Average Questions and Answers – Set 3

This set of Aptitude Questions and Answers (MCQs) focuses on “Average – Set 3”.

1. 5 shareholders of a company have 1290, 2302, 9802, 3202, 7480 shares of the company, respectively. If the company announces a bonus of 2 shares for every 4 shares held, what will be the new average of the shares held by these shareholders?
a) 7222.8
b) 722.28
c) 7288.2
d) 728.82
View Answer

Answer: a
Explanation: Let the shareholder name be a, b, c, d and e, respectively.
Shares owned by each after the bonus:
A: 1290 + 1290 / 4 * 2 = 1290 + 645 = 1935
B: 2302 + 2302 / 4 * 2 = 2302 + 1151 = 3453
C: 9802 + 9802 / 4 * 2 = 9802 + 4901 = 14703
D: 3202 + 3202 / 4 * 2 = 3202 + 1601 = 4803
E: 7480 + 7480 / 4 * 2 = 7480 + 3740 = 11220
Total shares held = 1935 + 3453 + 14703 + 4803 + 11220 = 36114
Average shares held = 36114 / 5 = 7222.8

2. Out of 20 students first 10 students scored 40% of the marks of the next 10 students. If the total score of the class was 700, find the average marks obtained by the first 10 students.
a) 10
b) 15
c) 18
d) 20
View Answer

Answer: d
Explanation: Total marks obtained by the class = 700
Let the marks obtained by the last 10 students be x.
Marks obtained by first 10 students = 40 % of x = 4x / 10
Total marks obtained by 20 students = x + 4x / 10 = 14x / 10
14x / 10 = 700
X = 700 / 14 * 10 = 500
Marks obtained by last 10 students = 500
Marks obtained by first 10 students = 500 / 10 * 4 = 200
Average marks of first 10 students = 200 / 10 = 20

3. What is the average of the prime numbers more than 60 but less then equal to 221?
a) 121.2
b) 123.2
c) 125.2
d) 127.2
View Answer

Answer: d
Explanation: Prime numbers less than 221 and more than 60 are 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163 ,167, 173, 179, 181, 191, 193, 197, 199 and 211.
Total of these numbers = 3817
Average = 3817 / 30 = 127.2
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4. What is the average of first 10 prime numbers, first 10 even numbers and first 10 odd numbers?
a) 10.3
b) 10.7
c) 11.3
d) 11.7
View Answer

Answer: c
Explanation: First 10 prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
First 10 even numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
First 10 odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Total of these numbers = 339
Average of these numbers = 339 / 30 = 11.3

5. Out of 25 students of a class, 12 failed the exam and has an average of 45 marks. From the passing students 3 got distinction. If the average of the class is 70 marks, passing marks are 50 and to get a distinction one must score 100 marks, find the average marks of the students who passed the exam without distinction. The exam was of 100 marks.
a) 85
b) 90
c) 91
d) 93
View Answer

Answer: c
Explanation: Total marks obtained by the students = average * number of students = 70 * 25 = 1750
Marks obtained by the students who failed the exam = 12 * 45 = 540
Marks obtained by the students with distinction = 100 * 3 = 300
Marks obtained by the students who passed without distinction = 1750 – 540 – 300 = 910
Average marks of the students = 910 / 10 = 91

6. There are 40 animals in a zoo and per animal 20 packets of milk is required. One pack of milk contains 200 ml of milk, and milk costs 46 rupees per litre. If the employees of the zoo fulfil only 80% of the animal’s intake of milk what will be the expenditure of the zoo?
a) 576.5
b) 587.8
c) 546.9
d) 579.6
View Answer

Answer: d
Explanation: Total amount of milk required = 40 * 20 * 200 ml = 16000 ml = 16 litres of milk
Milk given = 80% of required = 16 * 80 / 100 = 12.6 L
Cost of the milk = 12.6 * 46 = 579.6

7. Find the average of first 12 multiples of 7.
a) 42.5
b) 43.5
c) 44.5
d) 45.5
View Answer

Answer: d
Explanation: The first 12 multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 and 84.
Sum of these digits = 546
Average = 45.5
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8. Find the average of the following 7 digits up to 3 places of decimal 102.3, 293.5, 492.4, 483.7, 945.3, 164.4, 188.4.
a) 381.429
b) 381.428
c) 389.421
d) 389.422
View Answer

Answer: a
Explanation: Sum of the given digits = 2670
Average up to 3 places of decimal = 2670 / 7 = 381.4285 ≈ 381.429

9. In a group of 10 girls 5 have an average height of 5.2 feet. Rest of the 5 girls have an average height of 1.7 metres. Find the average height of the group in feet. (1 m = 3.281 feet).
a) 5.1
b) 5.2
c) 5.3
d) 5.4
View Answer

Answer: d
Explanation: Total height of first 5 girls of the group = 5.2 * 5 = 26 feet
Total height of next 5 girls = 1.7 * 5 = 8.5 metres
8.5 metres to feet = 8.5 * 3.281 ≈ 27.89
Total height of the girls = 27.89 + 26 = 53.89 feet
Average height = 53.89 / 10 = 5.389 ≈ 5.4 feet
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10. A class is consisted of 10 with weights 30, 31, 32, 33…39kg, respectively. If 3 new students are admitted to the class with weights 45, 50 and 60 kg what will be the average weight of boys of the newly formed class?
a) 37.5
b) 38.5
c) 39.5
d) 40.5
View Answer

Answer: b
Explanation: Total weight of the boys of existent class = 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 = 345
Weight of the boys admitted = 45, 50, 60 kg respectively.
Total weight of the boys of the newly formed class = 345 + 45 + 50 + 60 = 500 kg
Average weight of the boys of the newly formed class = 500 / 13 = 38.46 ≈ 38.5 kg

To practice all aptitude questions, please visit “1000+ Quantitative Aptitude Questions”, “1000+ Logical Reasoning Questions”, and “Data Interpretation Questions”.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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