Averages - CENTS INSIGHT

Everything below is designed for calculator-free aptitude tests, with the challenging examples reaching roughly SAT and GRE-style quantitative questions.

Before getting straight into the topic, lets test the level of your knowledge about averages with an example below:

The average of 8 numbers is 24. When one additional number is included, the average increases to 27. What is the additional number?

27

36

48

51

Don’t stress over a wrong answer. We’ve got you.

First let’s see where you got it wrong.

The first 8 numbers have a total of:

8(24)=192

After the new number is included, there are 9 numbers with an average of 27:

9(27)=243

Therefore, the additional number is:

243−192=51

So D is the right answer.

Note: In aptitude tests, average questions are rarely difficult because of the average formula. They become difficult because the total, number of observations, or composition of the group changes.

Most students are already familiar with the mathematical representation of the average. What they often haven’t learned is to use it strategically.

If $n$ observations have mean $\bar{x}$.

Finding mean, When you know the total and the number of observations:

\bar{x}=\frac{\sum x}{n}

For example:

\bar{x}=\frac{600}{8}=75

So the mean is 75.

Whereas, to find the sum use the following version of the same formula:

\sum x=n\bar{x}

Where $\bar{x}$ = mean
$\sum$ = sum of all observations
n = number of obersvations

For example:

The average weight of 12 boxes is 15 kg. Find their combined weight.

\sum{x}=12(15)=180

So the total weight is 180 kg.

The three forms of this formula to remember are:

What you need to find	Formula	What you need to know
Mean	$\displaystyle \bar{x}=\frac{\sum x}{n}$	Total + number of values
Total / Sum	$\displaystyle \sum x=n\bar{x}$	Mean + number of values
Number of values	$\displaystyle n=\frac{\sum x}{\bar{x}}$	Total + mean

Once an aptitude test moves beyond simply asking “find the average,” averages become a reasoning topic. Inside changes, groups, missing values, ratios, or constraints are merged together to test the student from multiple dimensions.

There are a few key concepts that we might discuss to develop the desired understanding of averages.

1. Finding a missing value from mean

A mean can be converted into a total, allowing an unknown observation to be recovered. Let’s do an example to know further:

The average of 6 numbers is 14. Five of the numbers are $8,11,13,17,$ and $19$ . Find the sixth number.

Total of all 6 numbers:

6(14)=84

Known total:

8+11+13+17+19=68

Therefore:

x=84−68=16

So the key idea here is: Convert the average into a total first.

2. Adding a new observation

When a new value is added, both the total and the number of observations change. For example:

The average of 8 numbers is 15. A ninth number is added, increasing the average to 17. Find the new number.

Old total:

8(15)=120

New total:

9(17)=153

Therefore, the added number is:

153−120=33​

Don’t get confuse by the new total to be much higher. The new value is much greater than 17 because it must pull the existing average upward.

3. Removing an observation

This is the reverse of adding a value. Compare the original total with the remaining total.

The average of 10 numbers is 24. After one number is removed, the average of the remaining 9 numbers becomes 22. Find the removed number.

Original total:

10(24)=240

Remaining total:

9(22)=198

Removed value:

240−198=42

Since removing the number caused the average to fall, the removed number must have been above the original average.

4. Change in average

A very common concept aptitude test asks how much the total must change when the average changes. For a fixed number of observations:

Change \ in \ total=n(Change\ in \ average)

For example:

The average score of 20 students increases from 64 to 67. By how much has their combined score increased?

You don’t need to calculate either total separately.

20(67−64)=20(3)=60​

5. Combined average

You cannot generally average the averages of two groups as their sizes matter. As a result, a combined mean might need an expression:

Combined \ mean=\frac{n_1\bar{x_1}+n_2\bar{x_2}}{n_1+n_2}

For example:

Class A contains 20 students with an average score of 70. Class B contains 30 students with an average score of 80. Find their combined average.

\frac{20(70)+30(80)}{50}

\frac{1400+2400}{50}

\frac{3800}{50}

=76

However, you can’t simply:

\frac{70+80}{2}=75

As that would work only if the two classes contained the same number of students.

The same expression can used to find the unknown group size as a technical question might ask you to find.

For example:

Group A has 20 students with an average score of 60. Group B has an average score of 75. The combined average is 65. How many students are in Group B?

Let Group B contain $n$ students.

\frac{20(60)+75n}{20+n}=65

Now multiply:

1200+75n=1320+65n

10n=100

n=\frac{100}{10}=10

6. Weighted average

This is an important and most commonly known concept in business, finance, and mathematics itself. Sometimes observations don’t contribute equally. Each value has a different weight. Weighted average takes all those varying contributions in account to give a simple average.

Weighted\ mean=\frac{\sum wx}{\sum w}

For example: A course grade consists of:

Assignments: 20%, score 80
Midterm: 30%, score 70
Final 50%, score 90

We can identify the score and weightage both given for each component.

Therefore:

\sum wx

\begin{aligned} \text{Assignments: } &(20\%)(80)=16 \\ \text{Midterm: } &(30\%)(70)=21 \\ \text{Final: } &(50\%)(90)=45 \end{aligned}

Now a simple average will be:

16+21+45=82

7. Average of consecutive numbers

For odd number of equally spaced terms, the average is simply the middle value. As below:

17,\ 21,\ \mathbf{25},\ 29,\ 33

The average is immediately 25.

But, for an even number of equally spaced terms:

12,\ 16,\ 20,\ 24,\ 28\ ,32

the mean is the average of the two middle terms:

\frac{20+24}{2}=22

However, if the average of consecutive integers with an unknown range is required, instead of listing every integer, you might just use the following expression:

Mean=\frac{First\ term+Last\ term}{2}

This is particularly useful for questions involving consecutive integers, consecutive odd/even numbers, or arithmetic sequences.

For example:

The average of all consecutive integers from $x$ through 46 is 31. Find $x$ .

\frac{x+46}{2}=31

x+46=62

x=16

This is much faster than constructing and summing the entire sequence.

8. Finding an unknown average

Instead of finding the combined mean, aptitude tests may give the combined mean and ask for one subgroup’s mean.

For example:

The average score of 30 students is 72. The average of 18 of them is 68. Find the average of the remaining 12.

Total of all students:

30(72)=2160

Total of first group:

18(68)=1224

Remaining total:

2160−1224=936

Remaining average:

\frac{936}{12}=78

This is essentially a combined-average problem in reverse.

9. Average of remaining numbers

Lets understand it with an example:

The average of 12 numbers is 18. The average of the first 7 numbers is 16. What is the average of the remaining 5 numbers?

Total of all 12 numbers:

12(18)=216

Total of the first 7:

7(16)=112

Therefore, the remaining 5 have a total of:

216−112=104

Their average is:

\frac{104}{5}=20.8

Practice questions

1. The average of 12 numbers is 18. If two numbers, 24 and 30, are removed, what is the average of the remaining 10 numbers?

A) 15.6
B) 16.2
C) 16.8
D) 17.4

View Solution

The total of the 12 numbers is 12 × 18 = 216.

The two removed numbers have a total of 24 + 30 = 54.

The remaining total is 216 − 54 = 162.

There are now 10 numbers, so the new average is 162 ÷ 10 = 16.2.

Answer: B) 16.2

2. The average score of 15 students is 24. One student’s score was incorrectly recorded as 18 instead of 33. What is the correct average?

A) 24.5
B) 25
C) 26
D) 27

View Solution

The incorrect total is 15 × 24 = 360.

The recorded score was 18, but it should have been 33. Therefore, the total must increase by 33 − 18 = 15.

The correct total is 360 + 15 = 375.

The correct average is 375 ÷ 15 = 25.

Answer: B) 25

3. A group of 20 students has an average score of 68. Another group has an average score of 80. If the average score of the two groups combined is 72, how many students are in the second group?

A) 8
B) 10
C) 12
D) 15

View Solution

Let the second group contain n students.

The first group’s total is 20 × 68 = 1360.

The second group’s total is 80n.

The combined total can also be written as 72(20 + n).

Therefore, 1360 + 80n = 72(20 + n).

Expanding gives 1360 + 80n = 1440 + 72n.

So 8n = 80, giving n = 10.

Answer: B) 10

4. The average of 10 numbers is 30. Nine of the numbers are each increased by 4, while the remaining number is decreased by 6. What is the new average?

A) 31
B) 32
C) 33
D) 34

View Solution

There is no need to determine the original numbers.

Nine numbers increase by 4, so the total increases by 9 × 4 = 36.

The remaining number decreases by 6, so the net increase in the total is 36 − 6 = 30.

Since there are 10 numbers, the average increases by 30 ÷ 10 = 3.

The new average is therefore 30 + 3 = 33.

Answer: C) 33

5. Five consecutive odd integers have an average of 37. What is the difference between the square of the largest integer and the square of the smallest integer?

A) 512
B) 560
C) 592
D) 624

View Solution

For five consecutive odd integers, the middle integer is equal to their average.

Therefore, the numbers are 33, 35, 37, 39 and 41.

We need 41² − 33².

Instead of calculating both squares, use the difference of squares: a² − b² = (a − b)(a + b).

So 41² − 33² = (41 − 33)(41 + 33) = 8 × 74 = 592.

None of the listed choices matches 592, so this question’s options need correction.

6. The average of 8 numbers is 25. If the largest number, 40, and the smallest number, 10, are removed, what is the average of the remaining six numbers?

A) 23
B) 24
C) 25
D) 26

View Solution

The original total is 8 × 25 = 200.

The two removed numbers total 40 + 10 = 50.

The remaining total is 200 − 50 = 150.

The average of the remaining six numbers is 150 ÷ 6 = 25.

Another way to see this is that the two removed numbers themselves have an average of (40 + 10) ÷ 2 = 25, which is the same as the original average. Removing a group whose average equals the overall average does not change the mean.

Answer: C) 25

7. A student has an average score of 72 on five tests. Two tests remain. What average must the student obtain on the final two tests to finish with an overall average of 76?

A) 82
B) 84
C) 86
D) 88

View Solution

The student’s current total is 5 × 72 = 360.

To average 76 over seven tests, the required total is 7 × 76 = 532.

Therefore, the final two tests must contribute 532 − 360 = 172 points.

The required average on those two tests is 172 ÷ 2 = 86.

Answer: C) 86

8. The average of five integers is an integer. Four of the integers are 8, 13, 17 and 24. Which of the following could be the fifth integer?

A) 21
B) 22
C) 23
D) 24

View Solution

The four known integers have a total of 8 + 13 + 17 + 24 = 62.

If the average of five integers is an integer, their total must be divisible by 5.

Since 62 leaves a remainder of 2 when divided by 5, the fifth integer must leave a remainder of 3 when divided by 5 so that the total becomes divisible by 5.

Among the choices, 23 leaves a remainder of 3 when divided by 5.

Indeed, 62 + 23 = 85, and 85 ÷ 5 = 17.

Answer: C) 23

9. Three numbers have an average of 24. The sums of every possible pair of these numbers are calculated. What is the average of these three pair sums?

A) 24
B) 36
C) 48
D) 72

View Solution

Let the three numbers be a, b and c.

Since their average is 24, their total is a + b + c = 3 × 24 = 72.

The three pair sums are a + b, a + c and b + c.

Adding these pair sums gives 2a + 2b + 2c = 2(a + b + c).

Therefore, their combined total is 2 × 72 = 144.

There are three pair sums, so their average is 144 ÷ 3 = 48.

Answer: C) 48

10. The average age of 16 employees is 30 years. Four employees whose average age is 24 years leave. Four new employees join, and the average age of all 16 employees becomes 32 years. What is the average age of the four new employees?

A) 32 years
B) 36 years
C) 38 years
D) 40 years

View Solution

The original total age is 16 × 30 = 480 years.

The four employees who leave have a total age of 4 × 24 = 96 years.

The total age of the 12 remaining employees is 480 − 96 = 384 years.

After four new employees join, the total required for 16 employees to have an average age of 32 is 16 × 32 = 512 years.

Therefore, the four new employees have a combined age of 512 − 384 = 128 years.

Their average age is 128 ÷ 4 = 32 years.

Answer: A) 32 years