**Average Questions and Answers** – **Formulas and Quick Tricks**

- Average =(Sum of Quantities / No.of Quantities)
- Sum of Quantities = Average * No.of Quantities
- The Average of first n natural Numbers = (n + 1) / 2
- The Average of Cubes of first n natural Numbers = n(n + 1)<sup>2</sup> / 4
- The Average of Squares of first n natural Numbers = ((n+1) * (2n+1)) / 6
- The Average of First n odd Numbers = (last odd number + 1) /2
- The Average of First n even Numbers = (last even number + 1) /2
- The Average of Squares of first n consecutive even numbers is (2(n+1) * (2n+1)) / 3
- The Average of Squares of consecutive even numbers till n = ((n+1) * (n+2)) /3
- The Average of Squares of Squares of consecutive odd numbers till n = (n(n + 2) / 3
- If the average of n consecutive numbers is m, then the difference between the smallest and the largest number = 2(m-1)
- 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>)
- 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
- 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.

95 | |

88 | |

98 | |

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.

30 Years | |

35 Years | |

40 Years | |

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.

7 | |

10 | |

8 | |

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

51.45 | |

48.6 | |

50.42 | |

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

58 | |

59 | |

68 | |

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.

32 | |

34 | |

36 | |

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

230 | |

220 | |

210 | |

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:

31 | |

36 | |

26 | |

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?

19 | |

10 | |

1 | |

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

35 | |

35.9
| |

36 | |

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.