## Number Solved Questions For IBPS, SBI and Other Bank Exams - Page 2

**Question 1**

Sum of seven consecutive odd numbers is 133. Find the sum of the first and the third number in the series.

a. 18

b. 38

c. 48

d. 28

e. 58

**Answer :** b. 38

Solution :

Since difference between any two successive odd numbers is 2, we can assume 7 consecutive odd numbers to be x-6,x-4,x-2,x,x+2,x+4,x+6. (In our assumed series, difference between any two numbers is 2.)

Sum of the numbers in the series = 133 = x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 7x

Therefore, x = 133 / 7 = 19

Sum of first term and last term = x - 6 + x + 6 = 2x

Substituting x = 19 (which we found earlier), we can get the answer to be 2 x 19 = 38.

**Question 2**

Sum of six consecutive numbers that are divisible by 3 is 99. Find the sum of the first two numbers.

a. 30

b. 21

c. 14

d. 15

e. 17

**Answer :** b. 21

Solution :

Difference between successive numbers divisible by 3 will be 3. Therefore, the series takes the form as below.

x-6,x-3,x,x+3,x+6,x+9

Sum of all terms in the series = 99 = x-6 + x-3 + x + x+3 + x+6 + x+9 = 6x+9

6x + 9 = 99

Or 6x = 90

Or x = 15

Sum of first two numbers = x-6 + x-3 = 2x-9 = 2(15) - 9 = 21

**Question 3**

Sum of five consecutive numbers that are divisible by 5 is 275. Find the last number in the series.

a. 65

b. 60

c. 75

d. 70

e. 85

**Answer :** a. 65

Solution :

Difference between two adjacent numbers divisible by 5 is 5.

Series takes the form x-10,x-5,x,x+5,x+10

Sum = 275 = x-10 + x-5 + x + x+5 + x+10 = 5x

5x = 275

Or x = 55

Last number in the series = x+10 = 55 + 10 = 65

**Question 4**

Sum of 7 consecutive natural numbers is 91. Find the last but one number.

a. 15

b. 19

c. 21

d. 25

e. 39

**Answer :** a. 15

Solution :

Difference between two successive natural numbers is 1.

Therefore, series takes the form, x-3,x-2,x-1,x,x+1,x+2,x+3

Sum = 91 = x-3 + x-2 + x-1 + x + x+1 + x+2 + x+3 = 7x

7x = 91

Or x = 13

Last but one number = x+2 = 13 + 2 = 15