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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-1 section-3

Each section contains 100 MCQs on Average .
1
If the arithmetic mean of seventy-five numbers us calculated, it is 35. If each number is increased by 5, then mean of new numbers is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Arithmetic mean of 75 members = 35
Sum of 75 numbers
= (75 × 35) = 2625
Total increase
= (75 × 5) = 375
Increased sum
= (2625 + 375) = 3000
Increased average
= $$\frac{3000}{75}$$
= 40
2
A student obtained 60, 75 and 85 marks respectively in three monthly examinations in Physics and 95 marks in the final examination. The three monthly examinations are equal weighted whereas the final examination is weighted twice as much as a monthly examination. His average marks in physics are-
Discuss
Answer & Solution
Answer: Option C
Solution:
Average marks in Physics
$$\eqalign{ & = \frac{{60 \times 1 + 75 \times 1 + 85 \times 1 + 95 \times 2}}{{1 + 1 + 1 + 2}} \cr & = \frac{{60 + 75 + 85 + 190}}{5} \cr & = \frac{{410}}{5} \cr & = 82 \cr} $$
3
The average age of all the students of a class is 18 years. The average age of boys of the class is 20 years and that of the girls is 15 years. If the number of girls in the class is 20, then find the number of boys in the class.
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the number of boys in the class be x
Then,
⇒ 18(x + 20) = 20x + 15 × 20
⇒ 18x + 360 = 20x + 300
⇒ 2x = 60
⇒ x = 30
4
In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the rate in the remaining 40 overs to each the target of 282 runs?
Discuss
Answer & Solution
Answer: Option A
Solution:
Required run rate
$$\eqalign{ & = \frac{{282 - \left( {3.2 \times 10} \right)}}{{40}} \cr & = \frac{{250}}{{40}} \cr & = 6.25 \cr} $$
5
If the average of 5 numbers is 10, the number which should be added to make the average 12 is-
Discuss
Answer & Solution
Answer: Option C
Solution:
Required number
= (12 × 6) - (10 × 5)
= 72 - 50
= 22
6
The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the highest score be x
Then, lowest score = (x - 172)
Then,
⇔ (50 × 40) - [x + (x - 172)] = 38 × 48
⇔ 2x = 2000 + 172 - 1824
⇔ 2x = 348
⇔ x = 174
7
The average marks of a student in 8 subjects is 87. Of these, the highest marks are 2 more than the one next in value. If these two subjects are eliminated, the average marks of the remaining subjects are 85. What are the highest marks now obtained by him?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let the highest marks obtained by the student be x
Then, second highest marks = x - 2
Sum of marks of these 2 subjects
= (87 × 8) - (85 × 6)
= 696 - 510
= 186
∴ x + (x - 2) = 186
⇒ 2x = 188
⇒ x = 94
8
The average of 80 boys in a class is 15. The average age of a group of 15 boys in the class is 16 and the average age of another 25 boys in the class is 14. What is the average age of the remaining boys in the class?
Discuss
Answer & Solution
Answer: Option C
Solution:
Required average
$$\eqalign{ & = \frac{{\left( {15 \times 80} \right) - \left[ {\left( {16 \times 15} \right) + \left( {14 \times 25} \right)} \right]}}{{80 - \left( {15 + 25} \right)}} \cr & = \frac{{1200 - \left( {240 + 350} \right)}}{{40}} \cr & = \frac{{610}}{{40}} \cr & = 15.25 \cr} $$
9
The average of 6 observations is 45.5. If one new observation is added to the previous observations, then the new average becomes 47. The new observation is-
Discuss
Answer & Solution
Answer: Option C
Solution:
New observation
= (47 × 7) - (45.5 × 6)
= 329 - 273
= 56
10
If the average temperature of first four days of the week was 39°C and the average temperature of the week was 40°C, then what was the average temperature of the last three days of the week?
Discuss
Answer & Solution
Answer: Option C
Solution:
Required average
$$\eqalign{ & = {\left[ {\frac{{\left( {40 \times 7} \right) - \left( {39 \times 4} \right)}}{3}} \right]^ \circ }{\text{C}} \cr & = {\left( {\frac{{124}}{3}} \right)^ \circ }{\text{C}} \cr & = {41.3^ \circ }{\text{C}} \cr} $$