Average Questions and Answers – Quantitative Aptitude

Average Questions and AnswersFormulas and Quick Tricks

  1. Average =(Sum of Quantities / No.of Quantities)
  2. Sum of Quantities = Average * No.of Quantities
  3. The Average of first n natural Numbers = (n + 1) / 2
  4. The Average of Cubes of first n natural Numbers = n(n + 1)<sup>2</sup> / 4
  5. The Average of Squares of first n natural Numbers = ((n+1) * (2n+1)) / 6
  6. The Average of First n odd Numbers = (last odd number + 1) /2
  7. The Average of First n even Numbers = (last even number + 1) /2
  8. The Average of Squares of first n consecutive even numbers is (2(n+1) * (2n+1)) / 3
  9. The Average of Squares of consecutive even numbers till n = ((n+1) * (n+2)) /3
  10. The Average of Squares of Squares of consecutive odd numbers till n = (n(n + 2) / 3
  11. If the average of n consecutive numbers is m, then the difference between the smallest and the largest number = 2(m-1)
  12. If the Number of Quantities in two groups be n<sub>1</sub> and n<sub>2</sub> and their Average is x and y respectively, the combined average is (n<sub>1</sub>x + n<sub>2</sub>y) / (n<sub>1</sub> + n<sub>2</sub>)
  13. The Average of n quantities is equal to x. When a quantity is removed, the average becomes y. The value of the removed quantity is n(x-y) + y
  14. The Average of n quantities is equal to x. When a quantity is added, the average becomes y. The value of the new quantity is n(y-x) + y

Average Questions and Answers With Detailed Explanation Quantitative Aptitude

Average

Question 1
The Average of Runs of a Cricket Player of 10 innings was 43. How Many runs must he make in his next innings so as to increase his Average of runs by 5.
A
95
B
88
C
98
D
87
Question 1 Explanation: 
Answer: Option C
Explanation:
Average = Total Runs / No.of.innings = 43
So, Total = Average * No.of.innings = 43 *10 = 430
Now increase in avg = 5 Runs.
So, New Avg = 43 + 5 = 48 Runs
Total Runs = New Avg * New No.of.innings = 48*11 =528
Runs Made in the 11th inning = 528-430 = 98
Question 2
The Average Age of Husband, Wife and their Child 3 Years Ago was 27 Years, and that of wife and the child 5 Years ago was 20 Years. The Present age of the Husband is.
A
30 Years
B
35 Years
C
40 Years
D
45 Years
Question 2 Explanation: 
Answer: Option C
Explanation:
Sum of the Present Ages of Husband, Wife & Child = (27 * 3 + 3 * 3) Years= 90 Years
Sum of the Present Ages of Wife & Child = (20 * 2 + 5 * 2) Years = 50 Years.
Husband's Present Age = (90 - 50)Years= 40 Years
Question 3
Average of 18 numbers is 6. If 3 is added to even number, the new average will be.
A
7
B
10
C
8
D
9
Question 3 Explanation: 
Answer: Option D
Explanation:
Average of 18 Numbers = 6
Then, Total of 18 Numbers = 18 * 6 = 108
When, 3 is added to every Number then Total of 18 Numbers increases by = 18 * 3 = 54
Then, New Total of 18 Numbers = 108 + 54 = 162
New Average of 18 Numbers = 162 / 18 = 9
Question 4
In a Room, there are 30 boys and girls, The Average Weight of 20 Girls is 51.50 in a Room, and the remaining 10 boys are 48.25kg in a Room. Find the average weights of all the Boys & Girls in the Room
A
51.45
B
48.6
C
50.42
D
49
Question 4 Explanation: 
Answer: Option C
Explanation:
Question 5
The Average of Senthil Marks in 6 Subject is 68. If his average in five Subject exclude English is 70, How many Marks he obtained in English
A
58
B
59
C
68
D
69
Question 5 Explanation: 
Answer: Option A
Explanation:
Total Marks Obtained in 6 Subject = 6 * 68 = 408
Total Marks in 5 Subjects Excluding English = 5 * 70 = 350
Therefore, Mark Obtained in English = 408 - 350 = 58
Question 6
If the Average of three consecutive even numbers is 36, Find the Largest of these numbers.
A
32
B
34
C
36
D
38
Question 6 Explanation: 
Answer: Option D
Explanation:
Let the First Number is X, then the next two even Numbers would be, X+2, X+4

3X + 6 = 108
3X = 108 - 6
3X = 102
X = 102/3
X = 34
Largest Number Would be = 34 + 4 = 38
Question 7
Find the Sum of First 20 Natural Numbers
A
230
B
220
C
210
D
205
Question 7 Explanation: 
Answer: Option C
Explanation:
Sum of n Natural Numbers = (n*(n+1))/2
=(20*(20+1))/2 =(20*21)/2 =420/2 =210
Question 8
The Average Weight of A, B & C is 50kg. If the Average weight of A and B be 45kg and that of B and C be 48kg, then the Weight of B is:
A
31
B
36
C
26
D
21
Question 8 Explanation: 
Answer: Option B
Explanation:
Let A, B, C represent their respective weights. Then, we have:
A+B+C = (50*3) = 150 .... (i)
A+B = (45*2) = 90 .... (ii)
B+C = (48*2) = 96 .... (iii)
Adding (ii) and (iii), We get: A+2B+C = 186 .... (iv)
Subtracting (i) from (iv), We get: B = 31.
Therefore, B's Weight =36kg
Question 9
The Average of 20 Numbers is Zero. Of them, at the most, How Many May be Greater than Zero?
A
19
B
10
C
1
D
0
Question 9 Explanation: 
Answer: Option A
Explanation:
Average of 20 Numbers = 0
Therefore, Sum of Numbers ( 0 * 20 ) = 0
It, is quite possible that 19 of these Numbers may be Positive and, if their Sum is a the 20th Number is (-a).
Question 10
Average of all Prime Numbers Between 20 to 50
A
35
B
35.9
C
36
D
36.9
Question 10 Explanation: 
Answer: Option B
Explanation:
Prime Numbers Between 20 & 50 are: 23, 29, 31, 37, 41, 43, 47
Average of Prime Numbers Between 20 to 50 will be : ((23+29+31+37+41+43+47)/7)
=251/7
=35.86 Rounded 35.9
There are 10 questions to complete.