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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-9 section-1

Each section contains 100 MCQs on Average .
81
A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
Discuss
Answer & Solution
Answer: Option A
Solution:
Total quantity of petrol consumed in 3 years
$$\eqalign{ & = \left( {\frac{{4000}}{{7.50}} + \frac{{4000}}{8} + \frac{{4000}}{{8.50}}} \right){\text{litres}} \cr & = 4000\left( {\frac{2}{{15}} + \frac{1}{8} + \frac{2}{{17}}} \right){\text{litres}} \cr & = {\frac{{76700}}{{51}}} {\text{ litres}} \cr & {\text{Total}}\,{\text{amount}}\,{\text{spent}} \cr & = Rs.\,\left( {3 \times 4000} \right) \cr & = Rs.\,12000 \cr & \therefore {\text{Average}}\,{\text{Cost}} \cr & = Rs.\, {\frac{{12000 \times 51}}{{76700}}} \cr & = Rs.\,\frac{{6120}}{{767}} \cr & = Rs.\,7.98 \cr} $$
82
In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
Discuss
Answer & Solution
Answer: Option A
Solution:
Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70
According to Arun's mother, X < 68
The values satisfying all the above conditions are 66, 67 and 68
$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{66 + 67 + 68}}{3}} \cr & = {\frac{{201}}{3}} \cr & = 67\,kg. \cr} $$
83
The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31
Therefore B's weight = 31 kg.
84
The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{50.25 \times 16 + 45.15 \times 8}}{{16 + 8}}} \cr & = {\frac{{804 + 361.20}}{{24}}} \cr & = \frac{{1165.20}}{{24}} \cr & = 48.55 {\text{ kg.}} \cr} $$
85
A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
Discuss
Answer & Solution
Answer: Option D
Solution:
Since the month begins with a Sunday, to there will be five Sundays in the month.
$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$
86
If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & \therefore {\text{Required}}\,{\text{average}} \cr & = {\frac{{55 \times 50 + 60 \times 55 + 45 \times 60}}{{55 + 60 + 45}}} \cr & = {\frac{{2750 + 3300 + 2700}}{{160}}} \cr & = \frac{{8750}}{{160}} \cr & = 54.68 \cr} $$
87
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 $$\frac{1}{2}$$ . The number of pupils in the class is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{there}}\,{\text{be}}\,x\,{\text{pupils}}\,{\text{in}}\,{\text{the}}\,{\text{class}}{\text{.}} \cr & {\text{Total}}\,{\text{increase}}\,{\text{in}}\,{\text{marks}} \cr & = {x \times \frac{1}{2}} = \frac{x}{2} \cr & \therefore \frac{x}{2} = \left( {83 - 63} \right) \cr & \Rightarrow \frac{x}{2} = 20 \cr & \Rightarrow x = 40 \cr} $$
88
The marks of six boys in a group are 48, 59, 87, 37, 78 and 57. What are the average marks of all six boys?
Discuss
Answer & Solution
Answer: Option A
Solution:
Total marks of six boys
= 48 + 59 + 87 + 37 + 78 + 57
= 366
Required Average
= $$\frac{{366}}{6}$$
= 61
89
Six numbers are arranged in decreasing order. The average of the first five numbers is 30 and the average of the last five numbers is 25. The difference of the first and the last numbers is
Discuss
Answer & Solution
Answer: Option B
Solution:
Numbers are
x > y > z > p > q > r
According to the question,
Average of first five numbers = 30
Sum of first five numbers
= a + y + z + p + q = 5 × 30 = 150.....(i)
Average of last five numbers
= y + z + p + q + r = 5 × 25 = 125.....(ii)
By equation (i) and (ii)
a - r = 150 - 125 = 25
90
The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the rest?
Discuss
Answer & Solution
Answer: Option B
Solution:
Average of 12 numbers = 15
Total of 12 numbers = 15 × 12 = 180
Average of first two number = 14
Total of first two number = 14 × 2 = 28
Total of remaining ten numbers = 180 - 28 = 152
Required average of remaining ten number
$$\eqalign{ & = \frac{{152}}{{10}} \cr & = \frac{{76}}{5} \cr & = 15\frac{1}{5} \cr} $$