# Average

This section comprises quantitative aptitude multiple choice questions with their solutions and Explanations on average. To browse other quantitative aptitude sections, please click here

1.

The age of A, B and C is 32 years, 36 years and 46 years. What is the average age of A, B and C:
[A]35
[B]32
[C]36
[D]38

38
The sum of age of A, B and C = 32 + 36 + 46 = 114 years
Hence The average age is 114 $\div$ 3 = 38 years.
∴ Option (d) is correct.

2.

Find the average of first 50 natural numbers?
[A]25
[B]26.5
[C]25.5
[D]25.7

25.5
Because we know that the sum of first n natural numbers is (1+2+3+…..+n) $= n\frac{\left ( n+1 \right )}{2}$
so we can say that (1+2+3+…..+50) $= 50\frac{\left ( 50+1 \right )}{2} = 50\times \frac{51}{2} = 1275$
and the average of first 50 natural numbers = $\frac{1275}{50} = 25.5$
Option [c] is the right answer.

3.

Find the average of all prime numbers between 20 and 40.
[A]29
[B]120
[C]31
[D]30

30
There are four prime numbers between 20 and 40 and they are 23, 29, 31, 37.
So now we have to find the average of these four numbers :
= $\frac{23 + 29 + 31 + 37}{4}= 30.$
Option (D) is the correct answer.

4.

Find the average of first 30 multiples of 5.
[A]75
[B]75.5
[C]76
[D]76.5

(b) 76.5
The average of first 30 multiples of 5 :
$5\times \frac{(1+2+3+.....+30)}{30} = \frac{50\times 30\times 31}{30\times 2} = 76.5$
Option (b) is the right answer.

5.

The average of six consecutive even numbers is 29. Find the largest of these numbers.
[A]32
[B]35
[C]34
[D]33

34
Let the numbers be x, x+2, x+4, x+6, x+8, x+10. Accoring to the question :
$\frac{x+\left ( x+2 \right )+\left ( x+4 \right )+\left ( x+6 \right )+\left ( x+8 \right )+\left ( x+10 \right )}{6}=29$
$=> 6x + 30 = 174$
$=> 6x = 174 - 30 = 144$
$=> x=\frac{144}{6}=24$
Now because the largest number is x+10 = 24+10 = 34
Hence option [C] is the right answer.

6.

There are two sections A and B of a class, consisting of 42 and 52 students respectively. If the average weight of section A is 40 kg and that of section B is 45 kg find the average weight of the whole class.
[A]40
[B]42.24
[C]42
[D]42.76

42.76
Total weight of (42+52) students = (42 $\times$ 40 + 52 $\times$ 45) = 1680 + 2340 = 4020
Average weight of the whole class = $\frac{4020}{94}$$= 42.76 kg$
Option (d) is the correct answer.

7.

The average of three numbers is 7 and that is of first two numbers is 4, then what is the third number?
[A]15
[B]13
[C]7
[D]4

13
The sum of three numbers is 7 $\times$ 3 = 21,
and the sum of first two numbers is 2 $\times$ 4 = 8
Then the third number is 21 – 8 = 13.
Hence option [B] is the right answer.

8.

If the average temprature of Monday, Tuesday and Wednesday was 40° c and that Tuesday, Wednesday and thrusday was 41° c, if the temprature of thrusday is 38° c, then what was the temprature on Monday?
[A]28° c
[B]35° c
[C]42° c
[D]48° c

35° c
The sum of temprature of (M+T+W) = 40 $\times$ 3 = 120° c. ……(1)
The sum of temprature of (T+W+Th) = 41 $\times$ 3 = 123° c. …(2)
and given that Thrusdau temprature is 38° c ………………(3)
To substract the equation (3) into (2): (T+W) = 123 – 38 = 85° c
Hence monday’s temprature is 120 – 85 = 35° c.

9.

Of the three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 66, find the largest number.
[A]16
[B]18
[C]15
[D]17

18
Let the the first number be x
Then, second number = 2x and third number = $\frac{2x}{3}$.
$=>x+2x+\frac{2x}{3}=66\times 3$
$=>3x+6x+2x=198$
$=>x=\frac{198}{11}=1$
Hence option [B] is the right answer.

10.

If the average of m numbers is $n^{2}$ and n numbers is $m^{2}$ then what will be the average of (m+n) numbers :
[A]$\frac{m}{n}$
[B](m + n)
[C](m – n)
[D]mn

mn
Because the average of m numbers is $n^{2}$,
and the average of n numbers is $m^{2}$
Since the sum of m numbers is = $mn^{2}$
and the sum of n numbers is = $nm^{2}$
Now we have to find the average of (m+n) nummbers :
$\frac{mn^{2}+nm^{2}}{\left ( m+n \right )}=\frac{mn\left ( n+m \right )}{\left ( m+n \right )}=mn$
Hence option [D] is the right answer.