Average Problems – Basic Concepts, Formulas, Tricks, Aptitude Questions and Solved Examples with Explanation (Video Lessons)

Formulas and Quick Tricks for Average Problems

1. Average = Sum of quantities/ Number of quantities
2. Sum of quantities = Average × Number of quantities
3. The average of first n natural numbers is (n +1) / 2
4. The average of the squares of first n natural numbers is (n +1)(2n+1 ) / 6
5. The average of cubes of first n natural numbers is n(n+1)² / 4
6. The average of first n odd numbers is given by (last odd number +1) / 2
7. The average of first n even numbers is given by (last even number + 2) / 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 is (n+1)(n+2) / 3
10. The average of squares of squares of consecutive odd numbers till n is n(n+2) / 3
11. If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2(m-1)
12. If the number of quantities in two groups be n₁ and n₂ and their average is x and y respectively, the combined average is (n₁x+n₂y) / (n₁+ n₂)
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

Questions and Solved Examples on Average Problems

Q.1: A batsman in his 17th innings makes a score of 85 and their by increasing his average by 3. What is his average after the 17th innings?
A. 34
B. 35
C. 36
D. 37
Answer: D. 37
Explanation: 16x + 85 = 17(x + 3)
x = 34 + 3 = 37

Q.2: The average of first 10 natural numbers is?
A. 5
B. 5.5
C. 6
D. 6.5
Answer: B. 5.5
Explanation: Sum of 10 natural no. = 110/2 = 55
Average = 55/10 = 5.5

Q.3: The average age of three boys is 15 years and their ages are in proportion 3:5:7. What is the age in years of the youngest boy?
A. 9
B. 12
C. 15
D. 18
Answer: A. 9
Explanation: 3x + 5x + 7x = 45
x = 3
3x = 9

Q.4: The average of 9 observations was 9, that of the 1st of 5 being 10 and that of the last 5 being 8. What was the 5th observation?
A. 6
B. 7
C. 8
D. 9
Answer: D. 9
Explanation: 1 to 9 = 9 × 9 = 81
1 to 5 = 5 × 10 = 50
5 to 9 = 5 × 8 = 40
5th = 50 + 40 = 90 – 81 = 9


Q.5: A team of eight entered for a shooting competition. The best marks man scored 85 points. If he had scored 92 points, the average scores for the team would have been 84. How many points altogether did the team score?
A. 625
B. 632
C. 656
D. 665
Answer: D. 665
Explanation: 8 × 84 = 672 – 7 = 665

Q.6: A batsman makes a score of 64 runs in the 16th innings and thus increased his average by 3. Find his average after the 16th inning?
A. 17
B. 18
C. 19
D. 20
Answer: C. 19
Explanation: Let the average after the 16th inning be P.
So, the average after the 15th inning will be (P-3)
Hence, 15(P-30) + 64 = 16P => P = 19

Q.7: The average of 10 numbers is 23. If each number is increased by 4, what will the new average be?
A. 23
B. 25
C. 27
D. 29
Answer: C. 27
Explanation: Sum of the 10 numbers = 230
If each number is increased by 4, the total increase = 4 × 10 = 40
The new sum = 230 + 40 = 270 The new average = 270/10 = 27

Q.8: Find the average of the series : 312, 162, 132, 142 and 122?
A. 162
B. 174
C. 186
D. 198
Answer: B. 174
Explanation: (312 + 162 + 132 + 142 + 122)/5 = 870/5 = 174


Q.9: In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 5.75
B. 6.25
C. 6.75
D. 7.25
Answer: B. 6.25
Explanation: Required run rate = 282 – (3.2 × 10) / 40 = 250 / 40 = 6.25

Q.10: The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
A. 75 kgs
B. 80 kgs
C. 85 kgs
D. Data inadequate
Answer: C. 85 kgs
Explanation: Total weight increased = (8 × 2.5) kg = 20 kgs
Weight of new person = (65 + 20) kg = 85 kgs

Q.11: The average weight of 39 Students in a class is 23. Among them Rony is the heaviest while Titu is the lightest. If both of them are excluded from the class still the average remains same. The ratio of weight of Rony to Titu is 15:8. Then what is the weight of the Titu?
A. 15 kgs
B. 16 kgs
C. 18 kgs
D. 19 kgs
Answer: B. 16 kgs
Explanation: The weight of Rony + Titu = 23 (39 - 37) = 46
Given, Rony / Titu = 15 / 8
Hence, the weigth of Titu = 16 kgs.

Q.12: There are 459 students in a hostel. If the number of students increased by 36, the expenses of the mess increased by Rs. 81/- per day while the average expenditure per head reduced by 1. Find the original expenditure of the mess?
A. Rs. 7314/-
B. Rs. 7324/-
C. Rs. 7334/-
D. Rs. 7344/-
Answer: D. Rs. 7344/-
Explanation: Given, total expenditure = 459x
When the number of students is increased by 36 then the total expenditure becomes = 459x + 81
Given, average = 459x + 81 / 495 = x - 1
So, x = 16
Hence, the original expenditure = 16 × 459 = Rs. 7344/-

Q.13: The average cost of 32 different mobiles is Rs. 9000/- Among them, Redmi which is the costliest, has 70% higher price than the cheapest mobile phone Realme. Excluding those both mobiles, the average of the mobiles is Rs. 8880/- Then what is the cost of Redmi Mobile?
A. Rs. 10,800/-
B. Rs. 11,600/-
C. Rs. 12,400/-
D. Rs. 13,600/-
Answer: D. Rs. 13,600/-
Explanation: Let the cost of Realme mobile be x.
As Redmi is 70% higher than Realme, the cost of Redmi = x + 0.7x = 1.7x
Excluding Realme and Redmi, there are 30 mobiles with an average cost of Rs. 8880/-
Total cost of these 30 mobiles = 30 × 8880 = Rs. 266400/-
Total cost of all 32 mobiles = 32 × 9000 = Rs. 288000/-
Cost of Realme + Cost of Redmi = Total cost of all mobiles - Total cost of 30 mobiles
x + 1.7x = 288000 - 266400
2.7x = 21600
x = 8000 (Cost of Realme)
Cost of Redmi = 1.7x = 1.7 × 8000 = Rs. 13,600/-

Q.14: Dhoni scored 8000 runs in a certain number of innings. In the next five innings, he was out of form and hence, could make only 85 runs, as a result his average reduced by 1 run. How many innings did he play in total?
A. 160
B. 165
C. 170
D. 175
Answer: B. 165
Explanation: Let's assume that Dhoni played 'n' innings before he was out of form and let's take 'a' as average.
Total runs scored by Dhoni in n innings = 8000.
a = 8000 / n
Average runs scored by Dhoni in (n+5) innings = 8085 / (n+5)
Now, we know that the average is reduced by 1 run. So, 8085 / (n+5) = a - 1
Now, after solving, we will get a quadratic equation => n² + 90n – 40000 = 0
(n + 250) (n - 160) = 0
n = 160
Total innings = n + 5 = 165.

Q.15: The average weight of 40 Students is 32. If the Heaviest and Lightest are excluded the average weight reduces by 1. If only the Heaviest is excluded then the average is 31. Then what is the weight of the Lightest?
A. 30 kgs
B. 31 kgs
C. 32 kgs
D. 33 kgs
Answer: B. 31 kgs
Explanation: Given, average weight of 40 students = 32 kgs, so total weight = 40 × 32 = 1280 kgs.
Given, average weight of 38 students except the heaviest and the lightest ones = 32 − 1 = 31 kgs.
Total weight = 38 × 31 = 1178 kgs.
Sum of weights of the heaviest and the lightest = 1280 − 1178 = 102 kgs.
Average weight of 39 students except the heaviest one = 31 kgs.
Total weight = 39 × 31 = 1209 kgs.
Weight of the heaviest = 1280 − 1209 = 71 kgs.
Weight of the lightest = 102 − 71 = 31 kgs.