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

# Average

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 $latex \div$ 3 = 38 years.

∴ Option (d) is correct.

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) $latex = n\frac{\left ( n+1 \right )}{2}&s=1$

so we can say that (1+2+3+…..+50) $latex = 50\frac{\left ( 50+1 \right )}{2} = 50\times \frac{51}{2} = 1275&s=1$

and the average of first 50 natural numbers = $latex \frac{1275}{50} = 25.5&s=1$

Option [c] is the right answer.

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 :

= $latex \frac{23 + 29 + 31 + 37}{4}= 30.&s=1$

Option (D) is the correct answer.

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 :

$latex 5\times \frac{(1+2+3+…..+30)}{30} = \frac{50\times 30\times 31}{30\times 2} = 76.5&s=1$

Option (b) is the right answer.

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 :

$latex \frac{x+\left ( x+2 \right )+\left ( x+4 \right )+\left ( x+6 \right )+\left ( x+8 \right )+\left ( x+10 \right )}{6}=29&s=1$

$latex => 6x + 30 = 174$

$latex => 6x = 174 – 30 = 144$

$latex => x=\frac{144}{6}=24&s=1$

Now because the largest number is x+10 = 24+10 = 34

Hence option [C] is the right answer.

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 $latex \times$ 40 + 52 $latex \times$ 45) = 1680 + 2340 = 4020

Average weight of the whole class = $latex \frac{4020}{94}&s=1$$latex = 42.76 kg$

Option (d) is the correct answer.

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 $latex \times$ 3 = 21,

and the sum of first two numbers is 2 $latex \times$ 4 = 8

Then the third number is 21 – 8 = 13.

Hence option [B] is the right answer.

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 $latex \times$ 3 = 120° c. ……(1)

The sum of temprature of (T+W+Th) = 41 $latex \times$ 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.

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 = $latex \frac{2x}{3}&s=1$.

$latex =>x+2x+\frac{2x}{3}=66\times 3$

$latex =>3x+6x+2x=198$

$latex =>x=\frac{198}{11}=1&s=1$

Hence option [B] is the right answer.

If the average of m numbers is $latex n^{2} $ and n numbers is $latex m^{2} $ then what will be the average of (m+n) numbers :

[A]$latex \frac{m}{n}&s=1$

[B](m + n)

[C](m – n)

[D]mn

**mn**

Because the average of m numbers is $latex n^{2} $,

and the average of n numbers is $latex m^{2} $

Since the sum of m numbers is = $latex mn^{2}$

and the sum of n numbers is = $latex nm^{2}$

Now we have to find the average of (m+n) nummbers :

$latex \frac{mn^{2}+nm^{2}}{\left ( m+n \right )}=\frac{mn\left ( n+m \right )}{\left ( m+n \right )}=mn&s=1$

Hence option [D] is the right answer.