Easy Average Aptitude Questions and Answers – Set 1

Let the average of Section B be x. Then the average of Section A is 20% more than B, so Section A average = 1.2x.

Total students = 50 + 60 = 110.
Total of Section A = 50 1.2x = 60x
Total of Section B = 60 x = 60x

Overall average = (Total of Section A + Total of Section B) / Total students = 75

(60x + 60x) / 110 = 75
120x / 110 = 75
120x = 75 * 110
120x = 8250
x = 8250 / 120
x = 68.75

Average of Section A = 1.2 * 68.75 = 82.5

Answer: 82.5

Total numbers = 16

Average = 48

⇒ Total sum = 16 × 48 = 768

First 6 numbers:

Average = 45

⇒ Sum = 6 × 45 = 270

Last 7 numbers:

Average = 53

⇒ Sum = 7 × 53 = 371

Sum of 7th, 8th, 9th numbers:

= 768 − (270 + 371)

= 768 − 641 = 127

Let 9th number = x

Then 7th = x + 3, 8th = x + 7

(x + 3) + (x + 7) + x = 127

3x + 10 = 127

3x = 117

x = 39

7th number = 42

Required average = (39 + 42) ÷ 2 = 40.5

Answer = 40.5

Sum of first 18 numbers = 37 × 18 = 666

Sum of 6 new numbers = 41 × 6 = 246

Total sum of all numbers = 666 + 246 = 912

Total number of numbers = 18 + 6 = 24

New average = 912 ÷ 24 = 38

Answer: 38

The sum of 100 observations with the wrong mean is 100 times 40 equals 4000.

The wrong value recorded was 83 instead of 53, so the sum needs to be corrected by subtracting 83 and adding 53.

Corrected sum equals 4000 minus 83 plus 53 equals 4000 minus 30 equals 3970.

The total number of observations is still 100.

Correct mean equals 3970 divided by 100 equals 39.7.

Answer: 39.7

Current Total Age: 10 members × 19 years = 190 years.

Total Age 10 Years Ago: Since the youngest is 10, you must subtract 10 years from every member of the group.

10 members × 10 years = 100 years to subtract. 190 years - 100 years = 90 years.

Average Age Before Birth: Just before the youngest was born, there were only 9 members in the group. You divide the total years from that time by these 9 members.

90 years ÷ 9 members = 10 years.

Answer: 10

Take numbers from 1 to 100.

Pair them like this:
1 + 100 = 101
2 + 99 = 101
3 + 98 = 101

Every pair gives 101.

There are 100 numbers, so total pairs = 50.

Total sum = 50 × 101 = 5050

Now average = total ÷ number of values
= 5050 ÷ 100 = 50.5

Answer = 50.5

Even numbers (7 terms) have middle d, odd numbers (5 terms) have middle l.

Total numbers = 12 (even count), so there is no single middle term.

Hence, the overall average lies between d and l, i.e., it is the average of the middle even and middle odd.

Answer = Average of middle even and odd

Ratio 4:5 means:

• Girls = 4 parts → 85 each

• Boys = 5 parts → 87 each

Total = (4 × 85) + (5 × 87) = 340 + 435 = 775
Total parts = 9

Average = 775 ÷ 9 ≈ 86.1

Answer = 86.1

Let Chemistry = C

Given:
P + C + M = C + 120

So:
P + M = 120

Average of P & M = 120 ÷ 2 = 60

Answer = 60

Number of machines = 7

Machine A = 2, Cost of each = ₹95,000
Machine B = 2, Cost of each = ₹75,000
Machine C = 7 − (2 + 2) = 3, Cost of each = ₹43,000

Total cost of machines
= (2 × 95,000) + (2 × 75,000) + (3 × 43,000)
= 1,90,000 + 1,50,000 + 1,29,000
= 4,69,000

Average cost
= Total cost / Total number of machines
= 4,69,000 / 7
= 67,000

Answer: ₹67,000

Total students = 75

Boys = 33⅓% of 75 = 25
Girls = 75 − 25 = 50

Let average marks of girls = x

Average marks of boys = 66⅔% more than girls
= x + (2/3)x = (5/3)x

Total marks of boys = 25 × (5/3)x = (125/3)x
Total marks of girls = 50x

Total marks = (125/3)x + 50x
= (125/3)x + (150/3)x
= (275/3)x

Average of all students = 66

(275/3)x ÷ 75 = 66
275x / 225 = 66
11x / 9 = 66
11x = 594
x = 54

Answer: 54

Total numbers = 33

Average = 74

⇒ Total sum = 33 × 74 = 2442

First 17 numbers:

Average = 72.8

⇒ Sum = 17 × 72.8 = 1237.6

Last 17 numbers:

Average = 80.3

⇒ Sum = 17 × 80.3 = 1365.1

The 17th number is included in both groups,

so subtract total once to avoid double counting:

17th number = 1237.6 + 1365.1 − 2442 = 160.7

Now remove this number:

New sum = 2442 − 160.7 = 2281.3

Remaining numbers = 32

New average = 2281.3 ÷ 32 = 74.6

Answer = 74.6

Let the number of wickets before the match be x.
Let the total runs conceded before the match be R.

Bowling average before the match = Total runs / Total wickets
So, R / x = 12.4
⇒ R = 12.4 × x

After the match, the cricketer takes 5 wickets for 26 runs.
Total wickets after the match = x + 5
Total runs after the match = R + 26 = 12.4 × x + 26

New average after the match = 12.4 − 0.4 = 12.0

New average formula:
(12.4 × x + 26) / (x + 5) = 12

Multiply both sides by (x + 5):
12.4 × x + 26 = 12 × (x + 5)
12.4 × x + 26 = 12 × x + 60

Subtract 12 × x from both sides:
0.4 × x + 26 = 60

Subtract 26 from both sides:
0.4 × x = 34

Divide both sides by 0.4:
x = 34 / 0.4 = 85

Number of wickets before the match = 85

Answer: 85

Let the tens digit be x.
Then the units digit is 2x (because the unit digit is twice the tens digit).

A two-digit number with tens digit x and units digit 2x is 10x + 2x = 12x.

Sum of digits = x + 2x = 3x

Use the condition: 2 subtracted from sum of digits equals 1/6th of the number.
So, 3x - 2 = (1/6) * (12x)

Simplify the right side.
(1/6) * 12x = 2x

Write the equation.
3x - 2 = 2x

3x - 2 = 2x
3x - 2x = 2
x = 2

Unit digit = 2x = 2 * 2 = 4

Number = tens digit 10 + unit digit = 2 10 + 4 = 24

Answer: 24

Let the sum of x numbers be S1. Since the average of x numbers is y squared, we have:

S1 divided by x equals y squared.

Multiplying both sides by x gives: S1 equals x times y squared.

Let the sum of y numbers be S2. Since the average of y numbers is x squared, we have:

S2 divided by y equals x squared.

Multiplying both sides by y gives: S2 equals y times x squared.

The combined sum of all numbers is S1 plus S2 equals (x times y squared) plus (y times x squared).

This can be written as x times y squared plus y times x squared equals x times y times (y plus x).

The total number of numbers is x plus y.

So the combined average is (x times y times (x plus y)) divided by (x plus y).

Simplifying, (x times y times (x plus y)) divided by (x plus y) equals x times y.

Answer: x times y

Mean of 9 observations = 16
So sum of 9 observations = 9 × 16 = 144

After adding 10th observation, mean = 17
So total sum of 10 observations = 10 × 17 = 170

10th observation = 170 − 144 = 26

Answer = 26

Average of 12 parcels = 1.8 kg
So total weight = 12 × 1.8 = 21.6 kg

Average decreases by 50 g = 0.05 kg
New average = 1.8 − 0.05 = 1.75 kg

Now total parcels = 13
New total weight = 13 × 1.75 = 22.75 kg

Weight of new parcel = 22.75 − 21.6 = 1.15 kg

Answer = 1.15kg

Total weight initially = 50 × 45 = 2250 kg

After one leaves:
New average = 45 − 0.1 = 44.9 kg
New total = 49 × 44.9 = 2200.1 kg

Weight of the student = 2250 − 2200.1 = 49.9 kg

Answer = 49.9kg

• Group A: 42 students, each ~25 kg → total = 42 × 25 = 1050

• Group B: 28 students, each ~40 kg → total = 28 × 40 = 1120

Now combine:

• Total weight = 1050 + 1120 = 2170

• Total students = 42 + 28 = 70

Finally:

• Average = 2170 ÷ 70 = 31 kg

• Answer = 31

Odd numbers up to 100 are:

1, 3, 5, 7, …, 99

This is a sequence of consecutive odd numbers.
For such numbers, the average = middle term.

The middle number between 1 and 99 is:

(1 + 99) ÷ 2 = 50

So the average is 50.

Answer = 50