Average - Aptitude MCQ Questions and Solutions with Explanations

31

The average of the first 100 positive integers is -

A) 100 B) 51 C) 50.5 D) 49.5

Discuss

Answer & Solution

Answer: Option C

Solution:

According to the question,
As we know that average of 'n' positive integer is
= $$\frac{n(n + 1)}{2 × n}$$
= $$\frac{(n + 1)}{2}$$
Here n = 100
∴ $$\frac{100 + 1}{2}$$
= $$\frac{101}{2}$$
= 50.5

32

The average age of 11 players of a cricket team decreases by 2 months when two new players are included in the team replacing two players of age 17 years and 20 years. The average age of new players is :

A) 17 years 1 month B) 17 years 7 month C) 17 years 11 month D) 18 years 3 month

Discuss

Answer & Solution

Answer: Option B

Solution:

Sum of age of two players = 17 + 20 = 37 years
Decreasing age of players = 11 × 2 = 22 months
Sum of age of two new players
= 37 years - 22 months
= 35 years 2 months
Average = 17 years 7 months

Alternate :
$$\frac{{\left( {17 + 20} \right) - \left( {{\text{Sum of two new players}}} \right)}}{{11}}$$
= $$\frac{2}{12}$$
= $$\frac{1}{6}$$
Sum of two new players = $$\frac{211}{6}$$
Average of two new players
= $$\frac{211}{12}$$
= 17 years 7 months

33

The mean of 50 numbers is 30. Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean.

A) 36.12 B) 30.66 C) 29.28 D) 38.21

Discuss

Answer & Solution

Answer: Option C

Solution:

According to the question,
The mean of 50 numbers is = 30
Sum of 50 numbers is = 50 × 30 = 1500
Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31.
∴ Difference = (82 + 13) - (28 + 31)
= 95 - 59
= 36 (extra)
∴ Actual sum of 50 numbers is = 1500 - 36 = 1464
∴ Actual average = $$\frac{1464}{50}$$ = 29.28

34

Average of n numbers is a. The first number is increased by 2, second one is increased by 4, the third one is increased by 8 and so on. The average of the new numbers is -

A) $$a + \frac{{2\left( {{2^n} - 1} \right)}}{n}$$ B) $$a + \frac{{{2^{n + 1}} - 1}}{n}$$ C) $$a + \frac{{{2^{n + 1}}}}{n}$$ D) $$a + 2\frac{{{2^{n - 1}}}}{n}$$

Discuss

Answer & Solution

Answer: Option A

Solution:

Series:- a, a + 2, a + 4.....
Sum = na + 2 + 4 + ..... upto n terms
Sum = na + S_n
$${S_n} = \frac{{2\left( {{2^n} - 1} \right)}}{{2 - 1}}$$
Average = $$a + \frac{{2\left( {{2^n} - 1} \right)}}{n}$$

35

The average age of 30 boys in a class in 15 years. One boy, aged 20 years, left the class, but two new boys came in his place whose age differs by 5 years. If the average age of all the boys now in the class becomes 15 years, the age of the younger newcomer is :

A) 20 years B) 15 years C) 10 years D) 8 years

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the age of younger boy = x years
Then the age of older boy = (x + 5)
According to the question,
⇒ (30 × 15) - 20 + x + x + 5 = 31 × 15
⇒ 430 + 2x + 5 = 465
⇒ 2x = 30
⇒ x = 15

36

The average of 8 numbers is 20. The average of first two numbers is $$15\frac{1}{2}$$ and that of the next three is $$21\frac{1}{3}$$ . If the sixth number be less than the seventh and eighth numbers by 4 and 7 respectively, them the eighth number is :

A) 18 B) 22 C) 25 D) 27

Discuss

Answer & Solution

Answer: Option C

Solution:

Let the sixth number = x
Then the seventh = x + 4
And the eighth = x + 7
According to the question,
$$ \Rightarrow 2 \times \frac{{31}}{2} + 3 \times \frac{{64}}{3} + x + x + 4 + x + 7$$ = 8 × 20
$$\eqalign{ & \Rightarrow 31 + 64 + 3x + 11 = 160 \cr & \Rightarrow 106 + 3x = 160 \cr & \Rightarrow 3x = 54 \cr & \Rightarrow x = 18 \cr} $$
∴ Eighth number x + 7 = 18 + 7 = 25

37

The average age of a class of 39 students is 15 years. If the age of the teacher is included, then the average increases by 3 months. Find the age of the teacher.

A) 30 years B) 25 years C) 35 years D) 40 years

Discuss

Answer & Solution

Answer: Option B

Solution:

Average age of the class = 15 years
Average age of the class including teacher = 15 years 3 months
Teacher's age
= 15 × 40 + $$\frac{3}{12}$$ × 40 - 15 × 39
= 610 - 585
= 25 years

38

The mean of 9 observation is 16. One more observation is included and the new mean becomes 17. The 10^th observation is :

A) 9 B) 16 C) 26 D) 30

Discuss

Answer & Solution

Answer: Option C

Solution:

According to the question mean of 9 observations is = 16
Sum of all observations is = 16 × 9 = 144
When one more observation included the new mean = 17
Sum of 10 observations = 10 × 17 = 170
∴ 10^th observation = 170 - 144 = 26

39

The average weight of first 11 persons among 12 persons is 95 kg. The weight of 12^th person is 33 kg more than the average weight of all the 12 persons. The weight of the 12^th person is :

A) 128.75 B) 131 C) 128 D) 97.45

Discuss

Answer & Solution

Answer: Option B

Solution:

Let the average weight of 12 person is = x and weight of 12 persons
According to the question,
$$\eqalign{ & \Rightarrow \frac{{11 \times 95 + x + 33}}{{12}} = x \cr & \Rightarrow 1045 + x + 33 = 12x \cr & \Rightarrow 11x = 1078 \cr & \Rightarrow x = 98 \cr} $$
∴ The weight of 12^th person is = 98 + 33 = 131 kg

40

In the afternoon, a student read 100 pages at the rate of 60 pages per hour. In the evening, when she was tired, she read 100 more pages at the rate of 40 pages per hour. What was her average rate of reading the pages per hours ?

A) 60 B) 70 C) 48 D) 50

Discuss

Answer & Solution

Answer: Option C

Solution:

According to the question,
Afternoon
100 pages read at the rate 60 pages per hour
Total time taken to read 100 pages
= $$\frac{100}{60}$$
= $$\frac{5}{3}$$ hours

Evening
100 pages read at the rate 40 pages per hour
Total time taken to read 100 pages
= $$\frac{100}{40}$$
= $$\frac{5}{2}$$ hours
Average rate of reading the pages per hour
$$\eqalign{ & = \frac{{200}}{{\frac{5}{3} + \frac{5}{2}}} \cr & = \frac{{200 \times 6}}{{10 + 15}} \cr & = \frac{{200 \times 6}}{{25}} \cr & = 48 \cr} $$