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-2 section-5

Each section contains 100 MCQs on Average .
11
A scored 73, 76, 20 and 7 runs in four out of five innings. What should be his score in fifth innings. If he has to make an average of 55 run in five innings.
Discuss
Answer & Solution
Answer: Option A
Solution:
Let average of 5th inning is 'x'.
Average of five innings = 55   (Given)
$$\eqalign{ & \frac{{73 + 76 + 20 + 7 + x}}{5} = 55 \cr & 176 + x = 275 \cr & x = 275 - 176 \cr & \therefore \,x = 99 \cr} $$
12
Sonali applied for a job of Science teacher in a school. In the test for job, she scored 8 in Physics, 8 in Chemistry, 6 in Biology, and 6.5 in the interview. For calculating the final score, weightage of 2, 3, 3 and 4 were assigned to Physics, Chemistry, Biology and interview, respectively. What is the weighted average score of Sonali?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Average score}} \cr & = \frac{{8 \times 2 + 8 \times 3 + 6 \times 3 + 6.5 \times 4}}{{2 + 3 + 3 + 4}} \cr & = \frac{{16 + 24 + 18 + 26}}{{12}} \cr & = \frac{{84}}{{12}} \cr & = 7 \cr} $$
13
The average weight of A, B and C is 65 kg. If the average weight of C and B is 61.5 kg, and the average weight of A and C is 68.5 kg, then the weight of C (in kg) is:
Discuss
Answer & Solution
Answer: Option A
Solution:
A + B + C = 65 × 3 = 195 . . .. . . . (i)
B + C = 61.5 × 2 = 123 . . . . . . . . (ii)
A + C = 68.5 × 2 = 137 . . . . . . . . (iii)
C = (123 + 137) - 195 = 65
14
The average of 5 consecutive odd numbers is 27. What is the product of the first and the last number?
Discuss
Answer & Solution
Answer: Option C
Solution:
Let first odd number = a
Five odd number = a, a + 2, a + 4, a + 6, a + 8
= 5a + 2(1 + 2 . . . 4)
= 5a + 2$$\frac{{\left( 4 \right)\left( 5 \right)}}{2}$$
= 5a + 20
According to the question,
$$\frac{{5a + 20}}{5}$$ = 27
a + 4 = 27
a = 23
First number = 23
Last number = 23 + 8 = 31
Product of first and last number = 23 × 31 = 713

Alternate:
\[\begin{array}{*{20}{c}} {23}&{25}&{\boxed{27}}&{29}&{31} \\ \downarrow &{}& \downarrow &{}& \downarrow \\ {{\text{First Number}}}&{}&{{\text{Average}}}&{}&{{\text{Last Number}}} \end{array}\]
Average is middle term if common difference between them is same.
Product = 23 × 31 = 713
15
The average age of a class of 6 girls is x years. Four new girls having ages x - 2, x + 2, x + 4 and x + 6 join the class. What is the new average age (in years) of the class?
Discuss
Answer & Solution
Answer: Option A
Solution:
Total age of a class of 6 girls = 6x
Four new girls ages = 4x + 10
Average age of 10 girls
$$\eqalign{ & = \frac{{10x + 10}}{{10}} \cr & = x + 1 \cr} $$
16
Find the average of cubes of first 49 positive integers.
Discuss
Answer & Solution
Answer: Option A
Solution:
Sum of cubes of first $$n$$ positive consecutive numbers is
$$\eqalign{ & = \frac{{{{\left\{ {n\left( {n + 1} \right)} \right\}}^2}}}{4} \cr & \therefore \,{\text{Average}} = \frac{{n{{\left( {n + 1} \right)}^2}}}{4} \cr & \Rightarrow n = 49 \cr & \Rightarrow \frac{{49{{\left( {50} \right)}^2}}}{4} = 30625 \cr} $$
17
The average revenues of 7 consecutive years is Rs. 83 lakhs. The average of first 4 years is Rs. 78 lakhs and that of last 4 years is Rs. 90 lakhs, find the revenue for the 4th year.
Discuss
Answer & Solution
Answer: Option A
Solution:
Average revenue for 4th year = [Sum of total revenue for first 4 years and last 4 years - Total revenue for 7 years]
= [4 × 78 + 4 × 90] - [7 × 83]
= 672 - 581
= 91 lakhs
18
The average of the marks of 25 students in a class, in an examination was calculated to be 19. Later, the teacher realized that the marks of two students were taken as 18 and 19 respectively, instead of 14 and 15. Find the new actual average marks of the class.
Discuss
Answer & Solution
Answer: Option C
Solution:
Total marks = 25 × 19 = 475
Error = (18 + 19) - (14 + 15)
= 37 - 29
= 8
Actual marks = 475 - 8 = 467
Average marks $$ = \frac{{467}}{{25}} = 18.68$$
19
If the average of s number is r4 and the average of r numbers is s4, then find the average of all r + s numbers?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Average}} = \frac{{{\text{Sum}}}}{{{\text{Number}}}} \cr & {\text{Sum}} = {r^4} \times s \cr & {\text{Again sum}} = r \times {s^4} \cr & {\text{Average of all }}r{\text{ and }}s{\text{ numbers}} \cr & = \frac{{{r^4} \times s + r \times {s^4}}}{{r + s}} \cr & = \frac{{rs\left( {{r^3} + {s^3}} \right)}}{{r + s}} \cr & = \frac{{rs\left( {r + s} \right)\left( {{r^2} + {s^2} - rs} \right)}}{{r + s}} \cr & = rs\left( {{r^2} + {s^2} - rs} \right) \cr} $$
20
The average of 22 number is 52. The average of the first 8 numbers is 48 and the average of next 11 number is 54 the 20th number is 7 less than the 21st number and 21st number is 4 more than 22nd number. What is the average of the 20th and 22nd number?
Discuss
Answer & Solution
Answer: Option D
No explanation is given for this question. Let's Discuss on Board