Practice MCQ Questions and Answer on Average

411.

The average of the two-digit numbers, which remain the same when the digits interchange their positions, is-

(A) 33
(B) 44
(C) 55
(D) 66

Solution: Average $= \frac{11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99}{9}$ $= \frac{(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55}{9}$ $\begin{aligned} = \frac{4 \times 110 + 55}{9} \\ = \frac{495}{9} \\ = 55 \end{aligned}$

412.

The average age of 9 students and their teacher is 16 years. The average age of the first four students is 19 years and that of the last five is 10 years. The teacher's age is -

(A) 36 years
(B) 34 years
(C) 30 years
(D) 28 years

413.

Total of ArunÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂs marks in Sanskrit and Mathematics together are 80 more than his marks in Science. His average marks in the three subjects are 100. What are his marks in Science?

(A) 80
(B) 110
(C) 120
(D) Data inadequate

414.

A student obtained the following marks in percentage in his semester examination English 50, Maths 65, Statistics 70, Economics 58 and Accountancy 63. The weights of these subjects are 2, 2, 1, 1 and 1 respectively. What is the weighted arithmetic mean?

(A) 60
(B) 61
(C) 62
(D) 63

Solution: Weighted arithmetic mean $\begin{aligned} = \frac{50 \times 2 + 65 \times 2 + 70 \times 1 + 58 \times 1 + 63 \times 1}{2 + 2 + 1 + 1 + 1} \\ = \frac{100 + 130 + 70 + 58 + 63}{7} \\ = \frac{421}{7} \\ = 60.14 \approx 60 \end{aligned}$

415.

19 people went to a hotel for combine dinner party 13 of them spent Rs. 79 each on their dinner and rest spent 4 more than the average expenditure of all the 19. What was the total money spent by them.

(A) 1628.4
(B) 1534
(C) 1492
(D) 1496

416.

Mean of 10 numbers is 30. Later on it was observed that numbers 15, 23 are wrongly taken as 51, 32. The correct mean is :

(A) 25.5
(B) 32
(C) 30
(D) 34.5

Solution: According to the question, Mean of 10 numbers is = 30 ∴ Sum of 10 numbers is = 300 It was observed that numbers 15, 23 are wrongly taken as 51, 32 Difference = (51 + 32) - (15 + 23) = 83 - 38 = 45 (more) ∴ Actual sum of 10 numbers = 300 - 45 = 255 ∴ Actual average of 10 numbers = $\frac{255}{10}$ = 25.5

417.

The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is-

(A) 35 years
(B) 40 years
(C) 50 years
(D) None of these

418.

A cricketer has a mean score of 60 runs in 10 innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62 ?

(A) 83
(B) 82
(C) 80
(D) 81

Solution: Let the score in the eleventh innings = x According to the question, ⇒ $\frac{60 \times 10 + x}{11}$ = 62 ⇒ 600 + x = 682 ⇒ x = 82

419.

A man travels equal distances of his journey at 40, 30 and 15 km/h. respectively. Find his average speed for whole journey.

(A) 24
(B) 25
(C) 27
(D) 28

Solution: Required average speed, $= \frac{(3 \times 40 \times 30 \times 15)}{(40 \times 30) + (40 \times 15) + (30 \times 15)}$ $= 24 km/hr$ Alternatively Time taken to traveled $\frac{1}{3}$ distance of journey with speed 40 kmph, $= \frac{\frac{1}{3}}{40} = \frac{1}{120}$ Time taken to traveled $\frac{1}{3}$ distance of journey with speed 30 kmph, $= \frac{\frac{1}{3}}{30} = \frac{1}{90}$ Time taken to traveled $\frac{1}{3}$ distance of journey with speed 15 kmph, $\begin{aligned} = \frac{\frac{1}{3}}{15} = \frac{1}{45} \\ Total time taken \\ = \frac{1}{120} + \frac{1}{90} + \frac{1}{45} \\ = \frac{45}{1080} \\ Average speed \\ = \frac{Total distance traveled}{Total time taken} \\ = \frac{1}{\frac{45}{1080}} \\ = 24 kmph \end{aligned}$

420.

The average temperature for the first four days of a week is 40.2ÃÂÃÂÃÂÃÂ° C and that of the last four days is 41.3ÃÂÃÂÃÂÃÂ° C. If the average temperature for the whole 40.6ÃÂÃÂÃÂÃÂ°C, then the temperature on the fourth day is-

(A) 40.8° C
(B) 41.3° C
(C) 41.8° C
(D) 38.5° C

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