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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-5 section-5

Each section contains 100 MCQs on Average .
41
The average weight of a certain number of students in a class is 68.5 kg. If 4 new students having weights 72.2 kg, 70.8 kg, 70.3 kg and 66.7 kg join the class, then the average weight of all the students increases by 300 g. The number of students in the class, initially, is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let total students}} = x \cr & \frac{{68.5x + 280}}{{x + 4}} = 68.8 \cr & 68.5x + 280 = 68.8x + 275.2 \cr & 0.3x = 4.8 \cr & x = 16 \cr} $$
42
Average age of 20 student is 21 years. 2 students leave the group and 1 new student joins the group. The average now becomes 20 years. If age of one of the student who left the group is 26 years and the one who joined is 20 years, then what is the age (in years) of other student who left the group?
Discuss
Answer & Solution
Answer: Option B
Solution:
Average age of 20 students = 21 years
Total age of 20 students = 21 × 20 = 420
2 students leave and one student come in group.
Total age of 19 students = 19 × 20 = 380 years
Student left
420 - A - B + N = 380
26 + B - 20 = 40
B = 40 - 6 = 34 years
43
The average of n number is 45. If 60% of the number are increased by 5 each and the remaining numbers are decreased by 10 each, then what is the average of the number so obtained?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Average of }}n{\text{ numbers}} = 45 \cr & {\text{Let, }}n = 100 \cr & {\text{New average}} \cr & = \frac{{100 \times 45 + 60 \times 5 - 40 \times 10}}{{100}} \cr & = \frac{{100 \times 45}}{{100}} + \frac{{300 - 400}}{{100}} \cr & = 45 - 1 \cr & = 44 \cr} $$
44
Rita buys 5 sarees at an average cost of Rs. 2250. If she buys three more at an average cost of Rs. 2750, what will be the average (in Rs.) of all the sarees she buys?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Average of all the sarees}} \cr & = \frac{{2250 \times 5 + 2750 \times 3}}{8} \cr & = \frac{{11250 + 8250}}{8} \cr & = \frac{{19500}}{8} \cr & {\text{Average of all the sarees}} = 2437.5 \cr} $$
45
The average of three consecutive odd numbers is 52 more than $${\frac{1}{3}^{{\text{rd}}}}$$ of the largest of these numbers. What is the smallest of these numbers?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let the three consecutive odd numbers are
$$\eqalign{ & x,\,x + 2,\,x + 4 \cr & \frac{{x + x + 2 + x + 4}}{3} = \frac{1}{3}\left( {x + 4} \right) + 52 \cr & \frac{{3x + 6}}{3} = \frac{{x + 4 + 156}}{3} \cr & 3x + 6 = x + 4 + 156 \cr & 2x = 154 \cr & x = 77 \cr} $$
46
The average of twelve numbers is 39. The average of the last five numbers is 35, and that of the first four numbers is 40. The fifth number is 6 less than the sixth number and 5 more than the seventh number. The average of the sixth and seventh numbers is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Sum of 12 numbers = 12 × 39 = 468
Sum of last 5 numbers = 5 × 35 = 175
Sum of first 4 numbers = 4 × 40 = 160
\[\begin{array}{*{20}{c}} {{5^{{\text{th}}}}}&{{6^{{\text{th}}}}}&{{7^{{\text{th}}}}} \\ {\left( {x - 6} \right)}&{\left( x \right)}&{\left( {x - 11} \right)} \end{array}\]
$$\eqalign{ & 3x - 17 = 468 - \left( {175 + 160} \right) \cr & 3x = 133 + 17 \cr & x = 50 \cr & {\text{Average of }}{{\text{6}}^{{\text{th}}}}{\text{ and }}{{\text{7}}^{{\text{th}}}}{\text{ number}} \cr & = \frac{{x + \left( {x - 11} \right)}}{2} \cr & = x - 5.5 \cr & = 50 - 5.5 \cr & = 44.5 \cr} $$
47
Find the average of the first 20 multiples of 9.
Discuss
Answer & Solution
Answer: Option B
Solution:
First 20 multiple of 9
Average mcq question image
$$\eqalign{ & {\text{Average}} = \frac{{90 + 99}}{2} \cr & = 45 + 49.5 \cr & = 94.5{\text{ Answer}} \cr & \cr & {\bf{Alternate \,Solution:}} \cr & {\text{Average}} = \frac{{{\text{First term}} + {\text{Last term}}}}{2} \cr & = \frac{{9 + 180}}{2} \cr & = 4.5 + 90 \cr & = 94.5 \cr} $$
48
Average temperature of a week is 30°C. If the average of first 4 days was 31°C, then average temperature of the remaining days will be:
Discuss
Answer & Solution
Answer: Option D
Solution:
Total temperature of the week = 30°C × 7 = 210°C
Total temperature first four day = 31°C × 4 = 124°C
Average temperature of remaining day
$$\eqalign{ & = \frac{{{{210}^ \circ }{\text{C}} - {{124}^ \circ }{\text{C}}}}{3} \cr & = \frac{{{{86}^ \circ }{\text{C}}}}{3} \cr & = {28.67^ \circ }{\text{C}} \cr} $$
49
The average weight of a certain number of students in a group is 72 kg. If 10 students having an average weight of 78 kg leave and 4 students having an average weight of 80 kg join the group, the average weight of the students in the group decreases by 0.7 kg. The number of students initially in the group is:
Discuss
Answer & Solution
Answer: Option B
Solution:
Let the number of students = x
Average of x students = 72
Total weight = 72x
Change in weight
= 10 × 78 - 4 × 80
= 460
Now, total students = x - 6
$$\eqalign{ & {\text{Average}} = \frac{{72x - 460}}{{x - 6}} \cr & 71.3 = \frac{{72x - 460}}{{x - 6}} \cr & 0.7x = 32.2 \cr & x = 46 \cr} $$
50
The average of prime numbers between 1 and 20 is-
Discuss
Answer & Solution
Answer: Option B
Solution:
The prime number between 1 and 20 is 2, 3, 5, 7, 11, 13, 17, 19
∴ Required average
$$\eqalign{ & = \frac{{{\text{sum of prime numbers}}}}{8} \cr & = \frac{{2 + 3 + 5 + 7 + 11 + 13 + 17 + 19}}{8} \cr & = \frac{{77}}{8} \cr & = 9\frac{5}{8} \cr} $$