Easy Average Aptitude Questions and Answers – Set 2

Let the average runs scored by Ajit in the first 9 innings be x.

As we know, the total runs is equal to the number of innings multiplied by the average. Therefore, the total runs scored in 9 innings = 9×x=9x.

In the 10th innings, he scores 100 runs.
So, the total runs after 10 innings = 9x+100.

Now, the new average for 10 innings is given as the old average increased by 8 runs.
New average = x+8.

We can also calculate the total runs for 10 innings by multiplying the number of innings by the new average.
Total runs = 10×(x+8)=10x+80.

Since both expressions represent the total runs scored in 10 innings, we can set them equal to each other:
9x+100=10x+80

Now, we solve for x:
Move the 9x to the right side and the 80 to the left side:
100−80=10x−9x
20=x

So, the average for the first 9 innings was 20 runs.

To find the new average after the 10th innings, we add the increase of 8 runs to the old average:
New average = x+8=20+8=28.

Answer: 28

Total members = 5

Average age = 22

Total age now = 5 * 22 = 110

Youngest age = 10

At the time of youngest's birth, ALL 5 members were 10 years younger

Total decrease = 5 * 10 = 50

Total age at that time = 110 - 50 = 60

Members at that time = 4

Average at that time = 60 / 4 = 15

Answer: 15

Let original number of students = n

Original average = 40

Original total = 40n

12 new students join with average = 32

Total of new students = 12 * 32 = 384

New average = 40 - 4 = 36

New total = (n + 12) * 36

So,

40n + 384 = 36(n + 12)

40n + 384 = 36n + 432

4n = 48

n = 12

Answer: 12

Average of 6 days = 15640

Total sales for 6 days = 6 * 15640 = 93840

Average from Tuesday to Saturday (5 days) = 14124

Total sales (Tue–Sat) = 5 * 14124 = 70620

Sunday sales = 93840 - 70620 = 23220

Answer: 23220

Initial eggs = n, average weight = k gm

Total weight (1st generation) = n * k

Each egg produces n eggs, so:

2nd generation eggs = n * n = n^2

3rd generation eggs = n^3

rth generation eggs = n^r

Each egg still has average weight = k gm

Total weight of rth generation = n^r * k

Answer: k * n^r

First five multiples of 3:
3, 6, 9, 12, 15

Sum = 3+6+9+12+15=45
Average = 45÷5=9

Answer: 9

Total people = 12 + 3 = 15

Initial total salary = 15 X 600 = 9000

After replacing one manager, new average = 580
New total salary = 15 X 580 = 8700

Decrease in total = 9000 - 8700 = 300

This decrease is because:
Old manager 720 replaced by new manager x

720 - x = 300

720-300 = x
x = 420

Answer: 420

Sales for 5 months = 6435, 6927, 6855, 7230, 6562

Sum of 5 months = 6435 + 6927 + 6855 + 7230 + 6562 = 34009

Required average = 6500

Total for 6 months = 6 * 6500 = 39000

Required sale in 6th month = 39000 - 34009 = 4991

Answer: 4991

Average of 12 numbers = 15

Total of 12 numbers = 12 * 15 = 180

Average of first 2 numbers = 14

Sum of first 2 = 2 * 14 = 28

Remaining numbers = 12 - 2 = 10

Sum of remaining = 180 - 28 = 152

Average of remaining = 152 / 10 = 15.2

Answer: 15.2

Marks: 76, 65, 82, 67, 85

Sum = 76 + 65 + 82 + 67 + 85 = 375
Average = 375 \div 5 = 75

Answer: 75

Average of 4 subjects = 120
Total (wrong) = 4×120=4804 \times 120 = 4804×120=480

But one mark 33 was misread as 13
Difference = 33−13=2033 - 13 = 2033−13=20

So correct total = 480+20=500480 + 20 = 500480+20=500

Correct average = 500÷4=125500 \div 4 = 125500÷4=125

Answer: 125

Total people = 19

13 people spent = 78 each

Money spent by them = 13 * 78 = 1014

Let overall average = x

Remaining people = 19 - 13 = 6

Each of them spent = x + 4

Money spent by them = 6 * (x + 4)

Total money = 19x

So,

1014 + 6(x + 4) = 19x

1014 + 6x + 24 = 19x

1038 + 6x = 19x

1038 = 13x

x = 1038 / 13 = 79

Total money spent = 19 * 79 = 1501

Answer = 1501

Student leaves = 45 kg

Remaining students = 59

Average increases by 0.2

Increase in total weight = 59 * 0.2 = 11.8

This increase happens because a lower weight (45) is removed

So, original average must be less than 45

But exact original average is not given

Hence, exact new average cannot be determined

Answer: Cannot be determined

Wrong mark = 83

Correct mark = 63

Increase in total = 83 - 63 = 20

Increase in average = 1/2

Let number of pupils = n

Increase in total = n * (1/2)

So,

n * (1/2) = 20

n = 40

Answer: 40

Let highest number = H

Average of lowest two = 8

Sum of lowest two = 2 * 8 = 16

Total of three numbers = H + 16

Average = (H + 16) / 3

One-third of average = (H + 16) / 9

Given,

(H + 16) / 9 = H - 8

H + 16 = 9H - 72

88 = 8H

H = 11

Answer: 11

Let third number = x

Second number = 2x

First number = 4x

Reciprocals = 1/(4x), 1/(2x), 1/x

Average of reciprocals = 7/72

So,

(1/(4x) + 1/(2x) + 1/x) / 3 = 7/72

(1/(4x) + 2/(4x) + 4/(4x)) / 3 = 7/72

(7/(4x)) / 3 = 7/72

7/(12x) = 7/72

12x = 72

x = 6

Numbers = 4x, 2x, x = 24, 12, 6

Answer: 24, 12, 6

Group sizes = 25, 50, 25

Means = 60, 55, x

Total students = 25 + 50 + 25 = 100

Combined mean = 58

Total sum = 100 * 58 = 5800

Sum of first group = 25 * 60 = 1500

Sum of second group = 50 * 55 = 2750

Sum of third group = 25 * x = 25x

So,

1500 + 2750 + 25x = 5800

4250 + 25x = 5800

25x = 1550

x = 62

Answer: 62

Students = 10, 20, 30, 40

Pass % = 20%, 30%, 60%, 100%

Passed in each class =

10 * 20% = 2

20 * 30% = 6

30 * 60% = 18

40 * 100% = 40

Total students = 10 + 20 + 30 + 40 = 100

Total passed = 2 + 6 + 18 + 40 = 66

Overall pass % = (66 / 100) * 100 = 66%

Answer: 66%

Average of A, B, C = 40

Sum of A + B + C = 3 * 40 = 120

C = A + 24

B = C + 3

So,

A + (C + 3) + C = 120

A + 2C + 3 = 120

Replace C = A + 24

A + 2(A + 24) + 3 = 120

A + 2A + 48 + 3 = 120

3A + 51 = 120

3A = 69

A = 23

C = 23 + 24 = 47

B = 47 + 3 = 50

D = C - 15 = 47 - 15 = 32

Sum of A + B + C + D = 23 + 50 + 47 + 32 = 152

Average = 152 / 4 = 38

Answer: 38

Grandparents = 2, average = 67

Total = 2 * 67 = 134

Parents = 2, average = 34

Total = 2 * 34 = 68

Grandchildren = 3, average = 5

Total = 3 * 5 = 15

Total members = 2 + 2 + 3 = 7

Total sum = 134 + 68 + 15 = 217

Family average = 217 / 7 = 31

Answer: 31