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Average Aptitude Questions

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.



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