Practice MCQ Questions and Answer on Average

Average age of 6 sons of a family is 8 years. Average age of sons together with their parents is 22 years. If the father is older than the mother by 8 years, the age of mother (in years) is -

(A) 44
(B) 52
(C) 60
(D) 68

There is a number consisting of two digits, the digit in the unit's place is twice than digit in the ten's place and if 2 subtracted from the sum of the digits, the difference is equal to ${\frac{1}{6}}^{th}$ of the number. The number is-

(A) 23
(B) 25
(C) 26
(D) 24

Solution: Shortcut method: Do by option Let the number be 24 Sum of digits = 2 + 4 = 6 ⇒ 6 - 2 = $\frac{1}{6}$ × 24 = 4 = 4 matched So, 24 is answer

The average annual income (in Rs.) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then the average monthly income (in Rs.) of all the workers is-

(A) $\frac{S + T}{2}$
(B) $\frac{S + 11 T}{2}$
(C) $\frac{1}{11 S} + 1$
(D) $\frac{11 S + T}{12}$

Solution: Let the number of other workers be x Then, number of agricultural workers = 11x Total number of workers = 12x ∴ Average monthly income $\begin{aligned} = \frac{S \times 11 x + T \times x}{12 x} \\ = \frac{11 S + T}{12} \end{aligned}$

The sum of three consecutive even numbers is 28 more than the average of these three number. Then the smallest of these number is -

(A) 6
(B) 16
(C) 12
(D) 14

Solution: Let consecutive even number x, x + 2, x + 4 According to the question, (x + x + 2 + x + 4) - $\frac{(x + x + 2 + x + 4)}{3}$ $\frac{3 (3 x + 6) - (3 x + 6)}{3}$ = 28 2 (3x + 6) = 28 × 3 3x + 6 = 42 3x = 36 x = 12 So, smallest number = x = 12

The weight of a person was entered incorrectly as 83 kg instead of 63 kg. As a result, the average weight of a group of people increased by 500 gm. What is the total number of people in the group?

(A) 44
(B) 48
(C) 40
(D) 50

Solution: Let, number of peoples in group = x Average = n ⇒ nx - 63 + 83 = n(x + 500 gm) ⇒ nx + 20 = nx + $\frac{n}{2}$ kg ⇒ n = 20 × 2 ⇒ n = 40 OR $\begin{aligned} Average = \frac{Sum}{Number} \\ 500 gm = \frac{20}{n} \\ \frac{1}{2} = \frac{20}{n} \\ n = 40 \end{aligned}$

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

The average of the three numbers x, y and z is 45. x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is :

(A) 3
(B) 5
(C) 7
(D) 8

Solution: According to the question, $\begin{aligned} \Rightarrow \frac{x + y + z}{3} = 45 \\ \Rightarrow x + y + z = 135. . . . . (i) \\ \Rightarrow x = \frac{y + z}{2} + 9 \\ \Rightarrow 2 x - y - z = 18. . . . . (i i) \\ x + y + z = 135 \\ \underset{―}{2 x - y - z = 18} \\ 3 x = 153 \\ x = 51 \end{aligned}$ From (i) y + z = 135 - 51 = 84.....(iii) Also, $\begin{aligned} \Rightarrow \frac{y + z}{2} = y + 2 \\ \Rightarrow y + z = 2 y + 4 \\ \Rightarrow z - y = 4 \\ + y + z = 84 \\ \underset{―}{- y + z = 4} \\ 2 z = 88 \\ z = 44 \end{aligned}$ Required difference = 51 - 44 = 7

The average of X₁, X₂ and X₃ is 14. Twice the sum of X₂ and X₃ is 30. What is the value of X₁?

(A) 12
(B) 16
(C) 20
(D) 27

Solution: $\begin{aligned} X_{1} + X_{2} + X_{3} \\ = (14 \times 3) \\ = 42 \end{aligned}$ According to the question, $\begin{aligned} 2 (X_{2} + X_{3}) = 30 \\ \Rightarrow X_{2} + X_{3} = 15 \\ ∴ X_{1} = (42 - 15) = 27 \end{aligned}$

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

10.

The average weight of 47 balls is 4 g. if the weight of the bag (in which the balls are kept) be included; the calculated average weight per ball increases by 0.3 g. What is the weight of the bag?

(A) 14.8 g
(B) 14.4 g
(C) 15 g
(D) 18.6 g

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Practice MCQ Questions and Answer on Average

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