AMCAT 3

Decimal Number

• In which of the system, decimal number 184 is equal to 1234?

• 184= 1*x^3 + 2*x^2 + 3*x + 4• X3 + 2x2 + 3x – 180 = 0. • Has no integral solution. • Hence answer choose none of these.

Numbers

• The citizens of planet nigiet are 6 fingered and have thus developed their decimal system in base 6.

• A certain street in nigiet contains 1000 (in base 6) buildings numbered 1 to 1000.

• How many 3s are used in numbering these buildings? a) 256 b) 54 c) 192 d) 108

• (N-1)*(BASE)^(N-2)N=NUMBER OF LAST DIGIT=4(1000 contains 4 digits)base=6

• (4-1)*62 = 3*36 = 108.

Numbers

• The citizens of planet nigiet are 8 fingered and have thus developed their decimal system in base 8. A certain street in nigiet contains 1000 (in base 8) buildings numbered 1 to 1000. How many 3s are used in numbering these buildings? a) 54 b) 64 c) 265 d) 192

• (N-1)*(BASE)^(N-2)• N=NUMBER OF LAST DIGIT=4(1000 contains 4 digits)• base=8• (4-1)*8^(4-2) =192

NUMBER

• If a and b are odd numbers, then which of the following is even ?

• A. a + b • B. a + b + 1• C. ab • D. ab + 2 • E. None of these• • always add two odd number it gives even number

Divisiblity

• 12 divides, ab313ab (in decimal notation, where a,b are digits>0, the smallest value of a+b is a)7 b)6 c)2 d) 4

• If a number is divisible by 12 then it should be divisible by 4&3

• for divisible by 4 last [2 digit]no's should be divisible by 4

• Also sum of digits should be divisible by 3.• Hence 2(a+b)+1 should be 3k and ab = 4m.• That is a + b = (3k-1)/2 . from the given choices only

first choice is possible k = 5 and smallest should be 7.

Divisibility • If a number 774958A96B is to be divisible by 8 and 9, the

values of A and B, respectively, will be:• Using the divisibility rules, • For 8, the last three digits have to divisible by 8, therefore the

number 774958A96B is divisible by 8 if 96B is divisible by 8. • 96B is divisible by 8 if it is 960 or 968 thus B is either 0 or 8.• For 9, the sum of the number has to be divisible by 9,

therefore (55 + A + B) is divisible by 9 if (A + B) is 8.• Now either of A or B could be 8, but the other has to be zero.

Divisibility

• 311311311311311311 is divisible by:a)3 and 11. b)11 but not 3. c)3 but not 11.d)none of the above.

• a.3 and 11• the sum of digits is 30 which is divisible by 3• the differences of odd place digits - even place

is divisible by 11

LCM

• Find the greatest number of five digits, which is exactly divisible by 7, 10, 15, 21 and 28.

• The number should be exactly divisible by 15 (3, 5), 21 (3, 7), 28 (4, 7).

• Hence, it is enough to check the divisibility for 3, 4, 5 and 7.

• Lcm of 3,4,5 and 7 is 420. • 105/420 = 238 is quotient• 238* 420 = 99960 is the only number which satisfies the

given condition.

HCF

• The ratio of two numbers is 3 : 4 and their H.C.F. is 4.

• Their L.C.M. is: A. 12 B. 16 C. 24 D. 48• Let the numbers be 3x and 4x. • Then, their H.C.F. = x. • So, x = 4(given)• the numbers are 12 and 16.• LCM of 12,16 is 48.• So, ans is OPT (D)

Fractions

• If the numerator of a fraction is increased by 200% and the denominator is increased by 250% the resultant fraction is 3/14. What is the original fraction? a)1/2 b) 3/4 c) 1/4 d) 1/8

• Let the number be . • 200% of x is 2x and 250% of y is y• Hence we have 2x + x =3x and y + y = y• Hence resultant fraction should be = =• Hence the original fraction will be .

Fractions

• If the numerator of a fraction is increased by 25% and denominator decreased by 20%, the new value is 5/4. What is the original value? a) 3/5 b) 4/5 c) 7/8 d) 3/7

• let numerator is x nd denominator is y.• if we increase x by 25% then it will be 125x/100 and

if u decrease y by 20% then value will be 80y/100. • new value is 125x/80y=5/4 .or x/y=4/5. • hence b)4/5

Factorization• The prime factorization of integer N is A*A*B*C where A, B and

C are all distinct prime integers. How many factors does N have? a)12 b)24 c)4 d)6

• n is A*A*B*C =A^2*B*Cno. of factors=(2+1)(1+1)(1+1)=12

• A HAS POWER OF 2,B HAS POWER OF 1,C HAS POWER 1SO PRIME FACTORIZATION CAN BE CALCULATED AS:IF A^P+B^Q+C^R,THENPRIME FACTORIZATION IS (P+1)*(Q+1)*(R+1)SO IN THIS CASE P=2,Q=1,R=1SO (2+1)(1+1)(1+1)=12 FACTORS

ALGEBRA• A person drives with constant speed and

after some time he sees a milestone with 2 digits. Then he travels for 1 hour and sees the same 2 digits in reverse order. 1 hour later he sees that the milestone has the same 2 digits with a 0 between them. What is the speed of the car (in mph)? (a) 54 (b) 45 (c) 27 (d) 36

• s=speed "Then travels for 1 hour and sees the same 2 digits in reverse order."10x + y + 1s(speed *time) = 10y + x10x - x + s = 10y - y9x + s = 9y

• 1 hours later he sees that the milestone has the same 2 digits with a 0 between them."

10y + x + s(again same) = 100x + y10y - y + s = 100x - x9y + s = 99x

• Rearrange the above two equations for elimination9x - 9y + s = 0-99x+9y + s = 0----------------adding eliminates y-90x + 2s = 02s = 90xs = 45x

• x has to be equal to 1, then s = 45 mph • (If x=2 and s=90 mph, when added to a

two digit milestone, could not be at another two digit milestone.)

SPEED, TIME & DISTANCE• A train travels a certain distance

taking 7 hrs in forward journey, during the return journey increased speed 12km/hr takes the times 5 hrs.

• What is the distance traveled A.) 210 kms B.) 30 kms C.) 60 kms D.) 90 kms

• let the speed in the forward journey be x km/hr, also given time taken in fwd journey = 7 hrs

• during return journey speed=(x+12)km/hr

time=5hrs• now distance travelled in both the

cases is same therefore• forward journey distance=return

journey distance7x=5(x+12)7x-5x=602x=60x=30km/hr

• therefore distance travelled= 30*7=210km

• or =5*(30+12)=210km

SPEED, TIME & DISTANCE• An athlete decides to run the same distance in 1/4th less time that she

usually took. By how much percent will she have to increase her average speed?

• let original speed be s1 and time be t1 • then s1 = ---eqn 1• Let new speed be s2 and time given is 3• therefore s2= -----eqn 2• dividing eqn 2 by eqn 1 s2=XXs1= • increased speed = - s1 = • percent increase= =33.33%

SPEED, TIME & DISTANCE• a train travels a distance of 300km at a constant speed. • if the speed of the train is increased by 5km/hr,the journey would have

taken two hours less.• find the initial speed of the train?• we know that, s = d / t• distance = 300 km• initial speed ( s) , new speed = s + 5• initial time taken ( t ) = 300 / s and new time taken = 300 / s + 5• so now, according to the ques :• 300 / (s + 5) + 2 = 300 / s• 2 = 300 / s - 300 / (s + 5)• s = 25 or -30

answer: 25 km / hr..

SPEED, TIME & DISTANCE

• A person drives with constant speed and after some time he sees a milestone with 2 digits. Then he travels for 1 hour and sees the same 2 digits in reverse order. 1 hour later he sees that the milestone has the same 2 digits with a 0 between them. What is the speed of the car (in mph)? (a) 54 (b) 45 (c) 27 (d) 36

SPEED, TIME & DISTANCE• s=speed

"Then travels for 1 hour and sees the same 2 digits in reverse order."10x + y + 1s(speed *time) = 10y + x10x - x + s = 10y - y9x + s = 9y

• 1 hours later he sees that the milestone has the same 2 digits with a 0 between them."10y + x + s(again same) = 100x + y10y - y + s = 100x - x9y + s = 99x

• Rearrange the above two equations for elimination9x - 9y + s = 0-99x+9y + s = 0----------------adding eliminates y-90x + 2s = 02s = 90xs = 45x

• x has to equal 1, then s = 45 mph (If x=2 and s=90 mph, when added toa two digit milestone, could not be at another two digit milestone.)

SPEED, TIME & DISTANCE• let during the 1st hr men sees two digit number be(1hr):10x+y• now after an hr he sees reverse of the digit(2 hr) :10y+x• again he sees a 3-digit number which is same as previous but with 0 added

so(3hr):100x+0+y• since,speed is constant then

distance covered in an hr=distance covered in 2 hr• (10y+x)-(10x+y)=(100x+y)-(10y+x)

-9x+9y=99x-9y-108x+18y=018y=108x9y=54xso,x=1 y=6

• ie:16 , 61 AND 106• so to get the speed when we add 45 to 16 ie:45+16=61 • similarly 61+45=106

so speed is 45mph

SPEED, TIME & DISTANCE• A train travels a certain distance taking 7 hrs in forward journey, during the return

journey increased speed 12km/hr takes the times 5 hrs.What is the distance traveled? A.) 210 kms B.) 30 kms C.) 60 kms D.) 90 kms

• let the speed in the forward journey be x km/hr, also given time taken in fwd journey = 7 hrs

• during return journey speed=(x+12)km/hrtime=5hrs

• now distance travelled in both the cases is same • Therefore forward journey distance=return journey distance

7x=5(x+12)7x-5x=602x=60x=30km/hr

• therefore distance travelled= 30*7=210km or 5*(30+12)=210km

PROFIT / LOSS

• In a shop 80% of the articles are sold at a profit of 10% and the remaining at a loss of 40%.what is the overall profit/loss?a.10% profit b.10% loss c.15% profit d. no profit, no loss

• Option d no profit,no loss• =1

PROFIT / LOSS

• If a pen is being sold at 4% profit instead of 4% loss the actual profit is Rs 16. What is the actual cost price of the pen ?

• Let x be the CP.• = 16• Solving we get x = Rs.200.• Shortcut • More amount is 16 and difference in profit =8• Using shortcut the cost price =

PROFIT / LOSS• 100 oranges are bought at the rate of Rs. 350 and sold at the rate of

Rs. 48 per dozen. The percentage of profit or loss is: A. 100/7% gain B. 15% gain C. 100/7% loss D. 15 % loss

• c.p of 100 oranges = 350• 100 oranges = 8 dozens + 4 oranges

1 dozen s.p = 48• 8 dozens s.p = 48*8=384• 4 0ranges s.p = *4 = 16• 100 oranges sp = 400;• profit = 50• profit% =100 = 100/7%;• SHORTCUT = =%

PROFIT / LOSS• John buys a cycle for 31 dollars and given a cheque of amount 35 dollars. Shop

Keeper exchanged the cheque with his neighbor and gave change to John. After 2 days, it is known that cheque is bounced. Shop keeper paid the amount to his neighbour. The cost price of cycle is 19 dollars. What is the profit/loss for shop keeper? a) loss 23 b) gain 23 c) gain 54 d) Loss 54

• CP of cycle = 19$ • SP of cycle = 31$• Profit = 31$-19$ = 12$• Again, shopkeeper gave 35$ to neighbour.• Loss = 35$• So, net loss = 35$-12$ = 23$• Other way there is no profit or loss in exchanging the cheque and change with

neighbour.• The shop keeper loses the cycle worth 19$ and 4$ change. Hence net loss = 23$.

PROBABILITY• Mr. Decimal enters a lucky draw that requires him to pick five different integers

from 1 through 37 inclusive. • He chooses his five numbers in such a way that the sum of their logarithms to base

10 is an integer. • It turns out that the winning 5 numbers have the same property, i.e. the sum of

their logarithms to base10 is also an integer. • What is the probability that Mr. Decimal holds the winning numbers? • We know log(a)+log(b)+log(c)+log(d)+log(e)=log(a*b*c*d*e).• From the above question there are only two no.s 1000 and 10000 within range

having correct decimal value of 3 & 4.• Now the factors of 1000 is = 2,2,2,5,5,5• If the no.s are arranged then we get One sample spaces 2*4*5*25*1.• Similarly for 10000 we get 2*4*5*10*25 , 5*8*10*25*1 & 4*5*20*25*1.• totally there are 4 sample spaces and Mr Decimal is choosing one of the combination

than the probability will be equal to 1/4.

PERMUTATION

• How many 13 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if repetition of digits is allowed?

• to be divisible by 4, last two digits must be divisible by 4. which are 12, 24, 32, 44, 52.

• so 5 combinations are possible for last two digits • also 5 combinations each for remaining 11 places. • so the answer is 5^12.

Permutation with restriction

LOGARITHM

• If find the value of a.• = a = ()6 = 23 = 8.

LOGARITHM

• Find the logarithm of 5832 to the base 3.• Let =x.• (3)x = 5832 = 8 x 729 • = 23 x 36 = ()6(3)6=(3)6

• X = 6.

Logarithm

• Find the logarithm of 64 to the base 2.• = 2 • = = 4 x 1 = 4.

POWER

• Which is greater? 2300 or 3200 • Method 1:• 2300 = 23*100 = 8100 and 3200 = 32*100 = 9100.• Hence 3200 is greater.• Method 2:• Using log, • 300log2 = 300*0.3010 = 90.3 approx and

200log3 = 200*0.4771 = 95.42 approx

POWER

• find out last two digits of 29573661 + 30813643. • A) 42 B) 38 C) 98 D) 22• 29573661= 29573660*2957 • so last two digits be 57.• 30813643=30813640*30813.• 811= 81; 812 = 6561; 813 = 6561*81 = 531441.• so last two digit will be 41• sum 57+41=98.so 98 ans.