ExamVeda
Login
Home

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-4

Each section contains 100 MCQs on Average .
1
The average age of 9 students and their teacher is 16 years. The average age of the first four students is 19 years and that of the last five is 10 years. The teacher's age is -
Discuss
Answer & Solution
Answer: Option B
Solution:
According to the question,
Average age of nine students and teacher = 16 years
Then, the total average age of students and teacher = 16 × 10 = 160
And, average age of first 4 students = 19 × 4 = 76
Average age of last 5 students = 10 × 5 = 50
∴ Teacher's age = 160 - 76 - 50 = 34 years
2
Total weekly emoluments of the workers of a factory is Rs. 1534. Average weekly emolument of a worker is Rs. 118. The number of workers in the factory is :
Discuss
Answer & Solution
Answer: Option C
Solution:
According to the question,
Total weekly emoluments the workers = Rs. 1534
Number of workers
= $$\frac{1534}{118}$$
= 13
3
B was born when A was 4 years 7 months old and C was born when B was 3 years 4 months old. When C was 5 years 2 months old, then their average age was :
Discuss
Answer & Solution
Answer: Option D
Solution:
According to the question,
A - B = 4y 7m . . . . . . . (i)
B - C = 3y 4m . . . . . . . (ii)
  (+)  (+)   (+)
A - C = 7y 11m . . . . . . . (ii)

Given :
When, C = 5 years 2 months
∴ A = 13 years 1 month
B = 8 years 6 months
∴ Average of $$\frac{A + B + C}{3}$$
$$ = \frac{{{\text{26 years 9 months}}}}{3}$$
= 8 years 11 months
4
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:
Max - Min = 172.....(i)
Max + Min = 50 × 40 - 48 × 38
Max + Min = 2000 - 1824
Max + Min = 176.....(ii)
From equation (i) and (ii)
Max = 174
5
The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number the average becomes 17.5 then the number of natural numbers is -
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the number of natural numbers = n
∵ The average of some natural numbers = 15
⇒ Sum of these natural number = 15 × n = 15n
∵ 30 is added and 5 is subtracted
So, now addition of these number
= 15n + 30 - 5
= 15n + 25
According to the question,
⇒ $$\frac{15n + 25}{n}$$ = 17.5
⇒ 15n + 25 = 17.5n
⇒ 2.5n = 25
⇒ n = 10
Therefore, the numbers of natural numbers n = 10
6
A student finds the average of ten 2 digits numbers. While copying numbers, by mistake, he writes in number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of digits of the number, in which he made mistake is ?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let us consider by mistake he writes 10th number with its digits interchanged.
$$\therefore \frac{{10x + y - \left( {10y + x} \right)}}{{10}} = 18$$
(In this remaining nine numbers are same and they cancel out)
∴ 10x + y - 10y - x = 18
⇒ 9x - 9y = 18
⇒ x - y = 2
7
The average marks obtained by 22 candidate in an examination are 45. The average marks of the first 10 candidates is 55 and those of the last eleven is 40. The number of marks obtained by the eleventh candidate is ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Given that average marks obtained by 22 students = 45
And average of first 10 students = 55
Average of last 11 students = 40
So using, average = $$\frac{{{\text{sum}}}}{{{\text{number of elements}}}}$$
Sum of marks of all 22 students
= 22 x 45
= 990
Sum of students of first 10 students
= 55 x 10
= 550
Sum of last 11 students
= 11 x 40
= 440
So marks obtained by 11th student
= 990 - (550 + 440)
= 990 - 990
= 0
8
Average of first five odd multiples of 3 is -
Discuss
Answer & Solution
Answer: Option C
Solution:
According to the question,
First five odd multiples of 3 is = 3, 9, 15, 21, 27
∴ Average
= $$\frac{3 + 9 + 15 + 21 + 27}{5}$$
= $$\frac{75}{5}$$
= 15
9
The average of 50 numbers is 38. If two numbers, namely 45 and 55 are discarded, the average of the remaining numbers is ?
Discuss
Answer & Solution
Answer: Option A
Solution:
According to the question,
Average of 50 numbers is = 38
Sum of 50 numbers is = 38 × 50 = 1900
Two numbers discarded = 45 + 55 = 100
Sum of 48 numbers = 1900 - 100 = 1800
∴ Average = $$\frac{1800}{48}$$   = 37.5
10
The average of a collection of 20 measurements was calculated to tb 56 cm. But later it was found that a mistake had occurred in one of the measurement which was recorded as 64 cm., but should have been 61 cm. The correct average must be -
Discuss
Answer & Solution
Answer: Option C
Solution:
According to the question,
Correct average
= $$\frac{20 × 56 - 64 + 61}{20}$$  
= $$\frac{1120 - 3}{20}$$  
= $$\frac{1117}{20}$$  
= 55.85 cm