Practice MCQ Questions and Answer on Average

151.

If a, b, c, d, e are five consecutive odd numbers, their average is

(A) 5 (a + 4)
(B) $\frac{abcde}{5}$
(C) 5 (a + b + c + d + e)
(D) None of these

Solution: Clearly, b = a + 2 c = a + 4 d = a + 6 e = a + 8 ∴ Average $\begin{aligned} = \frac{a + (a + 2) + (a + 4) + (a + 6) + (a + 8)}{5} \\ = (\frac{5 a + 20}{5}) \\ = (a + 4) \end{aligned}$

152.

The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the rest?

(A) 15
(B) $15 \frac{1}{5}$
(C) 14
(D) $14 \frac{1}{5}$

Solution: Average of 12 numbers = 15 Total of 12 numbers = 15 × 12 = 180 Average of first two number = 14 Total of first two number = 14 × 2 = 28 Total of remaining ten numbers = 180 - 28 = 152 Required average of remaining ten number $\begin{aligned} = \frac{152}{10} \\ = \frac{76}{5} \\ = 15 \frac{1}{5} \end{aligned}$

153.

The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?

(A) 13.5
(B) 14
(C) 15
(D) 12.5

Solution: At present the total age of the family = 5 × 20 =100 The total age of the family at the time of the birth of the youngest member, = 100 - 10 - (10 × 4) = 50 Therefore, average age of the family at the time of birth of the youngest member, $= \frac{50}{4} = 12.5$

154.

If a, b, c, d, e are five consecutive odd numbers, their average is :

(A) $5 (a + 4)$
(B) $\frac{abcde}{5}$
(C) $5 (a + b + c + d + e)$
(D) $a + 4$

Solution: Let five consecutive odd numbers are 1, 3, 5, 7, 9 Here, a = 1, b = 3, c = 5, d = 7, e = 9 According to the question, Average $\begin{aligned} = \frac{1 + 2 + 5 + 7 + 9}{5} \\ = \frac{25}{5} \\ = 5 \end{aligned}$ Now check the option Option (D) a + 4 Here a = 1 1 + 4 = 5 satisfy

155.

A shop of electronic goods is closed on Monday. The average daily sales for remaining six days of a week is Rs. 15,640 and the average sale of Tuesday to Saturday is Rs. 14,124. The sales on Sunday is -

(A) Rs. 20,188
(B) Data inadequate
(C) Rs. 23,220
(D) Rs. 21,704

156.

Find the weighted arithmetic mean of the first 'n' natural numbers, the weights being the corresponding numbers.

(A) $\frac{{n (n + 1) (2 n + 1)}}{6}$
(B) $\frac{{n (n + 1)}}{2}$
(C) $\frac{{(2 n + 1)}}{3}$
(D) $n$

Solution: $\begin{aligned} Weighted arithmetic mean \\ = \frac{1 \times 1 + 2 \times 2 + 3 \times 3 + . . . . . n \times n}{1 + 2 + 3 + . . . . . + n} \\ = \frac{n (n + 1) (2 n + 1) \times 2}{6 \times n (n + 1)} \\ = \frac{2 n + 1}{3} \end{aligned}$

157.

The average age of an adult class is 40 years. 12 new students with an average age of 32 years join the class, thereby decreasing the average by 4 years. The original strength of the class was-

(A) 10
(B) 11
(C) 12
(D) 15

Solution: Let the original strength of the class be x Sum of ages of the whole class = 40x years Sum of ages of 12 now students = (12 × 32) years= 384 years ∴ $\frac{40 x + 384}{x + 12}$ = 36 ⇒ 40x + 384 = 36x + 432 ⇒ 4x = 48 ⇒ x = 12 Hence, the original strength of the class = 12

158.

The average of 22 number is 52. The average of the first 8 numbers is 48 and the average of next 11 number is 54 the 20^th number is 7 less than the 21^st number and 21^st number is 4 more than 22^nd number. What is the average of the 20^th and 22^nd number?

(A) 52
(B) 52.5
(C) 53
(D) 53.5

159.

Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average of this eight-member joint family is nearest to-

(A) 21 years
(B) 22 years
(C) 23 years
(D) 24 years

Solution: Sum of the ages of 8 members, 10 years ago = 231 years Sum of the ages of all members, 7 years ago = (231 + 8 × 3 - 60) years = 195 years Sum of the ages of all members, 4 years ago = (195 + 8 × 3 - 60) years = 159 years Sum of the present ages of all 8 members = (159 + 8 × 4) years = 191 years ∴ Current average age = $\frac{191}{8}$ years = 23.8 years $\approx$ 24 years

160.

The average age of a cricket team of 11 player is 27 year. If two more players are included in the team the average become 26 years. Then the average age (in year) of the two included player is:

(A) 14.5
(B) 20.5
(C) 26
(D) 27

Solution: Sum of the age of 11 player = 27 × 11 = 297 Sum of age of 11 player + 2 more included player = 13 × 26 = 338 ∴ Average of the included player $= \frac{338 - 297}{2} = \frac{41}{2} = 20.5$

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Practice MCQ Questions and Answer on Average

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