Practice MCQ Questions and Answer on Average

411.

The ratio of the number of players in the three cricket teams A, B and C is 2 : 5 : 3. If the ratio of number of runs scored per player for each of the three teams A, B and C, is 30 : 17 : 25 respectively, then what is the average number of runs scored per player across all the three teams collectively?

(A) 20
(B) 21
(C) 22
(D) 23

Solution: $\begin{aligned} A : B : C = 2 : 5 : 3 \\ Average = \frac{30 \times 2 + 17 \times 5 + 25 \times 3}{10} \\ = \frac{60 + 85 + 75}{10} \\ = 22 \end{aligned}$

412.

The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

(A) 0
(B) 1
(C) 10
(D) 19

413.

The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.

(A) 25
(B) 26
(C) 27
(D) 28

414.

The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is

(A) 21 years
(B) 18 years
(C) 15 years
(D) 9 years

415.

A student finds the average of ten 2 digits numbers. While copying numbers, by mistake, he writes in number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of digits of the number, in which he made mistake is ?

(A) 2
(B) 3
(C) 4
(D) 6

Solution: Let us consider by mistake he writes 10th number with its digits interchanged. $∴ \frac{10 x + y - (10 y + x)}{10} = 18$ (In this remaining nine numbers are same and they cancel out) ∴ 10x + y - 10y - x = 18 ⇒ 9x - 9y = 18 ⇒ x - y = 2

416.

The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then find the average for the last four matches.

(A) 33.25
(B) 33.5
(C) 34.25
(D) 35

Solution: Total sum of last 4 matches, = (10 × 38.9) - (6 × 42) = 389 - 252 = 137 Average = $\frac{137}{4}$ = 34.25

417.

The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then find the average for the last four matches.

(A) 33.25
(B) 33.5
(C) 34.25
(D) 35

Solution: Total sum of last 4 matches, = (10 × 38.9) - (6 × 42) = 389 - 252 = 137 Average = $\frac{137}{4}$ = 34.25

418.

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

419.

Six numbers are arranged in decreasing order. The average of the first five numbers is 30 and the average of the last five numbers is 25. The difference of the first and the last numbers is

(A) 20
(B) 25
(C) 5
(D) 30

420.

The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:

(A) 17 kg.
(B) 20 kg.
(C) 26 kg.
(D) 31 kg.

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Practice MCQ Questions and Answer on Average

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