Practice MCQ Questions and Answer on Average

41.

There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.

(A) 30 kg.
(B) 35 kg.
(C) 42.5 kg.
(D) 37.25 kg.

Solution: Total weight of (36 + 44) Students, = (36 × 40 + 44 × 35) kg = 2980 kg Therefore average weight of the whole class, = $\frac{2980}{80}$ kg. Therefore average weight = 37.25 kg.

42.

The average of five numbers is 306.4. The average of the first two numbers is 431 and the average of the last two numbers is 214.5. What is the third number?

(A) 52
(B) 108
(C) 321
(D) Cannot be determined

43.

Average of n numbers is a. The first number is increased by 2, second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is -

(A) $a + \frac{2 (2^{n} - 1)}{n}$
(B) $a + \frac{2^{n + 1} - 1}{n}$
(C) $a + \frac{2^{n + 1}}{n}$
(D) $a + 2 \frac{2^{n - 1}}{n}$

Solution: Series:- a, a + 2, a + 4..... Sum = na + 2 + 4 + ..... upto n terms Sum = na + Sn $S_{n} = \frac{2 (2^{n} - 1)}{2 - 1}$ Average = $a + \frac{2 (2^{n} - 1)}{n}$

44.

The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number the average becomes 17.5 then the number of natural numbers is -

(A) 20
(B) 30
(C) 15
(D) 10

Solution: Let the number of natural numbers = n ∵ The average of some natural numbers = 15 ⇒ Sum of these natural number = 15 × n = 15n ∵ 30 is added and 5 is subtracted So, now addition of these number = 15n + 30 - 5 = 15n + 25 According to the question, ⇒ $\frac{15 n + 25}{n}$ = 17.5 ⇒ 15n + 25 = 17.5n ⇒ 2.5n = 25 ⇒ n = 10 Therefore, the numbers of natural numbers n = 10

45.

A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs in the next innings and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was

(A) 64
(B) 72
(C) 90
(D) 85

46.

The average of 7 consecutive numbers is 20. The largest of these numbers is-

(A) 20
(B) 22
(C) 23
(D) 24

Solution: Let the number be x, x + 1, x + 2, x + 4, x + 5 and x + 6 Then, $\Rightarrow \frac{x + (x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6)}{7} = 20$ $\begin{aligned} \Rightarrow 7 x + 21 = 140 \\ \Rightarrow 7 x = 119 \\ \Rightarrow x = 17 \end{aligned}$ ∴ Largest number = x + 6 = 17 + 6= 23

47.

Three numbers are such that if the average of any two of them is added to the third number, the sums obtained are 164, 158 and 132 respectively. What is the average of the original three numbers?

(A) $75 \frac{2}{3}$
(B) 74
(C) 76
(D) $75 \frac{1}{3}$

Solution: $\begin{aligned} \frac{a + b}{2} + c = 164 \\ \frac{b + c}{2} + a = 158 \\ \frac{c + a}{2} + b = 132 \\ \overset{―}{\frac{2 (a + b + c)}{2} + a + b + c = 454} \\ a + b + c = 227 \\ Average = \frac{a + b + c}{3} = \frac{227}{3} = 75 \frac{2}{3} \end{aligned}$

48.

Find the average of all the numbers between 6 and 34 which are divisible by 5 -

(A) 18
(B) 20
(C) 24
(D) 30

Solution: Average $\begin{aligned} = (\frac{10 + 15 + 20 + 25 + 30}{5}) \\ = \frac{100}{5} \\ = 20 \end{aligned}$

49.

The average of the first nine prime numbers is-

(A) 9
(B) 11
(C) $11 \frac{1}{9}$
(D) $11 \frac{2}{9}$

Solution: Average $= (\frac{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23}{9})$ $\begin{aligned} = \frac{100}{9} \\ = 11 \frac{1}{9} \end{aligned}$

50.

If the average of 5 consecutive integers is x then, find the average of next to next 5 consecutive integers.

(A) x + 5
(B) x + 15
(C) x + 10
(D) x + 25

Solution: $\begin{aligned} Let numbers be a, a + 1, a + 2, a + 3, a + 4 \\ Next, a + 5, a + 6, a + 7, a + 8, a + 9 \\ Next to Next, a + 10, a + 11, a + 12, a + 13, a + 14 \\ I^{st} condition = \frac{5 a + 10}{5} = x \\ 5 a + 10 = 5 x . . . . . . (i) \\ I I^{nd} condition \\ = \frac{a + 10 + a + 11 + a + 12 + a + 13 + a + 14}{5} \\ = \frac{5 a + 50 + 10}{5} \\ From equation (i) \\ = \frac{5 x + 50}{5} \\ = x + 10 \end{aligned}$

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