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Average - Aptitude MCQ Questions and Solutions with Explanations

Average MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Average with easy and logical explanations. Average MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-10 section-2

Each section contains 100 MCQs on Average .
91
The average weight of 3 men A, B, and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. The weight if A is-
Discuss
Answer & Solution
Answer: Option C
Solution:
Let A, B, C, D and E represent their respective weights.
Then,
A + B + C = (84 × 3) = 252 kg
A + B + C + D = (80 × 4) = 320 kg
∴ D = (320 - 252) kg = 68 kg
E = (68 + 3) kg = 71 kg
B + C + D + E = (79 × 4) = 316 kg
Now,
= (A + B + C + D) - (B + C + D + E)
= (320 - 316) kg
= 4 kg
∴ A - E = 4
⇒ A = (4 + E)
⇒ A = 75 kg
92
A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was-
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the total score be x
$$\eqalign{ & \therefore \frac{{x + 92 - 85}}{8} = 84 \cr & \Rightarrow x + 7 = 672 \cr & \Rightarrow x = 665 \cr} $$
93
A batsman makes a score of 84 runs in the 21st inning and thus increases his average by 2 runs. His average after 21st inning is-
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the average for 20 innings be x
$$\eqalign{ & \therefore \frac{{20x + 84}}{{21}} = x + 2 \cr & \Rightarrow 20x + 84 = 21x + 42 \cr & \Rightarrow x = 42 \cr} $$
∴ Average after 21st innings
= 42 + 2
= 44
94
When 15 is included in a list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let there be n numbers in the original list and let their mean be x.
Then, sum of n numbers = nx
$$\eqalign{ & \therefore \frac{{nx + 15}}{{n + 1}} = x + 2 \cr & \Rightarrow nx + 15 = \left( {n + 1} \right)\left( {x + 2} \right) \cr & \Rightarrow nx + 15 = nx + 2n + x + 2 \cr & \Rightarrow 2n + x = 13.....(i) \cr} $$
And,
$$\eqalign{ & \therefore \frac{{nx + 16}}{{n + 2}} = \left( {x + 2} \right) - 1 \cr & \Rightarrow nx + 16 = \left( {n + 2} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 16 = nx + n + 2x + 2 \cr & \Rightarrow n + 2x = 14.....(ii) \cr} $$
Solving (i) and (ii), we get:
n = 4, x = 5
95
The average weight of 8 men is increased by 1.5 kg when one of the men, who weight 65 kg is replaced by a new man. The weight of the new man is-
Discuss
Answer & Solution
Answer: Option D
Solution:
Total weight increased
= (8 × 1.5) kg
= 12 kg
Weight of the new man
= (65 + 12) kg
= 77 kg
96
The average temperature of the town in the first four days of month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperature of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day?
Discuss
Answer & Solution
Answer: Option A
Solution:
Sum of temperature on 1st, 2nd, 3rd and 4th days
= (58 × 4)
= 232 degrees
Sum of temperature on 2nd, 3rd, 4th and 5th days
= (60 × 4)
= 240 degrees
Subtracting (i) from (ii), we get :
Temperature on 5th day - Temperature on 1st day = 8 degrees
Let the temperature on 1st and 5th days be 7x and 8x degrees respectively.
Then,
⇒ 8x - 7x = 8
⇒ x = 8
∴ Temperature on the 5th day
= 8x
= 8 × 8
= 64 degrees
97
The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle number is :
Discuss
Answer & Solution
Answer: Option D
Solution:
Middle number
= [(10.5 × 6 + 11.4 × 6) - 10.9 × 11]
= (131.4 - 119.9)
= 11.5
98
A pupil’s marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is-
Discuss
Answer & Solution
Answer: Option C
Solution:
Let there be x pupils in the class
Total increase in marks
= x × $$\frac{1}{2}$$
= $$\frac{x}{2}$$
∴ $$\frac{x}{2}$$ = (83 - 63)
⇒ $$\frac{x}{2}$$ = 20
⇒ x = 40
99
The average marks in Science subject of a class of 20 students is 68. If the marks of two students were misread as 48 and 65 instead of the actual marks 72 and 61 respectively, what would be the correct average?
Discuss
Answer & Solution
Answer: Option C
Solution:
Correct sum
= (68 × 20 + 72 + 61 - 48 - 65)
= 1380
∴ Correct average
= $$\frac{1380}{20}$$
= 69
100
Average of ten positive numbers is $$\overline x $$. If each number is increased by 10%, then $$\overline x $$ -
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & \Rightarrow \frac{{{x_1} + {x_2} + ..... + {x_{10}}}}{{10}} = \overline x \cr & \Rightarrow {x_1} + {x_2} + ..... + {x_{10}} = 10\overline x \cr & \Rightarrow \frac{{110}}{{100}}{x_1} + \frac{{110}}{{100}}{x_2} + ..... + \frac{{110}}{{100}}{x_{10}} = \frac{{110}}{{100}} \times 10\overline x \cr & \Rightarrow \frac{{\frac{{110}}{{100}}{x_1} + \frac{{110}}{{100}}{x_2} + ..... + \frac{{110}}{{100}}{x_{10}}}}{{10}} = \frac{{11}}{{10}}\overline x \cr} $$
⇒ Average is increased by 10%