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1. NUMBERS

IMPORTANT FACTS AND FORMULAE

1.Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 calleddigits to represent any number.

A group of digits, denoting a number is called a numeral.

We represent a number, say 689745132 as shown below :

TenCrores(108)

Crores(107) Ten Lacs(Millions)(106)

Lacs(105) TenThousands(104)

Thousands(103)

Hundreds(102)

Tens(101) Units(100)

6 8 9 7 4 5 1 3 2

We read it as : 'Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred andthirty-two'.

II 2. Place Value or Local Value of a Digit in a Numeral :

In the above numeral :

Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;

Place value of 1 is (1 x 100) = 100 and so on.

Place value of 6 is 6 x 108 = 600000000

III3.Face Value: The face value of a digit in a numeral is the value of the digit itself atwhwhatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3isais 3 and so on.

IV4.Types of Numbers:

I. Natural Numbers: Counting numbers 1, 2, 3, 4, 5 ,……are called natural

numbers.

II. Whole Numbers: All counting numbers together with zero form the set of whole

numbers. Thus, (i) 0 is the only whole number which is not a natural number.

(ii) Every natural number is a whole number.

III. Integers : All natural numbers, 0 and negatives of counting numbers i.e.,{…, - 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of integers.

(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.

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(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.

(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nornegative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while{0, - 1 , - 2 , -3 , …..} represents the set of non-positive integers.

IV. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, etc.

V. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9,11, etc.

VI. Prime Numbers : A number greater than 1 is called a prime number, if it has exactlytwo factors, namely 1 and the number itself.

Prime numbers up to 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,47, 53, 59,61, 67, 71, 73, 79, 83, 89, 97.

Prime numbers Greater than 100 : Let p be a given number greater than 100. To find outwhether it is prime or not, we use the following method :

Find a whole number nearly greater than the square root of p. Let k > *jp. Test whether pis divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p isprime.e.g. We have to find whether 191 is a prime number or not. Now, 14 > V191.

Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.

191 is not divisible by any of them. So, 191 is a prime number.

VII.Composite Numbers : Numbers greater than 1 which are not prime, are known ascomposite numbers, e.g., 4, 6, 8, 9, 10, 12.

Note : (i) 1 is neither prime nor composite.

(ii) 2 is the only even number which is prime.

(iii) There are 25 prime numbers between 1 and 100.

V. 5.TESTS OF DIVISIBILITY:

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

Ex. 84932 is divisible by 2, while 65935 is not.

2. Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.

Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which isdivisible by 3.

But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32, which isnot divisible by 3.

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

Ex. 892648 is divisible by 4, since the number formed by the last two digits is 48, which isdivisible by 4. But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is82, which is not divisible by 4.

4. Divisibility By 5: A number is divisible by 5, if its unit's digit is either 0 or 5. Thus, 20820 and50345 are divisible by 5, while 30934 and 40946 are not.

5. Divisibility By 6: A number is divisible by 6, if it is divisible by both 2 and 3.

Ex. The number 35256 is clearly divisible by 2.Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, whichis divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.

6. Divisibility By 8 : A number is divisible by 8, if the number formed by the last Three digits ofthe given number is divisible by 8.

Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which isdivisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is418, which is not divisible by 8.

7. Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible by 9.

Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9.

But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is notdivisible by 9.

8. Divisibility By 10 : A number is divisible by 10, if it ends with 0.

Ex. 96410, 10480 are divisible by 10, while 96375 is not.

9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits atodd places and the sum of its digits at even places, is either 0 or a number divisible by 11.

Ex. The number 4832718 is divisible by 11, since :(sum of digits at odd places) - (sum of digits ateven places)

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

10. Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and3.

Ex. Consider the number 34632.

(i) The number formed by last two digits is 32, which is divisible by 4,

(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4as well as 3. Hence, 34632 is divisible by 12.

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11. Divisibility By 14: A number is divisible by 14, if it is divisible by 2 as well as 7.

12. Divisibility By 15: A number is divisible by 15, if it is divisible by both 3 and 5.

13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits isdivisible by 16.

Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, which isdivisible by 16.

14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.

15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.

16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.

Note : If a number is divisible by p as well as q, where p and q are co-primes, then the givennumber is divisible by pq. If p and q are not co-primes, then the given number need not bedivisible by pq, even when it is divisible by both p and q.

Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4 and 6 are not co-primes.

6 MULTIPLICATION BY SHORT CUT METHODS:

1. Multiplication By Distributive Law :(i) a * (b + c) = a * b + a * c (ii) a*(b-c) = a * b-a * c.

Ex. (i) 567958 x 99999 = 567958 x (100000 - 1)

= 567958 x 100000 - 567958 x 1

= (56795800000 - 567958)

= 56795232042.

(ii) 978 x 184 + 978 x 816 = 978 x (184 + 816)

= 978 x 1000

= 978000.

2. Multiplication of a Number By 5n : Put n zeros to the right of the multiplicand and divide thenumber so formed by 2n

Ex. 975436 x 625 = 975436 x 54

= 9754360000

= 609647600

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7. BASIC FORMULAE

1. (a + b)2 = a2 + b2 + 2ab 2. (a - b)2 = a2 + b2 - 2ab

3. (a + b)2 - (a - b)2 = 4ab 4. (a + b)2 + (a - b)2 = 2 (a2 + b2)

5. (a2 - b2) = (a + b) (a - b)

6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)

7. (a3 + b3) = (a +b) (a2 - ab + b2) 8. (a3 - b3) = (a - b) (a2 + ab + b2)

9. (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)

10. If a + b + c = 0, then a3 + b3 + c3 = 3abc.

8. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM

If we divide a given number by another number, then :

Dividend = (Divisor x Quotient) + Remainder

(i) (xn - an ) is divisible by (x - a) for all values of n.

(ii) (xn - an) is divisible by (x + a) for all even values of n.

(iii) (xn + an) is divisible by (x + a) for all odd values of n.

9. PROGRESSION

A succession of numbers formed and arranged in a definite order according to certain definiterule, is called a progression.

1. Arithmetic Progression (A.P.) : If each term of a progression differs from its preceding termby a constant, then such a progression is called an arithmetical progression. This constantdifference is called the common difference of the A.P.

An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d),(a + 3d),.....

The nth term of this A.P. is given by Tn =a (n - 1) d. The sum of n terms of this A.P.

Sn = n/2 [2a + (n - 1) d] = n/2 (first term + last term).

SOME IMPORTANT RESULTS :

(i) (1 + 2 + 3 +…. + n) =n(n+1)/2

(ii) (l2 + 22 + 32 + ... + n2) = n (n+1)(2n+1)/6

(iii) (13 + 23 + 33 + ... + n3) =n2(n+1)2

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2. Geometrical Progression (G.P.) : A progression of numbers in which every term bears aconstant ratio with its preceding term, is called a geometrical progression. The constant ratio iscalled the common ratio of the G.P. A G.P. with first term a and common ratio r is : a, ar, ar2,

In this G.P. Tn = arn-1

sum of the n terms, Sn= a(1-rn)(1-r)

2.DECIMAL FRACTIONS

IMPORTANT FACTS AND FORMULAE

I. Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimalfractions. Thus ,1/10=1 tenth=.1;

1/100=1 hundredth =.01;

9/100=99 hundreths=.99;

7/1000=7 thousandths=.007,etc

II. Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under thedecimal point and annex with it as many zeros as is the number of digits after the decimal point.Now, remove the decimal point and reduce the fraction to its lowest terms.

Thus, 0.25=25/100=1/4;2.008=2008/1000=251/125.

III. 1. Annexing zeros to the extreme right of a decimal fraction does not change its valueThus, 0.8 = 0.80 = 0.800, etc.

2. If numerator and denominator of a fraction contain the same number of decimalplaces, then we remove the decimal sign.

Thus, 1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584=5

IV. Operations on Decimal Fractions :

1. Addition and Subtraction of Decimal Fractions : The given numbers are soplaced under each other that the decimal points lie in one column. The numbersso arranged can now be added or subtracted in the usual way.

2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimalpoint to the right by as many places as is the power of 10.Thus, 5.9632 x 100 = 596,32; 0.073 x 10000 = 0.0730 x 10000 = 730.

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3.Multiplication of Decimal Fractions : Multiply the given numbers consideringthem without the decimal point. Now, in the product, the decimal point is markedoff to obtain as many places of decimal as is the sum of the number of decimalplaces in the given numbers. Suppose we have to find the product (.2 x .02 x .002).

Now, 2x2x2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.

4.Dividing a Decimal Fraction By a Counting Number : Divide the givennumber without considering the decimal point, by the given counting number.Now, in the quotient, put the decimal point to give as many places of decimal asthere are in the dividend.

Suppose we have to find the quotient (0.0204 + 17). Now, 204 ^ 17 = 12. Dividend contains 4places of decimal. So, 0.0204 + 17 = 0.0012.

5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend andthe divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.Thus, 0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V

V. Comparison of Fractions : Suppose some fractions are to be arranged in ascending ordescending order of magnitude. Then, convert each one of the given fractions in the decimal form,and arrange them accordingly. Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 indescending order.

now, 3/5=0.6, 6/7 = 0.857, 7/9 = 0.777.... since 0.857>0.777...>0.6, so 6/7>7/9>3/5

VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeatedcontinuously, then such a number is called a recurring decimal. In a recurring decimal, if a singlefigure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it isexpressed by putting a bar on the set

Thus 1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857

Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point arerepeated, is called a pure recurring decimal.

Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures onlyonce in the numerator and take as many nines in the denominator as is the number of repeatingfigures. thus ,0.5 = 5/9; 0.53 = 53/59 ;0.067 = 67/999;etc...

Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some ofthem are repeated, is called a mixed recurring decimal. e.g., 0.17333= 0.173.

Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take thedifference between the number formed by all the digits after decimal point (taking repeated digitsonly once) and that formed by the digits which are not repeated, In the denominator, take thenumber formed by as many nines as there are repeating digits followed by as many zeros as is thenumber of non-repeating digits.

Thus 0.16 = (16-1) / 90 = 15/19 = 1/6;

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SOLVED EXAMPLES

Ex. 1. Simplify : (i) 8888 + 888 + 88 + 8 (RRB 2000)(ii) 11992 - 7823 - 456

Sol. i ) 8888 ii) 11992 - 7823 - 456 = 11992 - (7823 + 456)888 = 11992 - 8279 = 3713-88 7823 11992

+ 8 + 456 - 82799872 8279 3713

Ex. 2, What value will replace the question mark in each of the following equations ?(i) ? - 1936248 = 1635773 (ii) 8597 - ? = 7429 - 4358 (BANK 1998)

Sol. (i) Let x - 1936248=1635773.Then, x = 1635773 + 1936248=3572021.(ii) Let 8597 - x = 7429 - 4358.Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.

Ex. 3. What could be the maximum value of Q in the following equation5P9 + 3R7 + 2Q8 = 1114 (CAT 2004)

Sol. We may analyse the given equation as shown : 1 2Clearly, 2 + P + R + Q = ll. 5 P 9So, the maximum value of Q can be 3 R 7(11 - 2) i.e., 9 (when P = 0, R = 0); 2 Q 8

11 1 4

Ex. 4. Simplify : (i) 5793405 x 9999 (ii) 839478 x 625 (GATE 1998)

Sol. i)5793405x9999 =5793405(10000-1) =57934050000-5793405 =57928256595.

ii) 839478 x 625 = 839478 x 54 = 8394780000 = 524673750.16

Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 - 983 x 107 (TECH MAHIN 98)

Sol.(i) 986 x 137 + 986 x 863 = 986 x (137 + 863)= 986 x 1000 = 986000.

(ii) 983 x 207 - 983 x 107 = 983 x (207 - 107)= 983 x 100 = 98300.

Ex. 6. Simplify : (i) 1605 x 1605 ii) 1398 x 1398 (TCS 2002)Sol.i) 1605 x 1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2 + 2 x 1600 x 5

= 2560000 + 25 + 16000 = 2576025.(ii) 1398 x 1398 - (1398)2 = (1400 - 2)2= (1400)2 + (2)2 - 2 x 1400 x 2

=1960000 + 4 - 5600 = 1954404.

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Ex. 7. Evaluate : (313 x 313 + 287 x 287). (CTS 1998)

Sol. (a2 + b2) = 1/2 [(a + b)2 + (a- b)2](313)2 + (287)2 = 1/2 [(313 + 287)2 + (313 - 287)2] = ½[(600)2 + (26)2]= 1/2 (360000 + 676) = 180338.Ex. 8. Convert the following into vulgar fraction:(i) 3.004 (ii) 0.0056 (INFOSYS)

Sol. (i) 3.004 = 3004/1000 = 751/250

(ii) 0.0056 = 56/10000 = 7/1250

Ex.9. Arrange the fractions 5/8, 7/12, 13/16, 16/29 and 3/4 in ascending order of magnitude.

(CAT 1999)

Sol. Converting each of the given fractions into decimal form, we get :

5/8 = 0.624, 7/12 = 0.8125, 16/29 = 0.5517, and 3/4 = 0.75

Now, 0.5517<0.5833<0.625<0.75<0.8125

16/29 < 7/12 < 5/8 < 3/4 < 13/16

Ex.10. arrange the fractions 3/5, 4/7, 8/9, and 9/11 in their descending order.(BANK 2000)

Sol. Clearly, 3/5 = 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/11= 0.818.

Now, 0.88 > 0.818 > 0.6 > 0.571

8/9 > 9/11 > 3/4 > 13/ 16

Ex. 11. Evaluate : 5.064 + 3.98 + 0.7036 + 7.6 + 0.3 + 2 (RRB 1998)

Sol. 5.0643.980.70367.60.32.0___19.6476

Ex. 12. Evaluate : 13 – 5.1967 (ACCENTURE 2004)

Sol. 13.0000– _5.1967

7.8033

Ex. 13. What value will replace the question mark in the following equations ?5172.49 + 378.352 + ? = 9318.678 (GATE 2005)

Sol. Let 5172.49 + 378.352 + x = 9318.6

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Then , x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836

Ex. 14. Find the products: (i) 6.3204 * 100 (ii) 0.069 * 10000 (GATE 1997)

Sol. (i) 6.3204 * 1000 = 632.04

(ii) 0.069 * 10000 = 0.0690 * 10000 = 690

Ex. 15. Find the product: (i) 2.61 * 1.3 (ii) 2.1693 * 1.4 (BANK 2000)

Sol.(i)261 * 13 = 3393. Sum of decimal places of given numbers = (2+1) = 3.

so 2.61 * 1.3 = 3.393.

(ii)21693 * 14 = 303702. Sum of decimal places = (4+1) = 5.

so 2.1693 * 1.4 = 3.03702.

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EXERCISE PROBLEMS:

Ex. 1. Which of the following are prime numbers ? (i) 241 (ii) 337 (iii) 391

(CAT 1998)A)241,337 B)337,391 C)all the above D)none

Sol. (i)Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, and 13. 241 is notdivisible by any one of them. 241 is a prime number.

(ii)Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11, 13, and 17. 337 is notdivisible by any one of them. 337 is a prime number.

(iii)Clearly, 20 > Ö39l". Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, and 19.We find that 391 is divisible by 17. 391 is not prime.

Ex. 2. Find the unit's digit in (264)102 + (264)103 (TCS 2000)

A) 0 B)2 C)4 D)20

Sol. Required unit's digit = unit's digit in (4)102 + (4)103.

Now, 42 gives unit digit 6.(4)102 gives unjt digit 6.

(4)103 gives unit digit of the product (6 x 4) i.e., 4.Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.

Ex. 3. Find the total number of prime factors in the expression (4)11 x (7)5 x (11)2.(CTS)

A)30 B)40 C)29 D)15

Sol. (4)11x (7)5 x (11)2 = (2 x 2)11 x (7)5 x (11)2 = 211 x 211 x75x 112 = 222 x 75 x112

Total number of prime factors = (22 + 5 + 2) = 29.

Ex.4. Simplify : (i) 896 x 896 - 204 x 204 (ii) 387 x 387 + 114 x 114 + 2 x 387 x 114(RRB 1997)

A)761000,251000 B)761200,251001 C)761200,251000 D)76100,25100

Sol. (i) Given exp = (896)2 - (204)2 = (896 + 204) (896 - 204) = 1100 x 692 = 761200.(ii) Given exp = (387)2+ (114)2+ (2 x 387x 114)

= a2 + b2 + 2ab, where a = 387,b=114

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= (a+b)2 = (387 + 114 )2 = (501)2 = 251001.

Ex.5. Which of the following numbers is divisible by 3 ? (i) 541326 (ii) 5967013(RRB 2004)

A) i & ii B)only (ii) C)only (i) D)none

Sol.(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.Hence, 541326is divisible by 3.(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.Hence, 5967013 is not divisible by 3.

Ex.6 which one of the following is not a prime number? (TECH MAHIN 2001)

A)31 B)61 C)71 D)91

Sol:91 is divisible by 7.so,it is a prime number

Ex.7 (112*54)=? (CAT 2003)

A)67000 B)50055 c)70000 D)75000

Sol: (112 x 54) = 112 x10 4

=112 x 104

=1120000

= 700002 24 16

Ex.8.What least value must be assigned to * so that the number 197*5462 is r 9 ?(CTS 2005)

A)1 B)0 C)2 D)3Sol: Let the missing digit be x.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +2) = (34 + x).

For (34 + x) to be divisible by 9, x must be replaced by 2 .

Hence, the digit in place of * must be 2.

Ex. 9. Which of the following numbers is divisible by 4 ? (i) 6792059 (ii) 618703572

A) i & ii B)only (ii) C)only (i) D)none (SBI PO 2000)

Sol. (i) The number formed by the last two digits in the given number is 94, which is not

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divisible by 4. Hence, 67920594 is not divisible by 4.(ii) The number formed by the last two digits in the given number is 72, which is divisible

by 4. Hence, 618703572 is divisible by 4.

Ex. 10. Which digits should come in place of * and $ if the number 62684*$ is divisible byboth 8 and 5?

(TECH MAHINDRA 2005)A)4,0 B)0,4 C)4,4 D)0,0

Sol. Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a numberending with 5 is never divisible by 8. So, 0 will replace $. Now, the number formed by the lastthree digits is 4*0, which becomes divisible by 8, if * is replaced by 4. Hence, digits in place of *and $ are 4 and 0 respectively.

Ex. 11. Is 4832718 is divisible by 11. (RRB 1999)A)yes B)no C)noneSol.(Sum of digits at odd places) - (Sum of digits at even places) = (8 + 7 + 3 + 4) - (1 + 2 + 8)= 11, which is divisible by 11. Hence, 4832718 is divisible by 11.

Ex.12. what least number must be added to 3000 to obtain a number exactly divisible by19?A) 5 B) 3 C) 2 D) 0 (CTS 2004)

Sol. On dividing 3000 by 19, we get 17 as remainder.Number to be added = (19 - 17) = 2.

Ex.13. what least number must be subtracted from 2000 to get a number exactly divisibleby 17? (INFOSYS 2005)A) 11 B) 2 C) 5 D) 1Sol. On dividing 2000 by 17, we get 11 as remainder.Required number to be subtracted = 11.

Ex. 14. Find the smallest number of 6 digits which is exactly divisible by 111. (GATE 98)

A)100000 B)100011 C)100111 D)100110

Sol. Smallest number of 6 digits is 100000.On dividing 100000 by 111, we get 100 as remainder.

Number to be added = (111 - 100) - 11. Hence, required number = 100011.

Ex. 15. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37.Find the divisor. (SBI PO 97)

A)180 B)179 C)184 D)25

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Dividend - Remainder 15968-37Sol. Divisor = -------------------------- = ------------- = 179.

.Quotient 89

Ex. 16. A number when divided by 342 gives a remainder 47. When the same number isdivided by 19, what would be the remainder? (INFOSYS 2000)

A) 5 B) 9 C) 4 D) 0

Sol. On dividing the given number by 342, let k be the quotient and 47 as remainder.Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.

The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex.17. Find the remainder when 231 is divided by 5. (VIRTUSA 2006)

A)4 B)5 C)3 D)7

Sol. 210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].Unit digit of 231 is 8. Now, 8 when divided by 5, gives 3 as remainder.

Hence, 231 when divided by 5, gives 3 as remainder.

Ex.18. Find the sum of all odd numbers up to 100. (BANK PO 2008)

A)2000 B)2500 C)2800 D)3000

Sol. The given numbers are 1, 3, 5, 7, ..., 99. This is an A.P. with a = 1 and d = 2.Let it contain n terms. Then, 1 + (n - 1) x 2 = 99 or n = 50.

Required sum = n (first term + last term)2

= 50 (1 + 99) = 2500.2

Ex.19. Find the sum of all 2 digit numbers divisible by 3. (CAT 2005)

A)1700 B)1665 C)1600 D)1605

Sol. All 2 digit numbers divisible by 3 are: 12, 51, 18, 21... 99. This is an A.P. with a = 12 andd = 3. Let it contain n terms. Then,12 + (n - 1) x 3 = 99 or n = 30.

Required sum = 30 x (12+99) = 1665.2

Ex.20. It is being given that (232 + 1) is completely divisible by a whole number. Which ofthe following numbers is completely divisible by this number?(ACCENTURE 98)

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A)(216+1) B)(216-1) C)(7*223) D)(296+1)

Sol: Let 232 = x. Then, (232 + 1) = (x + 1). Let (x + 1) be completely divisible by the naturalnumber N. Then, (296 + 1) = [(232)3 + 1] = (x3 + 1) = (x + 1)(x2 - x + 1), which is completelydivisible by N, since (x + 1) is divisible by N.

Ex:21. How many prime numbers are less than 50 ?

A)16 B)15 C)14 D)18

Sol: Prime numbers less than 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Their number is 15

Ex.22.Which natural number is nearest to 9217, which is completely divisible by 88 ?

A)9152 B)9240 C)9064 D)9184

Sol: On dividing we get,88) 9217 (104

8841735265

Therefore, Required number = 9217 + (88 -65) = 9217 + 23 = 9240.

Ex.23. 3251 + 587 + 369 - ? = 3007

A) 1250 B) 1300 C) 1375 D) 1200

Sol: 3251 Let 4207 - x = 3007+ 587 Then, x = 4207 - 3007 = 1200+ 3694207

Ex.24. If the product 4864 x 9 P 2 is divisible by 12, then the value of P is:A) 2 B) 5 C) 6 D) noneSol: Clearly, 4864 is divisible by 4. So, 9P2 must be divisible by 3. So, (9 + P + 2) must bedivisible by 3. P = 1.

Ex.25. In a division sum, the divisor is 10 times the quotient and 5 times the remainder. Ifthe remainder is 46, what is the dividend ?A)4236 B)4306 C)4336 D)5336

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Sol: Divisor = (5 x 46) = 230 10 x Quotient = 230230

= 2310

Dividend = (Divisor x Quotient) + Remainder = (230 x 23) + 46 = 5290 + 46 = 5336.

Ex.26. (489 + 375)2 - (489 - 375)2

= ?(489 x 375)

A)144 B)864 C)2 D)4

Sol: Given Exp. =(a + b)2 - (a - b)2

=4ab

= 4ab ab

Ex.27. If 1400 * x = 1050. Then, x = ?

A)1/4 B)3/5 C)2/3 D)3/4

Sol: 1400 * x = 1050 x =1050

=3

1400 4

Ex. 28. Given that 268 * 74 = 19832, find the values of 2.68 * 0.74.

A)198.32 B)19.832 C)1.9832 D)1983.2

Sol. Sum of decimal places = (2 + 2) = 4

2.68 * 0.74 = 1.9832.

Ex. 29. Find the quotient: 0.0204 / 17

A)0.0012 B)0.012 C)0.00012 D)none

Sol: 204 / 17 = 12. Dividend contains 4 places of decimal. 0.2040 / 17 = 0.0012.

Ex. 30. Evaluate : 2.5 + 0.0005

A)500 B)0.500 C)5000 D)none

Sol: 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000

Ex. 31. What value will come in place of question mark in the following equation

0.006 +? = 0.6

A)0.01 B)0.1 C)0.001 D)0.0001

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Sol. Let 0.006 / x = 0.6, Then, x = (0.006 / 0.6) = (0.006*10) / (0.6*10) = 0.06/6 = 0.01

Ex. 32. If (1 / 3.718) = 0.2689, Then find the value of (1 / 0.0003718).

A)2698 B)2689 C)2968 D)9286

Sol. (1 / 0.0003718 ) = ( 10000 / 3.718 ) = 10000 * (1 / 3.718) = 10000 * 0.2689 = 2689.

Ex. 33. Express as vulgar fractions : 0.37A)99/37 B)0.37/99 C)37/99 D)99/0.37

___Sol. 0.37 = 37 / 99 .

--------Ex.34.Express as vulgar fractions: 0.053A)999/53 B)53/999 C)999/0.53 D)0.53/999

-------Sol. 0.053 = 53 / 999.

Ex.35. Simplify: 0.05 * 0.05 * 0.05 + 0.04 * 0.04 * 0.040.05 * 0.05 – 0.05 * 0.04 + 0.04 * 0.04

A)0.9 B)0.09 C)9 D)0.009

Sol. Given expression = (a3 + b3) / (a2 – ab + b2), where a = 0.05 , b = 0.04

= (a +b ) = (0.05 +0.04 ) =0.09

Ex. 36. Evaluate : 35 + 0.07

A)50 B)55 C)500 D)5000

Sol. 35/0.07 = ( 35*100) / (0.07*100) = (3500 / 7) = 500

Ex.37 Find the quotient: 0.63 / 9

A)0.7 B)0.07 C)7 D)none

Sol. 63 / 9 = 7. Dividend contains 2 places decimal. 0.63 / 9 = 0.7.

Ex.38 Convert the following into vulgar fraction: 0.75

A)3/4 B)4/3 C)0.75 D)0.075

Sol: 0.75 = 75/100 = 3/4

Ex.39 Evaluate : 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025

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A)6891.222 B)6918.888 C)6891.59775 D)6891.57975

Sol: 6202.5620.25

62.0256.2025

+ 0.620256891.59775

Ex.40 Evaluate : 31.004 – 17.2368A)13.7654 B)24.3452 C)12.2345 D)19.6753Sol: 31.0040

– 17.238613.7654

Ex.41 What value will replace the question mark in the following equation? – 7328.96 + 5169.38

A)12498.34 B)12948.43 C)12849.33 D)89421.44

Sol: Let x – 7328.96 = 5169.38. Then, x = 5169.38 + 7328.96 = 12498.34.

Ex.42 Find the product of 0.4 * 0.04 * 0.004 * 40

A)0.002650 B)0.002560 C)0.02560 D)0.05260

Sol: 4 * 4 * 4 * 40 = 2560. Sum of decimal places = (1 + 2+ 3) = 6so 0.4 * 0.04 * 0.004 * 40 = 0.002560.

Ex.43 How many digits will be there to the right of the decimal point in the product of95.75 and .02554A)5 B)6 C)7 D)none

Sol: Sum of decimal places = 7. Since the last digit to the extreme right will be zero (since 5 x 4= 20), so there will be 6 significant digits to the right of the decimal point.

Ex.44.which of the following is equal to 3.14x106?

A)314 B)314000 C)3140 D)none

Sol: 3.14 x 106 = 3.14 x 1000000 = 3140000.

Ex.45 5 x 1.6 - 2 x 1.4 = ?1.3

A)1.4 B)0.4 C)1.2 D)4

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Sol: Given Expression =8 - 2.8

=5.2

=52

= 4.1.3 1.3 13

QUESTION BANK:

1.The largest 4 digit number exactly divisible by 88 is:A.9944 B.9768

C.9988 D.8888

2.Which of the following is a prime number ?A.33 B.81

C.93 D.97

3. (?) + 3699 + 1985 - 2047 = 31111A.34748 B.27474

C.30154 D.27574

4.The difference between the local value and the face value of 7 in the numeral 32675149 isA.75142 B.64851

C.5149 D.69993

5.What will be remainder when (6767 + 67) is divided by 68 ?A.1 B.63

C.66 D.67

6.What least number must be subtracted from 13601, so that the remainder is divisible by 87 ?A.23 B.31

C.29 D.37

7. If the number 97215 * 6 is completely divisible by 11, then the smallest whole number inplace of * will be:A.3 B.2

C.1 D.5

8. Which of the following numbers will completely divide (4915 - 1) ?A.8 B.14

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C.48 D.50

9.The sum all even natural numbers between 1 and 31 is:A.16 B.128

C.240 D.512

10.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

11.The sum of how may tersm of the series 6 + 12 + 18 + 24 + ... is 1800 ?A.16 B.24

C.20 D.18

12.1904 x 1904 = ?A.3654316 B.3632646

C.3625216 D.3623436

13. 854 x 854 x 854 - 276 x 276 x 276= ?

854 x 854 + 854 x 276 + 276 x 276A.1130 B.578

C.565 D.1156

14.The sum of first 45 natural numbers is:A.1035 B.1280

C.2070 D.2140

15.What is the unit digit in 7105 ?A.1 B.5

C.7 D.9

16.106 x 106 - 94 x 94 = ?A.2400 B.2000

C.1904 D.1906

17.287 x 287 + 269 x 269 - 2 x 287 x 269 = ?A.534 B.446

C.354 D.324

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18.Which one of the following can't be the square of natural number ?A.30976 B.75625

C.28561 D.143642

19. .M and N are only two odd numbers with M > N. The largest even integer which divides is

A) 12 B) 4

C) 6 D) 8

20. How many three-digit numbers are divisible by 6 in all?

A) 149 B) 150

C) 166 D) 151

21. The least perfect square number which is divisible by 3, 4, 5, 6 and 8 is

A. 1600 B 900

C. 2500 D 3600

22.How many of the following numbers are divisible by 132 ? { 264, 396, 462, 792, 968, 2178,5184, 6336 }A.4 B.5

C.6 D.7

23.Which one of the following is the common factor of (4743 + 4343) and (4747 + 4347) ?A.(47 - 43) B.(47 + 43)

C.(4743 + 4343) D.None of these

24.A number when divided by 296 leaves 75 as remainder. When the same number is divided by37, the remainder will be:A.1 B.2

C.8 D.11

25.What smallest number should be added to 4456 so that the sum is completely divisible by 6 ?A.4 B.3

C.2 D.1

26.If 2994 ÷ 14.5 = 172, then 29.94 ÷ 1.45 = ?A.0.172 B.1.72

C.17.2 D.172

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27.The expression (11.98*11.98 + 11.98 * x + 0.02 x 0.02) will be a perfect square for x equal to:A.0.02 B.0.2

C.0.04 D.0.4

28.3889 + 12.952 - ? = 3854.002A.47.095 B.47.752

C.47.932 D.47.95

29.0.009= .01

?A..0009 B..09

C..9 D.9

30.The value of (0.96)3 - (0.1)3

is:(0.96)2 + 0.096 + (0.1)2

A.0.86 B.0.95

C.0.97 D.1.06

31.Evaluate : (2.39)2 - (1.61)2

2.39 - 1.61A.2 B.4

C.6 D.8

32.The value of 0.1 x 0.1 x 0.1 + 0.02 x 0.02 x 0.02is:

0.2 x 0.2 x 0.2 + 0.04 x 0.04 x 0.04A.0.0125 B.0.125

C.0.25 D.0.5

33.0.04 x 0.0162 is equal to:A.6.48 x 10-3 B.6.48 x 10-4

C.6.48 x 10-5 D.6.48 x 10-6

34. 4.2 x 4.2 - 1.9 x 1.9 is equal to:2.3 x 6.1

A.0.5 B.1.0

C.20 D.22

35.If 144=

14.4, then the value of x is:0.144 x

A.0.0144 B.1.44

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C.14.4 D.144

36. 617 + 6.017 + 0.617 + 6.0017 = ?A.6.2963 B.62.965

C.629.6357 D.None of these

37.0.002 x 0.5 = ?A.0.0001 B.0.001

C.0.01 D.0.1

38.Which of the following is equal to 3.14 x 106 ?A.314 B.3140

C.3140000 D.None of these

39.The least among the following is:A.0.2 B.1 ÷ 0.2

C.3.2 D.(0.2)2

40. 5 x 1.6 - 2 x 1.4= ?

1.3A.0.4 B.1.2

C.1.4 D.4

41.How many digits will be there to the right of the decimal point in the product of 95.75 and0.02554 ?A.5 B.6

C.7 D.None of these

42.4.036 divided by 0.04 gives :A.1.009 B.10.09

C.100.9 D.None of these

43. 0.0203 x 2.92= ?

0.0073 x 14.5 x 0.7A.0.8 B.1.45

C.2.40 D.3.25

44.The fraction 101 27 in decimal for is:100000

A.01027 B..10127

C.101.00027 D.101.000027

45.The value of(4.7*13.26+4.7*9.43+4.7*77.31) is

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A)0.47 B)47

C)470 D)4700

46.Which one of the following is equal to 3.14*106?

A)314 B)3140

C)3140000 D)none

47. 0.014*0.014=?

A)0.000196 B)0.00196

C)19.6 D)196

48. 4.036 divided by 0.04 gives

A)1.009 B)100.09

C)10.9 D)none

49.If 2994/14.5=172,then 29.94/1.45=?

A)172 B)0.172

C)1.72 D)17.2

50.12.1212+17.0005-9.1102=?

A)20.0015 B)20.0105

C)20.0115 D)20.1015