Practice MCQ Questions and Answer on Average

When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students?

(A) 57
(B) 56.8
(C) 58.2
(D) 52.2

Solution: Let the average weight of the 59 students be X Therefore, the total weight of the 59 of them will be 59X The questions states that when the weight of this student who left is added, the total weight of the class = 59X + 45 When this student is also included, the average weight decreases by 0.2 kgs 59X + $\frac{45}{60}$ = X - 0.2 => 59X + 45 = 60X - 12 => 45 + 12 = 60X - 59X => X = 57

A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:

(A) 250
(B) 276
(C) 280
(D) 285

Solution: Since the month begins with a Sunday, to there will be five Sundays in the month. $\begin{aligned} ∴ Required average \\ = \frac{510 \times 5 + 240 \times 25}{30} \\ = \frac{8550}{30} \\ = 285 \end{aligned}$

The body weight of seven students of a class is recorded as 54 kg, 78 kg, 43 kg, 82 kg, 67 kg, 42 kg and 75 kg. What is the average body weight of call the seven students?

(A) 63 kg
(B) 69 kg
(C) 741 kg
(D) 73 kg

Solution: Average body weight $\begin{aligned} = \frac{54 + 78 + 43 + 82 + 67 + 42 + 75}{7} kg \\ = \frac{441}{7} kg \\ = 63 kg \end{aligned}$

A man bought 13 articles at Rs. 70 each, 15 at Rs. 60 each and 12 at Rs. 65 each. The average price per article is -

(A) Rs. 60.25
(B) Rs. 64.75
(C) Rs. 65.75
(D) Rs. 62.25

Solution: According to the question, $\begin{aligned} = \frac{13 \times 70 + 15 \times 60 + 12 \times 65}{40} \\ = \frac{910 + 900 + 780}{40} \\ = \frac{2590}{40} \\ = 64.75 \end{aligned}$

A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

(A) $28 \frac{4}{7} years$
(B) $31 \frac{5}{7} years$
(C) $32 \frac{1}{7} years$
(D) None of these

Solution: Required average $\begin{aligned} = \frac{67 \times 2 + 35 \times 2 + 6 \times 3}{2 + 2 + 3} \\ = \frac{134 + 70 + 18}{7} \\ = \frac{222}{7} \\ = 31 \frac{5}{7} years \end{aligned}$

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

Ras Bihari, a plumber, earned on an average Rs. 925 per day in the month of January. He earned on an average Rs. 881 per day during the first 20 days and Rs. 915 per day during the last 20 days. What was his average income (in Rs.) per day from 12^th January to 20^th January?

(A) 792
(B) 875
(C) 800
(D) 805

Solution: Total earning in January = 925 × 31 = Rs. 28675 Earning of first 20 days = 881 × 20 = Rs. 17620 Earning of last 20 days = 915 × 20 = Rs. 18300 Earning of 12th January to 20th January = (17620 + 18300) - 28675 = 35920 - 28675 = Rs. 7245 ∴ Average of earning of 12th January to 20th January $\begin{aligned} = \frac{7245}{9} \\ = Rs . 805 \end{aligned}$

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, $\begin{aligned} = \frac{\frac{1}{3}}{15} = \frac{1}{45} \\ Total time taken \\ = \frac{1}{120} + \frac{1}{90} + \frac{1}{45} \\ = \frac{45}{1080} \\ Average speed \\ = \frac{Total distance traveled}{Total time taken} \\ = \frac{1}{\frac{45}{1080}} \\ = 24 kmph \end{aligned}$

The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?

(A) 83
(B) 92
(C) 90
(D) 97

10.

Distance between two stations A and B is 778km. A train covers the journey from A to B at 84km per hour and returns back to A with a uniform speed of 56km per hour. Find the average speed of train during the whole journey.

(A) 60 km/hr
(B) 30.5 km/hr
(C) 57 km/hr
(D) 67.2 km/hr

Solution: $\begin{aligned} Average speed \\ = \frac{2 x y}{x + y} km/hr \\ = \frac{2 \times 84 \times 56}{84 + 56} km/hr \\ = \frac{2 \times 84 \times 56}{140} km/hr \\ = 67.2 km/hr \end{aligned}$

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