A/3 = A/60 + A/36 + 11A/90 + L (as B=5A/9 and G=11A/18) 2A/3 = A/6 + L A/2 = L A = 2L

Also, it is given that Guran had ten more sheep than Lakha. G = L + 10 11A/18 = A/2 + 10 A/9 = 10 A = 90 sheep

Thus, Arjan had 90 sheep, Bhuvan had 5A/9 i.e. 50 sheep, Guran had 11A/18 i.e. 55 sheep and Lakha had A/2 i.e. 45 sheep.


Vipul was studying for his examinations and the lights went off. It was around 1:00 AM. He lighted two uniform candles of equal length but one thicker than the other. The thick candle is supposed to last six hours and the thin one two hours less. When he finally went to sleep, the thick candle was twice as long as the thin one.

For how long did Vipul study in candle light?

Answer

Vipul studied for 3 hours in candle light.

Assume that the initial lenght of both the candle was L and Vipul studied for X hours.

In X hours, total thick candle burnt = XL/6 In X hours, total thin candle burnt = XL/4

After X hours, total thick candle remaining = L - XL/6 After X hours, total thin candle remaining = L - XL/4

Also, it is given that the thick candle was twice as long as the thin one when he finally went to sleep. (L - XL/6) = 2(L - XL/4) (6 - X)/6 = (4 - X)/2 (6 - X) = 3*(4 - X) 6 - X = 12 - 3X 2X = 6 X = 3

Hence, Vipul studied for 3 hours i.e. 180 minutes in candle light.


My friend collects antique stamps. She purchased two, but found that she needed to raise money urgently. So she sold them for Rs. 8000 each. On one she made 20% and on the other she lost 20%.

How much did she gain or lose in the entire transaction?

Answer

She lost Rs 666.67

Consider the first stamp. She mades 20% on it after selling it for Rs 8000.

So the original price of first stamp is = (8000 * 100) / 80 Rs 6666.67

Similarly, consider second stamp. She lost 20% on it after selling it for Rs 8000

So the original price of second stamp is = (8000 * 100) / 80 Rs 10000 Total buying price of two stamps = Rs 6666.67 + Rs 10000 = Rs 16666.67

Total selling price of two stamps = Rs 8000 + Rs 8000 = Rs 16000

Hence, she lost Rs 666.67


Imagine that you have 26 constants, labelled A through Z. Each constant is assigned a value in the following way: A = 1; the rest of the values equal their position in the alphabet (B corresponds to the second position so it equals 2, C = 3, etc.) raised to the power of the preceeding constant value. So, B = 2 ^ (A's value), or B = 2^1 = 2. C = 3^2 = 9. D = 4^9, etc.

Find the exact numerical value to the following equation: (X - A) * (X - B) * (X - C) * ... * (X - Y) * (X - Z)

Answer

(X - A) * (X - B) * (X – C) * ... * (X - Y) * (X - Z) equals 0 since (X - X) is zero.


If three babies are born every second of the day, then how many babies will be born in the year 2001?

Answer

9,46,08,000 babies

The total seconds in year 2001 = 365 days/year * 24 hours/day * 60 minutes/hours * 60 seconds/minute = 365 * 24 * 60 * 60 seconds = 3,15,36,000 seconds

Thus, there are 3,15,36,000 seconds in the year 2001. Also, three babies born are every second. Hence, total babies born = 3 * 3,15,36,000 seconds = 9,46,08,000


A terrorist, after having comitted a crime wanted to cross the border. He knew that a pick up vehicle had been arranged at the border for him. But the driver would identify him by a code word which the terrorist didn't know. So he mailed to his group leader using a certain code language. This is what he mailed:

TWDK XJ KWQ GIPQ TIFP X HQQP KI LJQ DK KWQ YIFPQF?

His group leader replied back on mail, saying:

ZIL HQQP KI JDZ KWDK "XR TXJWQJ TQFQ WIFJQJ YQVVDFJ TILCP FXPQ"

Can you decipher the code and make out as to what the terrorist mailed and what was the reply sent back to him?

HINT : In the code language,QUEEN ELIZABETH can be written as ELQQH QCXADYQKW

Answer

The terrorist mailed : WHAT IS THE CODE WORD I NEED TO USE AT THE BORDER? The Leader replied : YOU NEED TO SAY THAT "IF WISHES WERE HORSES BEGGARS WOULD RIDE"

Use the hint. It says that "In the code language, QUEEN ELIZABETH can be written as ELQQH QCXADYQKW." It means that in code language E means Q, L means U, Q means E, H means N and so on. Q U E E N E L I Z A B E T H

E L Q Q H Q C X A D Y Q K WUse the above hint and put decode the corresponding characters. The terrorist:

T W D K X J K W Q G I P Q T I F P X

H A T I T H E E I

H Q Q P K I L J Q D K K W Q Y I F P Q F ?

N E E T U E A T T H E B E

The leader:

G I L H Q Q P K I J D Z K W D K

U N E E T A T H A T

" X R T X J W Q J T Q F Q W I F J Q J

I I H E E E H E

Y Q V V D F J T I L C P F X P Q "

E A U L I ENow try to figure out some words like WHAT, NEED, TO, USE, IS etc. and decode the sentences further using the characters you get from that.

Countinue the above procedure, you can figure out more words like CODE, WORD, YOU, SAY, IF, WERE, HORSES, WOULD etc. and finally you will get the answer.



S E N D

+ M O R E

----------

M O N E YNote that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all

other S in the puzzle must be 3.

Answer

It is obvious that M=1.

If S=9 and if there is a carry, the maximum value of O will be 1. But M=1. Hence, O has to be 0.

Also, S has to be 9 as there is no other way of getting total 10 for S+M.

Now, N is (E+1) i.e. there must be a carry from (N+R). It means that R must be 9 which is already assigned to S. Hence, R has to be 8 and a carry from (D+E) will make R's value 9. Now, using trial-n-error: 9 5 6 7 O=0, M=1, Y=2, E=5,

+ 1 0 8 5 N=6, D=7, R=8, S=9

----------

1 0 6 5 2


In a sports contest there were m medals awarded on n successive days (n > 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded. 2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so

on. 3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

Answer

Total 36 medals were awarded and the contest was for 6 days.

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.


If one person sends the e-mail to two friends, asking each of them to copy the mail and send it to two of their friends, those in turn send it to two of their friends and so on.

How many e-mails would have been sent by the time it did 30 sets?

Answer

2147483646

First person sent the mail to 2 persons. Those 2 sent the mail to 2 persons each, total 4 persons. Now, those 4 person sent mail to total 8 persons, then 8 to 16 persons, 16 to 32 persons and so on.... Hence, it a series of 2, 4, 8, 16, 32 upto 30 numbers

It is a Geometric series with common ratio 2 and first number is also 2. Summation of such series is given by A * (Rn - 1) / (R - 1) where A = First term R = Common Ratio n = total numbers

So total number of times mail sent by the time it did 30 sets = 2 * (230 - 1) / (2 - 1) = 2 * (1073741824 - 1) = 2 * 1073741823 = 2147483646


A frog starts climbing 15 feet wall. Each hour he climbs 3 feet and rests for 30 minutes. During rest, he slips back 2 feet.

How many hours does the frog take to reach the top?

Answer

19 hours

A frog climbs 1 foot per 1 1/2 hours as during 30 minutes rest he slips back 2 feet. This way he will climb 12 feet in 18 hours. In next hour he will climb 3 more feet i.e. he will complete 15 feet in 19 hours and will reach the top of the wall.


Here is the 7 step proof which says 2=1.

a and b are non-zero integers.

1. a = b

2. a*a = a*b

3. a*a - b*b = a*b - b*b

4. (a + b) * (a - b) = b * (a - b)

5. a + b = b

6. 2*b = b

7. 2 = 1Is it Correct? If not, then what is wrong with the proof?

Answer

The proof is false. It is obvious that 2 is not equal to 1. The step 5 is wrong although it is just a dividing both the sides by (a - b).

It is given in the first step that a = b which means that (a - b) = 0. You get step 5 after dividing both the sides by (a -b). But you can not divide by (a - b) as in mathematics division by 0 is not a valid operation. (Division by Zero)


How long would it take you to count 1 billion orally if you could count 200 every minute and were given a day off every four years?

Assume that you start counting on 1 January 2001.

Answer

9 Years, 187 Days, 5 Hours, 20 minutes

As you can count 200 per minute, to count 1 billion you require = 1,000,000,000/200 minutes = 5,000,000 minutes = 83,333.3333 hours = 3,472.2222 days = 9.512937 years = 9 Years, 187 Days, 5 Hours, 20 minutes

Note that a day off every four year will be a Leap day. Hence, no need to consider leap year.


In Column-I below, are given some words. These have been translated into a code language. The code equivalents of the words in Column-I are given in Column-II, not necessarily opposite to the corresponding words. Also, the codes for the different letters in each word have also not been given the same order as these letter occur in the original word.

COLUMN-I COLUMN-II

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

DELIBERATION aemrqs

CONSIDERATE ccehlmo

GHOSTLIKE cfhmoqqrx

WORLDLY cdgmqrsxz

KNOWLEDGE adefmopqqsz

ROCKET cefkmopqqszzCan you decode the individual letter codes?

Answer

We first find the exact codes of the each given words.

DELIBERATION is a 12-letter word. So its code is cefkmopqqszz. CONSIDERATE is a 11-letter word. So its code is adefmopqqsz. GHOSTLIKE and KNOWLEDGE both are 9-letter words. But KNOWLEDGE has 2 E's and so its code is cfhmoqqrx and hence the code for GHOSTLIKE is cdgmqrsxz. WORLDLY is a 7-letter word. So its code is ccehlmo. ROCKET is a 6-letter word. So its code is aemrqs.

Thus, the words and their codes are: DELIBERATION cefkmopqqszz

CONSIDERATE adefmopqqsz

GHOSTLIKE cdgmqrsxz

WORLDLY ccehlmo

KNOWLEDGE cfhmoqqrx

ROCKET aemrqsThe common letter in the given words is O and the common code letter is m. So m stands for O.

In GHOSTLIKE and WORLDLY, the other common code letter c stands for L. In WORLDLY and ROCKET, the other common code letter e stands for R. In DELIBERATION and WORLDLY, the remaining common code letter (other than c, e and m) o stands for D.

Working similarly, we get a-C, c-L, d-S, e-R, f-N, g-H, h-W, k-B, l-Y, m-O, o-D, p-A, q-E, r-K, s-T, x-G, z-I


The secret agent X emailed a code word to his head office. They are "AIM DUE OAT TIE MOD". But four of these five words are fake and only one contains the information.

The agent X also mailed a sentence as a clue - if I tell you any one character of the code word, you would be able to tell the number of vowels in the code word.

Can you tell which is the code word?

Answer

The code word is TIE.

If you were told any one character of MOD, then you would not be able to determine whether the number of vowels are one or two. e.g. if you were told M, there are two words with M - AIM with 2 vowels and MOD with 1 vowel. So you would not be able to say the number of vowels. Same arguments can be given for characters O and D.

Hence, the word with any one of M, O or D is not a code word i.e. AIM, DUE, OAT and MOD are not the code word. Thus, TIE is the code word. T : two words - TIE and OAT, both with 2 vowels I : two words - TIE and AIM, both with 2 vowels E : two words - TIE and DUE, both with 2 vowels.


A polygon has 1325 diagonals. How many vertices does it have?

Answer

The formula to find number of diagonals (D) given total number of vertices or sides (N) is

N * (N - 3)

D = -----------

2

Using the formula, we get 1325 * 2 = N * (N - 3) N2 - 3N - 2650 = 0

Solving the quadratic equation, we get N = 53 or -50

It is obvious that answer is 53 as number of vertices can not be negative.

Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so on......

Hence the series is 0, 0, 0, 2, 5, 9, 14, ........ (as diagram with 1,2 or 3 vertices will have 0 diagonals).

Using the series one can arrive to the formula given above.


Grass in lawn grows equally thick and in a uniform rate. It takes 24 days for 70 cows and 60 days for 30 cows to eat the whole of the grass.

How many cows are needed to eat the grass in 96 days?

Answer

20 cows

g - grass at the beginning r - rate at which grass grows, per day y - rate at which one cow eats grass, per day n - no of cows to eat the grass in 96 days

From given data, g + 24*r = 70 * 24 * y ---------- A g + 60*r = 30 * 60 * y ---------- B g + 96*r = n * 96 * y ---------- C

Solving for (B-A), (60 * r) - (24 * r) = (30 * 60 * y) - (70 * 24 * y) 36 * r = 120 * y ---------- D

Solving for (C-B), (96 * r) - (60 * r) = (n * 96 * y) - (30 * 60 * y) 36 * r = (n * 96 - 30 * 60) * y 120 * y = (n * 96 - 30 * 60) * y [From D] 120 = (n * 96 - 1800) n = 20

Hence, 20 cows are needed to eat the grass in 96 days.


There are 3 persons X, Y and Z. On some day, X lent tractors to Y and Z as many as they had. After a month Y gave as many tractors to X and Z as many as they have. After a month Z did the same thing. At the end of this transaction each one of them had 24.

Find the tractors each originally had?

Answer

One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using Backtracing.

It's given that at the end, each had 24 tractors (24, 24, 24) i.e. after Z gave tractors to X & Y as

many as they had. It means that after getting tractors from Z their tractors got doubled. So before Z gave them tractors, they had 12 tractors each and Z had 48 tractors. (12, 12, 48)

Similarly, before Y gave tractors to X & Z, they had 6 & 24 tractors respectively and Y had 42 tractors i.e. (6, 42, 24)

Again, before X gave tractors to Y & Z, they had 21 & 12 tractors respectively and X had 39 tractors i.e. (39, 21, 12)

Hence, initially X had 39 tractors, Y had 21 tractors and Z had 12 tractors.


PRIMAL SERIES Decide what the next 5 figures in this series should be:

0110101000101000101000

Answer

10000

The title holds a hint, although this is still a tough puzzle. The series begins with the number 1, & continues through 22, giving a 1 for each prime number, & a 0 for each number that is not prime.

Of the last 5 numbers (23-27), only 23 is prime.


A soldier looses his way in a thick jungle. At random he walks from his camp but mathematically in an interesting fashion.

First he walks one mile East then half mile to North. Then 1/4 mile to West, then 1/8 mile to South and so on making a loop.

Finally how far he is from his camp and in which direction?

Answer

The soldier is 0.8944 miles away from his camp towards East-North.

It is obvious that he is in East-North direction.

Distance travelled in North and South directions = 1/2 - 1/8 + 1/32 - 1/128 + 1/512 - 1/2048 + and so on... (a geometric series with r = (-1/4) )

(1/2) * ( 1 - (-1/4)n ) = --------------------------- ( 1 - (-1/4) )

= 1 / ( 2 * ( 1 - (-1/4) ) ) = 2/5

Similarly in East and West directions = 1 - 1/4 + 1/16 - 1/64 + 1/256 - and so on... (a geometric series with r = (-1/4) )

(1) * ( 1 - (-1/4)n ) = --------------------------- ( 1 - (-1/4) )

= 1 / ( ( 1- (-1/4) ) = 4/5

So the soldier is 4/5 miles away towards East and 2/5 miles away towards North. So using right angled triangle, soldier is 0.8944 miles away from his camp.


If A+B=C, D-C=A and E-B=C, then what does D+F stands for? Provide your answer in letter terms as well as in number terms.

Answer

J or 10

A simple one.

Assume that each character represents the number equivalent to the position in the alphabet i.e. A = 1, B = 2, C = 3, D = 4 and so on. Now let's check our assumption.

A + B = C i.e. 1 + 2 = 3 D - C = A i.e. 4 - 3 = 1 E - B = C i.e. 5 - 2 = 3

Thus, our assumption was Correct. Hence, D + F = J i.e. 4 + 6 = 10


In Column-I below, are given some words. These have been translated into a code language. The code equivalents of the words in Column-I are given in Column-II, not necessarily opposite to the corresponding words. Also, the codes for the different letters in each word have also not been given the same order as these letter occur in the original word.

COLUMN-I COLUMN-II

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

TAPE moi

COP lhhpok

TIE nls

ROTATE nhpk

SAY nkpl

TREAT msr

YEAR khlph

SIP hrp

TYRE pmlhCan you decode the individual letter codes?

Answer

We first find the exact codes of the each given words.

ROTATE is a 6-letter word. So its code is lhhpok. And h is for T. TREAT is a 5-letter word. So its code is khlph.

The 4-letter words are TAPE, YEAR, TYRE and codes are nhpk, nkpl, pmlh. YEAR and TYRE have 3 letters in common (Y, E, R). They must be either nhpk or nkpl. Hence, the code for TAPE is pmlh and m is for P. Also the code for TYRE is nhpk (as h is for T) and the code for YEAR is nkpl.

The 3-letter words are COP, TIE, SAY, SIP and codes are moi, nls, msr, hrp. The code for TIE is hrp. The code for SIP is msr. The code for COP is moi. And the code for SAY is nls.

Thus, the words and their codes are: ROTATE lhhpok

TREAT khlph

TAPE pmlh

TYRE nhpk

YEAR nkpl

TIE hrp

COP moi

SIP msr

SAY nlsSo far we know that h is for T and m is for P.

In SAY and SIP, the common letter is S stands for s. In TYRE and SAY, the common letter Y stands for n. Thus, in SAY, the remaining letter A stands for l.

In TIE and SIP, the common letter I stands for r. Thus, in TIE, the remaining letter E stands for p.

In ROTATE and COP, the common letter O stands for o. Thus, in ROTATE, the remaining letter R stands for k. Also, in COP, the remaining letter C stands for i.

Summerizing h-T, i-C, k-R, l-A, m-P, n-Y, o-O, p-E, r-I, s-S



S T I L L

+ W I T H I N

--------------------

L I M I T SNote that the leftmost letter can't be zero in any word. Also, there must be a one-to-one mapping between digits and letters. e.g. if you substitute 3 for the letter S, no other letter can be 3 and all other S in the puzzle must be 3.

Answer

The value of L must be one more than W i.e. L=W+1 and there must be one carry from S+I=I. Also, the value of S must be 9 as S+I=I with one carry from T+T=M, which means that the value of T must be greater than 4.

From I+H=I, the value of H must be 0 as the value of S is 9.

Now, applying all those constraints and using trial-n-error, we get two possible answers. 9 7 1 6 6 9 8 5 3 3

+ 5 1 7 0 1 3 + 2 5 8 0 5 6

--------------- ---------------

6 1 4 1 7 9 3 5 6 5 8 9


An emergency vehicle travels 10 miles at a speed of 50 miles per hour.

How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?

Answer

While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles is = (60 * 10) / 50 = 12 minutes

Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes. So the speed of the vehicle must = (60 * 10) / 8 = 75 miles per hour

appt itude

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