IBPS Sample Problems On Average

Below are four problems based on average of consecutive integers.

Question 1

Find the average of 5 consecutive integers when a is the smallest of these numbers.

a) a b) a+1 c) a+2 d) a+4

Solution :

Given that a is the least of 5 consecutive numbers.
Then a, a+1, a+2, a+3 and a+4 are that consecutive numbers.
The average of that numbers = (a + a+1 + a+2 + a+3 + a+4) / 5
= (5a + 10) / 5
= 5(a + 2) / 5
= a+2

Hence, the required answer is a+2.

Question 2

If the average of 5 consecutive odd integers is 11 then the average of last 3 of them is:
a) 11 b) 13 c) 19 d) 9

Solution :

Let x, x+2, x+4, x+6 and x+8 be given 5 consecutive odd integers.
It is given that, their average is 11.
i.e., (x + x+2 + x+4 + x+6 + x+8) / 5 = 11
(5x + 20) / 5 = 11
(5x + 20) = 11x5
5x = 55-20
x = 35/5 = 7
Therefore x = 7.
We have to find the average of x+4=11, x+6=13 and x+8=15.
i.e, 11 + 13 + 15 / 3 = 39 / 3 = 13

Question 3

The average of four consecutive odd integers is 24 less than the sum of these integers. Then which will be the largest of them?
a) 27 b) 22 c) 24 d) 20

Solution :

Let the x, x+2, x+4 and x+6 be the 4 consecutive odd integers.
We have to find the largest integer x+6.
It is given that the average of these numbers is 24.
i,e., (x + x+2 + x+4 + x+6) / 4 = 24
4x + 12 = 24x4
4x = 84
x = 84 / 4 = 21
Then x + 6 = 21 + 6 = 27

Question 4
The arithmetic mean of the five consecutive numbers a, a+1, a+2, a+3 and a+4 is A, then the value of the median of these numbers is equal to:
a) +2 b) a+3 c) a d) a+4

Solution :

Let a, a+1, a+2, a+3 and a+4 be the given 5 consecutive numbers.

We have to find value of their median.
i.e, middle number of them = a+2

It is given that, their average = arithmetic mean = A = (a + a+1 + a+2 + a+3 + a+4) / 5
A = (5a+10) / 5 = 5(a+2) / 5 = a+2
a+2 = A.

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