It is the value of the digit itself. eg, in '3452', face value of 4 is four and face value of 2 is two.

It is the face value of the digit multiplied by the place value at which it is situated. eg, in 2586, place value of 5 is 5 x 10^{2}= 500

If N is the set of natural numbers, then we write N = {1, 2, 3, 4, 5, 6,......} The smallest natural number is 1.

If W is the set of Whole numbers, then we write W = {0, 1, 2, 3, 4, 5, 6,......} The smallest Whole number is 0.

If I is the set of integers, then we write I = {......, -3, -2, -1, 0, 1, 2, 3,.....} Where {1, 2, 3, ...} is the set of Positive integers. {-1, -2, -3,....} is the set of negative integers and 0 is neither positive nor negative.

Any number which can be expressed in the form of p/q, where both p and q are integers and q not equal to 0 is called rational number. eg, 2/3, -(7/9), 5, -2, 0, 3...... There exists infinite number of rational numbers between any two rational numbers.

Non-recurring and non-terminating decimals are called irrational numbers. These numbers cannot be expressed in the form of p/q. eg, √3, √5, √29,...... Square root of a prime number is always irrational

Example:- 1.) The sum first 10 natural number is =

Example:- 2.) The sum first 49 natural number is =

The sum of the squares of the first n natural numbers is =

Example:- 1.) The sum of the cubes of the first 10 natural numbers is =

= [(10*11)/2]

The sum of the first n odd natural numbers is =

Example:- 2.) The sum of the first 9 odd natural numbers is =

Example:- 2.) The sum of the first 12 even natural numbers is =

Numbers which are exactly divisible by 2 are called even numbers. eg, 0, 2, 4, 6, 8,..... Sum of first n even numbers = n(n + 1)

Numbers which are divisible by 2 and the remainder is 1 is called odd numbers. eg, 1, 3, 5,........ Sum of first n odd numbers = n^{2}How 1 is an odd number? 2)1(0 0 --- 1 --- 2 x 0 + 1 = 1 So, here remainder is 1.

Prime Numbers are divisible by only 1 and itself. Prime Numbers upto 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, 59, 61, 71, 73, 79, 83, 89, 97. 2 is the only even prime number. 1 is not a prime number because it has two equal factors. Every prime number greater than 3 can be written in the form of (6k + 1) or (6k - 1) where k is an integer.

Two numbers are said to be relatively prime if they do not have any common factor other than 1. eg, (3, 5), (4, 7), (9, 16), (13, 15) .......

Two prime numbers which differ by 2 are called Twin Primes. Example : (3, 5), (5, 7), (11, 13)....

Numbers which are not prime are called Composite Numbers. Example : 4, 6, 8, 9, 12, ..... 1 is neither prime nor composite.

A number is said to be a perfect number, if the sum of all its factors excluding itself is equal to the number itself. Example : Factors of 6 are 1, 2, 3 and 6. Sum of factors excluding 6 = 1 + 2 + 3 = 6 .: 6 is a perfect number. Other examples of perfect numbers are 28, 496, 8128 etc..

Divisibility by 2:- A number is divisible by 2 when the digit at ones place is 0, 2, 4, 6 or 8 Example:- 3582, 460, 28, 352, ......... Whereas 2345, 4299, are not divisible by 2. Divisibility by 3:- A number is divisible by 3 only when the sum of its digits is divisible by 3. Example:- 1.) 453 => 4 + 5 + 3 = 12 = 12/3 = 4 ==> 12 is divisible by 3. So, 453 is also divisible by 3. Example:- 2.) 72954 => 7 + 2 + 9 + 5 + 4 = 27/3 = 9 So, 72954 is also divisible by 3. Whereas 652 = 6 + 5 + 2 = 13, 13 is not divisible by 3. So, 652 is not divisible by 3. Divisibility by 4:- A number is divisible by 4, if the number formed with its last two digits is divisible by 4. Example:- 1.) 784936 is divisible by 4, since 36 is divisible by 4 423441 is not divisible by 4, since 41 is not divisible by 4. Divisibility by 5:- A number is divisible by 5 when the digit at ones place is 0 or 5. Example:- 1.) 34970 and both divisible by 5, Whereas, 25654, 9521, 3721, 3723 are not divisible by 5. Divisibility by 6:- A number is divisible by 6 as it is divisible both by 2 and 3. Example:- 1.) 3672 is divisible by 6 as it is divisible by both 2 and 3. Example:- 2.) 4586 is not divisible by 6 as it is not divisible 3. Divisibility by 7:- A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by the remaining digits is either 0 or a number divisible by 7. Example:- 1.) 658 is divisible by 7 as 65 - (2 x 8) = 65 - 16 = 49 is divisible by 7. 349 is not divisible by 7 as 34 - (2 x 9) = 34 - 18 = 16 is not divisible by 7. Divisibility by 8:- A number is divisible by 8 only when the number formed by its last 3 digits is divisible by 8. Example:- 1.) 485120 is divisible by 8, since 120 is divisible by 8. 364940 is not divisible by 8, since 940 is not divisible by 8. Divisibility by 9:- A number is divisible by 9, if the sum of its digits is divisible by 9. Example:- 1.) 786546 is divisible by 9, since the sum of its digits 7 + 8 + 6 + 5 + 4 + 6 = 36 = 36/9 = 4 36 is divisible by 9. 432512 is not divisible by 9, since the sum of its digits 4 + 3 + 2 + 5 + 1 + 2 = 17 is not divisible by 9 Divisibility by 10:- A number is divisible by 10 when the digit at ones place is 0. Example:- 42570 is divisible by 10 Whereas 39745 is not divisible by 10 Divisibility by 11:- A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a number divisible by 11. Example : 30426 is divisible by 11, since (3 + 4 + 6) - (0 + 2) = 13 - 2 = 11 is divisible by 11 345912 is not divisible by 11, since (4 + 9 + 2) - (3 + 5 + 1) = 15 - 9 = 6 is not divisible by 11. 1. If a number is divisible by two numbers a and b, where a and b are co-primes, then the number is divisible by ab. eg. if a number is divisible by 3 and 5, then it is also divisible by 14. 560 is divisible by 2 and 7 and hence 14. 2. (x^{n}- a^{n}) is divisible by (x - a) for all values of n. 3. (x^{n}- a^{n}) is divisible by (x + a) for even values of n. 4. (x^{n}+ a^{n}) is divisible by (x + a) for odd values of n.

In a sum of division, we have Dividend = (Divisor x Quotient) + Remainder

A number which divides a given number exactly is called a factor of the given number. eg, 24 = (1 x 24), (2 x 12), (3 x 8) and (4 x 6) Thus, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 1. 1 is a factor of every number. 2. A number is a factor of ifself. 3. The smallest factor of a given number is 1. 4. The greatest factor of a given number is the number itself. 5. If a number is divided by any of the factors, the remainder is always zero. 6. Every factor of a number is either less than or at the most equal to the given number. 7. Number of factors of a number are finite.

If N is a composite number such that N = a^{m}.....Where a,b,c, are prime factors of N and m,n,o, ..... are positive integer, then the number of factors of N is given by the expression. (m + 1)(n + 1)(0 + 1)..... 224 = 2^{5}x 7^{1}=> 224 has (5 + 1) = (6 x 2) = 12 factors

A multiple of a number is a number obtained by multiplying it by a natural number eg Multiples of 5 are 5, 10, 15, 20,........ Multiples of 12 are 12, 24, 36, 48, ...... Every number is a multiple of 1. The smallest multiple of a number is the number itself. We cannot find the greatest multiple of a number. Number of multiples of a number are infinite.

The lowest common multiple of two or more given numbers is the least of their common multiples. Multiples of 25 are 25, 50, 75, 100, 125, 150,..... Multiples of 30 are 30, 60, 90, 120, 150, 180,..... Lowest common multiple is 150.

The highest common factor of two or more given numbers is the largest of their common factors. Factors of 20 are 1, 2, 4, 5, 10, 20. Factors of 36 are 1, 2, 4, 6, 9, 12, 18, 36. common factors are 1, 2 and 4. Highest common factor is 4

Solution. To find LCM, 2|12, 52 ------ 2|6, 26 ------ 3|3, 13 ------ 13|1, 13 ------ 1, 1 Thus, LCM of 12 and 52 is (2 x 2 x 3 x 13) = 156 To find HCF Method 1: 2|12, 52 -------- 2|6, 26 ------- 3, 13 Now, 3 and 13 do not have any common factor. Hence, HCF of 12 and 52 = (2 x 2) = 4 Method 2: we have, 12)52(4 48 -- 4)12(3 12 -- 0 .: HCF is 4.

Solution: To find LCM, 5|15, 20, 30 ---------- 2|3, 4, 6 ---------- 3|3, 2, 3 --------- 2|1, 2, 1 -------- 1, 1, 1 Thus, LCM of 15, 20 and 30 = (5 x 3 x 2 x 2) = 60 To find HCF, Method 1 5|15, 20, 30 ---------- | 3, 4, 60 Now 3, 4 and 5 do not have any common factor. Hence, HCF of 15, 20 and 30 is 5. Method 2 15)20(1 15 ---- 5)15(3 15 -- x -- 5)30(6 30 -- x -- .: Required HCF is 5.

LCM and HCF of fractions (i) LCM of fractions = LCM of numerators/HCF of denominators eg, LCM of 3/5, 4/15 and 9/8 = LCM of 3, 4, 9/HCF of 5, 15, 8 = 36/1 = 36 (ii) HCF of fractions = HCF of numerators/LCM of denominators eg, HCF of 3/5, 4/15 and 9/8 HCF of 3, 4, 9/LCM of 5, 15, 8 = 1/120

(i) (x + y)^{2}= x^{2}+ y^{2}+ 2xy = (x - y)^{2}+ 4xy (ii) (x - y)^{2}= x^{2}+ y^{2}- 2xy = (x + y)^{2}- 4xy (iii)(x + y)^{2}- (x - y)^{2}= 4xy (iv) (x + y)^{2}+ (x - y)^{2}= 2(x^{2}+ y^{2}) (v) (x^{2}- y^{2}) = (x + y)(x - y) (vi) (x + y + z)^{2}= x^{2}+ y^{2}+ z^{2}+ 2(xy + yz + zx) (vii)(x^{3}+ y^{3}) = (x + y)(x^{2}- xy + y^{2}) (viii)(x^{3}- y^{3}) = (x - y)(x^{2}+ xy + y^{2}) (ix) (x^{3}+ y^{3}+ z^{3}- 3xyz) = (x + y + z)(x^{2}+ y^{2}+ z^{2}- xy - yz - zx) (x) (x + y + z) = 0 => x^{3}+ y^{3}+ z^{3}= 3xyz

Solution : (Local value of 8) - (Face value of 8) = (800 - 8) = 792

Solution : Divisor = [(Dividend) - (Remainder)]/Quotient = (258 - 3)/17 - 255/17 = 15

Solution : Let the quotient be q when 442 is divided by 43. => Given number = 442q + 43 = (17 x 26q) + (17 x 2) + 9 = 17 x (26q + 2) + 9 .: The required remainder = 9

Solution : Let the five odd numbers be (x - 4), (x - 2), x, (x + 2) and (x + 4) respectively. Then, (x - 4) + (x - 2) + x + (x + 2) + (x + 4) = 225 => 5x = 225 => x = 45 .: The numbers are 41, 43, 45, 47 and 49. The sum of the next set of five odd numbers = (51 + 53 + 55 + 57 + 59) = 275

Solution : Let the 4 consecutive even numbers be (x - 2), x, (x + 2) and (x + 4) respectively. Then, (x - 2) + x + (x + 2) + (x + 4) = 4x + 4 = 92 => x = 22 The four numbers are 20, 22, 24, and 26 respectively. Thus , the sum of their squares = (20^{2}+ 22^{2}+ 24^{2}+ 26^{6}) = (400 + 484 + 576 + 676) = 2136

Solution : Let the given prime numbers be a, b, c, d. Then, abc/bcd = 2431/4199 = 11/19 => a/d = 11/19 ==> a = 11 and d = 19 Hence, the first prime number = a = 11

Give answer (1) If the data given in statement A alone are sufficient to answer the question whereas the data given in statement B alone are not sufficient to answer the question.

Give answer (2) If the data given in statement B alone are sufficient to answer the question whereas the data given in statement A alone are not sufficient to answer the question.

Give answer (3) If the data in either statements A alone or in statement B alone are sufficient to answer the question.

Give answer (4) If the data in both the statements A and B are not sufficient to answer the question.

Give answer (5) If the data given in both the statements A and B are necessary to answer the question.

A. The number is a multiple of 62. B. The sum of the digit 'a' and 'b' is 8. Solution : From statement A, we get, the number to be 62, 124, 186,..... But as it is greater than 9 and a two digit number, it has to be 62 only. From statement B we get, a + b = 8 ie, the numbers can be 17, 26, 35, 44, 53, 62, 71, 80 of the these the only two digit number multiple of 62 is 62 itself. Hence, both the statements are independently sufficient to get the answer . .: The required answer is (3).

A. An odd number is obtained when x is divided by 3. B. (x + y) is an odd number. Solution : Statement A alone is sufficient to answer the question. We know that whenever any odd number is divided by any odd number, It gives an odd number. From statement B, we get, x is either even or odd as the sum of an even number and an odd number is odd. If y is odd, then x is even and if y is even, x is odd. .: The correct answer is (1).

A. x/7 is a positive integer. B. 7/x is a positive integer. Solution : From statement A, x can take values = 7, 14, 21, 28, ..... From statement B, x = 1 or 7. Thus, combining both statements x = 7 is a positive integer .: The correct answer is (5).