Average - Aptitude MCQ Questions and Solutions with Explanations

21

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

Discuss

Answer & Solution

Answer: Option A

Solution:

Weighted arithmetic mean
$$\eqalign{ & = \frac{{50 \times 2 + 65 \times 2 + 70 \times 1 + 58 \times 1 + 63 \times 1}}{{2 + 2 + 1 + 1 + 1}} \cr & = \frac{{100 + 130 + 70 + 58 + 63}}{7} \cr & = \frac{{421}}{7} \cr & = 60.14 \approx 60 \cr} $$

22

Visitor to a show were charged Rs. 15 each on the first day, Rs. 7.50 each on the second day and Rs. 2.50 each on the third day. The attendance on the three days was in the ratio 2 : 5 : 13. The average charge per person for the whole show was-

A) Rs. 5 B) Rs. 6.33 C) Rs. 7.50 D) Rs. 9

Discuss

Answer & Solution

Answer: Option A

Solution:

Let the attendance on the three days be 2x, 5x and 13x respectively.
Then, total charges
= Rs. (15 × 2x + 7.50 × 5x + 2.50 × 13x)
= Rs. (30x + 37.5x + 32.5x)
= Rs. 100x
∴ Average change per person
= Rs. $$\left( {\frac{{100x}}{{2x + 5x + 13x}}} \right)$$
= Rs. 5

23

A company produces on an average 4000 items per must for the first 3 months. How many items it must produce on an average per month over the next 9 months, to average 4375 items per month over the whole?

A) 4500 B) 4600 C) 4680 D) 4710

Discuss

Answer & Solution

Answer: Option A

Solution:

Required average
$$\eqalign{ & = \frac{{\left( {4375 \times 12} \right) - \left( {4000 \times 3} \right)}}{9} \cr & = \frac{{52500 - 12000}}{9} \cr & = \frac{{40500}}{9} \cr & = 4500 \cr} $$

24

In a cricket eleven, the average age of eleven players is 28 years. Out of these, the average age of three groups of three players each are 25 years, 28 years and 30 years respectively. If in these groups the captain and the youngest player are not included and the captain is eleven years older than the youngest player, what is the age of the captain?

A) 33 years B) 34 years C) 35 years D) 36 years

Discuss

Answer & Solution

Answer: Option C

Solution:

Sum of ages of the captain and the youngest player
= [(28 × 11) - (25 × 3) + (28 × 3) + (30 × 3)] years
= (308 - 249) years
= 59 years
Let the age of the youngest player be x years.
Then, age of the captain = (x + 11) years
∴ x + (x + 11) = 59
⇒ 2x = 48
⇒ x = 24
So, the age of captain is
= (x + 11)
= (24 + 11)
= 35 years

25

The mean monthly salary paid to 75 workers in a factory is Rs. 5680. The mean salary of 25 of them is Rs. 5400 and that of 30 others is Rs. 5700. The mean salary of the remaining workers is-

A) Rs. 5000 B) Rs. 6000 C) Rs. 7000 D) Rs. 8000

Discuss

Answer & Solution

Answer: Option B

Solution:

Required average
$$ = {\text{Rs}}{\text{.}}$$ $$\left[ {\frac{{\left( {5680 \times 75} \right) - \left\{ {\left( {5400 \times 25} \right) + \left( {5700 \times 30} \right)} \right\}}}{{75 - \left( {25 + 30} \right)}}} \right]$$
$$ = {\text{Rs}}{\text{. }}\left[ {\frac{{426000 - \left( {135000 + 171000} \right)}}{{20}}} \right]$$
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {\frac{{120000}}{{20}}} \right) \cr & = {\text{ Rs}}{\text{. 6000}} \cr} $$

26

There were 35 students in a hostel. If the number of the students is increased by 7, then the expenses of the mess increased by Rs. 42 per day, while the average expenditure per head diminishes by Rs. 1. The original expenditure of the mess per day was-

A) Rs. 400 B) Rs. 420 C) Rs. 432 D) Rs. 442

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the original expenditure of the mess per day be Rs. x
Then, new expenditure
= Rs. (x + 42)
$$\eqalign{ & \therefore \frac{x}{{35}} - \frac{{\left( {x + 42} \right)}}{{42}} = 1 \cr & \Rightarrow 6x - 5\left( {x + 42} \right) = 210 \cr & \Rightarrow x - 210 = 210 \cr & \Rightarrow x = 420 \cr} $$

27

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

Discuss

Answer & Solution

Answer: Option B

Solution:

Excluded number
= (18 × 5) - (16 × 4)
= 90 - 64
= 26

28

The average of the reciprocals of x and y is-

A) $$\frac{(x + y)}{(x - y)}$$ B) $$\frac{(x + y)}{2xy}$$ C) $$\frac{2(x + y)}{xy}$$ D) $$\frac{2xy}{(x + y)}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

Required average
$$\eqalign{ & = \frac{{\left( {\frac{1}{x} + \frac{1}{y}} \right)}}{2} \cr & = \frac{{x + y}}{{2xy}} \cr} $$

29

The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is-

A) 414 B) 420 C) 620 D) 621

Discuss

Answer & Solution

Answer: Option D

Solution:

Sum of numbers
= (41.4 × 15)
= 621

30

If the average of m numbers is n² and that of n numbers is m² , then the average of (m + n) numbers is-

A) m - n B) mn C) m + n D) $$\frac{m}{n}$$

Discuss

Answer & Solution

Answer: Option B

Solution:

Sum of m numbers = mn²
Sum of n numbers = nm²
∴ Average of (m + n) numbers
= $$\frac{mn(m + n)}{(m + n)}$$
= mn