**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 - 1

Question 1 |

Average of 3

^{rd}and 5^{th}number in 5 consecutive number is 12. Find the sum of 5 Numbers.10.25 | |

22.5 | |

50 | |

55 |

Question 1 Explanation:

**Answer: Option D**

__Explanation:__

**Average = (Sum of Observations / Number of Observations)**

Let first of the 5 consecutive number be N.

So, the 5 Numbers will be,

N, N+1, N+2, N+3, N+4

Also,

Average of 3rd and 5th = ((N + 2) + (N + 4)) / 2 = 12

Therefore, N = (18 / 2)

Sum = N + (N + 1) + (N + 2) + (N + 3) + (N + 4)

=5N + 10

**Sum = 55**

Question 2 |

Rajat Carried two types of pens with him. The ratio of cost of Pen A: Pen B was 10:4. There were two pens of type B with him and the average cost of all the three pens together was Rs 60/-. Find the cost of Pen A.

Rs : 72 | |

Rs : 75 | |

Rs : 100 | |

Rs : 150 |

Question 2 Explanation:

**Answer: Option C**

__Explanation:__

Total cost of 3 Pens = 3 * 60 = Rs : 180

Ratio of Cost of Pen A : Pen B = 10 : 4

Let K be common factor.

So, Cost of Pen A = 10K; Cost of both Pen B = 4K

Therefore, 10K + 4K + 4K = 180

Therefore, K = 10

**Cost of Pen A will be = 10K = Rs : 100**

Question 3 |

Rishabh's current average of runs is 64. He wanted it to e 66 and this could have been achieved if he had scored 22 runs more in all his innings put together. How many innings has he played till now?

10 | |

11 | |

12 | |

15 |

Question 3 Explanation:

**Answer: Option B**

__Explanation:__

Let number of innings be N

Total increase in runs = 2 * N = 2N

Also, Rishabh needs to score 22 runs more to increase average.

Therefore, 2N = 22

**Number of Innings = 11**

Question 4 |

A Pulpil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The Number of pupils in the class is:

45 | |

40 | |

39 | |

37 |

Question 4 Explanation:

**Answer: Option B**

__Explanation:__

Let there be x pupils in the class.

Total increase in marks = (x * (1/2)) = x / 2

(x/2) = (83 - 63) => (x/2) = 20

**x = 40**

Question 5 |

The Average weight of a class of 24 students is 35kg. If the weight of the teacher be included, the average rises by 400g. The weight of the teacher is:

45 kg | |

46 kg | |

47 kg | |

48 kg |

Question 5 Explanation:

**Answer: Option A**

__Explanation:__

Weight of the teacher = (35.4 * 25 - 35 * 24)kg

**Weight of the teacher = 45 kg**

Question 6 |

A certain factory employed 600 men and 400 women and the average wager was Rs. 25.50 per day, If a woman got Rs. 5 less than a man, then what are their daily wages?

M : 25.50, W : 27.50 | |

M : 27.50, W : 22.50 | |

M : 26.50, W : 27.50 | |

M : 24.50, W : 26.50 |

Question 6 Explanation:

**Answer: Option B**

__Explanation:__

Let the daily wages of a man be Rs. x

Then, daily wages of a woman = Rs (x - 5)

Now, 600x + 400 (x- 5) = 25.50 * (600 + 400) <=> 1000x = 27500

x = 27.50

Man's daily wages = Rs. 27.50

Woman's daily wages =(x - 5) = Rs.22.50

Question 7 |

A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If he had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was:

665 | |

657 | |

786 | |

678 |

Question 7 Explanation:

**Answer: Option A**

__Explanation:__

Let the total score be x.

(x + 92 - 85) / 8 = 84

So, x + 7 = 672

**x = 665**

Question 8 |

David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, Maths, Physics, Chemistry and Biology what are his average marks?

75 | |

68 | |

65 | |

60 |

Question 8 Explanation:

**Answer: Option A**

__Explanation:__

Average = (76 + 65 + 82 + 67 + 85) / 5 = 375 / 5

**Ans = 75**

Question 9 |

For 9 innings, Boman has an average of 75 runs. In the tenth innings, he scores 100 runs, thus increasing his average. His new average is

Rs.75 | |

Rs. 100 | |

Rs. 72 | |

Rs. 77.5 |

Question 9 Explanation:

**Answer: Option D**

__Explanation:__

Total score for 9 innings is 75 * 9 = 675

Total score after 10th innings = 675 + 100 = 775

**Average = 775 /10 = 77.5**

Question 10 |

The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is

45 | |

50 | |

54.6 | |

51.4 |

Question 10 Explanation:

**Answer: Option D**

__Explanation:__

Let the required means score be x. Then, (20 * 80) + (25 * 31) + (55 * x) = (52 * 100)

=> 1600 + 775 + 55x = 5200

=> 55x = 2825

=> x = 565/11

**Ans = 51.4**

There are 10 questions to complete.