## Problems on Numbers

**Basic Formulae:**(Must Remember)

1) (a - b)

^{2}= (a

^{2}+ b

^{2}- 2ab)

2) (a + b)

^{2}= (a

^{2}+ b

^{2}+ 2ab)

3) (a + b) (a – b) = (a

^{2}– b

^{2})

4) (a

^{3}+ b

^{3}) = (a + b) (a

^{2}– ab + b

^{2})

5) (a

^{3}- b

^{3}) = (a - b) (a

^{2}– ab + b

^{2})

6) (a + b + c)

^{2}= a

^{2}+ b

^{2}+ c

^{2}+ 2 (ab + bc + ca)

7) (a

^{3}+ b

^{3}+ c

^{3}– 3abc) = (a + b + c) (a

^{2}+ b

^{2}+ c

^{2 }– ab – bc – ac)

**Question Variety**

**Type 1: Find the numbers, if we are given a) Sum of squares/sum of productsb) Sum/average of consecutive numbers**

**Q 1.**If the sum two numbers is 31 and their product is 240, then find the absolute difference between the numbers.

a. 1

b. 3

c. 4

d. 5

View solution

Correct Option: (a)

Let two numbers be x and y

We are given that, sum of two numbers x + y = 31 and product = xy = 240

Therefore,

x – y = (x + y)^{2} – 4xy

Substituting the values, we get

x – y = (31)^{2} – 4 × y

= 961 – 960

= 1

= 1

The required difference between the numbers is 1.

**Q 2.**Find the three consecutive odd numbers whose sum of the squares is 2531.

a. 19, 21, 23

b. 23, 25, 27

c. 27, 29, 31

d. 31, 33, 35

View solution

Correct Option: (c)

Let three consecutive odd numbers be x, x+2, x+4.

x^{2}+(x+2)^{2}+(x+4)^{2}=2531

Simplifying we get,

x^{2}+4x-837=0

27 x 31 = 837 and also the difference between 27 and 31 is 4

Therefore,

x^{2}+31x-27x-837=0

(x + 31) (x – 27)

X = 27 or x = –31

Hence, the value of

x = 27

(x+2) = 27 + 2 = 29

(x+4) = 27 + 4 = 31

**Q 3.**The sum of squares of three numbers is 138 and the sum of their products taken two at a time is 131. Find their sum.

a. 35

b. 42

c. 20

d.18

View solution

Correct Option: (c)

Let the three numbers be x, y and z.**Given: **Sum of squares of three numbers is 138 and sum of their products taken two at a time is 131

Therefore,

x^{2}+y^{2}+z^{2}=138

xy+yz+zx=131

Formula: (a + b + c)^{2} = a^{2} + b^{2} + c^{2 }+ 2 (ab + bc + ca)

This formula can be used to easily find the sum of three numbers. Substituting the values, we get

(x + y + z) ^{2} = x^{2} + y^{2 }+ z^{2} + 2 (xy + yz + zx)

(x + y + z ) ^{2} = 138 + 2(131)

(x + y + z ) ^{2} = 400

Hence, (x + y + z) = 20

**Q 4.**The average of 3 consecutive even numbers is 18, find the largest of these numbers.

a. 15

b. 16

c. 20

d. 26

View solution

Correct Option: (c)

Let the numbers be x, x +2, and x + 4

Hence, sum of these numbers = (average x no. of numbers) = (18 x 3) = 54

x + (x +2) + (x + 4) = 54

3x = 48

x = 16

Largest number = (x + 4) = (16 + 4) = 20