Average: Aptitude Questions and Solutions, Formula, Examples

The concept of average is fundamental in Mathematics and our day-to-day lives. The average definition refers to the middle value or mean value of the given set of numbers. The average is commonly calculated by dividing the sum of the values by the number of values. We can also use an average calculator or average Excel functions to calculate it more easily.

The average speed formula which is the total distance divided by the total time, is used in daily life.

Understanding the average in maths meaning and using the average in maths formula allows for the simplification and interpretation of data. To master this concept, one should practice average in maths questions and review various average in maths examples.

Average: Concept

Average is a measure of central tendency that represents the central or mean value of a given set of data.

It is calculated by $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$.

Suppose $a_1, a_2, a_3, a_4, \ldots \ldots ., a_n$ are $n$ number of observations and their sum is

$a_1+a_2+a_3+a_4+\ldots \ldots+a_n$.

Then their average = $\frac{a_1+a_2+a_3+a_4+......+a_n}{n}$

Generally, average is denoted by $\bar x$. It is also represented by the symbol “mu” ($µ$).

How to calculate an average

There are three steps involved in calculating the average of a given set of numbers.

Step 1: Calculate the sum of the given set of numbers.

Step 2: Count the number of observations in that set.

Step 3: Divide the sum of the numbers by the number of observations to get the average.

Example: (5, 10, 15, 20, 25, 30) is a given set of numbers.

The sum of the numbers = 5 + 10 + 15 + 20 + 25 + 30 = 105

Number of observations = 6

Therefore, average = $\frac{105}{6}$ = 17.5

Average of negative numbers

Calculating the average of negative numbers involves the same steps as calculating the average of the positive or set of positive and negative numbers.

It also involves the same three steps.

Step 1: Calculate the sum of the given set of negative numbers.

Step 2: Count the number of negative numbers in that set.

Step 3: Divide the sum of the negative numbers by the number of observations to get the average.

Example:

Average of -2, -5, -5, -10, -15, and -11 is:

Here, the number of observations = 6

The sum of the observations = - 48

So, average = $\frac{-48}{6}$ = -8

Arithmetic Mean

Arithmetic Mean generally means the average value of a given set of numbers.

It is the easiest way to find the average of a given set of numbers.

Suppose you got 70, 78, 85, 90, and 43 in 5 subjects in an exam. You need to find your average score. This is when the Arithmetic Mean will be used.

Arithmetic Mean is always greater than Harmonic Mean and greater than or equal to the Geometric Mean.

For a set of n positive integers $a_1, a_2, a_3, a_4, \ldots \ldots \ldots a_n$

Arithmetic mean = $\frac{a_1+a_2+a_3+a_4+.......+a_n}{n}$

For an AP of n terms, AM = $\frac{2a_1+(n-1)d}{n}$ = $\frac{a_1+a_n}{2}$ where n is the number of terms, $a_1$ is the first term of AP and d is the common difference.

First, you have to add all the n observations, then divide that by n to get the Arithmetic Mean.

Example: Arithmetic mean of 70, 78, 85, 90, 43 = $\frac{70+78+ 85+ 90+ 43}{5}=\frac{366}{5}=73.2$

Geometric mean

In mathematics, the term mean is defined as the average of the given number of values.

Generally, we get the mean of the numbers by dividing the sum of the numbers by the count of the number.

However, to find the geometric mean, we have to multiply the numbers and then take the root of the corresponding count of numbers.

The geometric mean is very useful for understanding the central tendency of sets of numbers that are multiplicative.

The geometric mean is always less than or equal to the Arithmetic Mean and greater than or equal to the Geometric Mean.

For a set of n positive integers $a_1, a_2, a_3, a_4, \ldots \ldots \ldots a_n$

Geometric mean = $\sqrt[n]{a_1× a_2× a_3× a_4× .......× a_n}$

Example: Geometric mean of 2, 4, 8, and 16 = $\sqrt[4]{2×4×8×16}=\sqrt[4]{1024}=5.66$

Harmonic Mean

The Harmonic Mean is the reciprocal of the average of the reciprocals of a given set of numbers.

In simpler terms, you take the average of the reciprocals (1 divided by each number) and then take the reciprocal of that result.

The Harmonic Mean is always less than the Arithmetic Mean and less than or equal to the Geometric Mean.

In general, it is mostly used to find ratios and rates.

For a set of $n$ positive integers $a_1, a_2, a_3, a_4, \ldots \ldots . . a_n$

Harmonic mean = $\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+\frac{1}{a_4}+.......+\frac{1}{a_n}}$

Example:

If you go to a place at a speed of 60 km/hr and return with a speed of 40 km/hr. Then to find the average during this journey Harmonic Mean is used.

So, to go 1 km, you will take $\frac{1}{60}$

And to return 1 km, you will take $\frac{1}{40}$

So, average speed = $\frac{\frac{1}{60}+\frac{1}{40}}{2}=\frac{1}{48}$

Finally, we will take the reciprocal of that average i.e., 48 km/hr as the average speed.

Important formulae on Average

Weighted Average: Concept

Weighted Average measures central tendency that considers each value's relative importance, or weight, in a dataset. Unlike a simple average, which treats all values equally, a weighted average assigns different weights to different values, reflecting their varying significance in the calculation.

In weighted average, we multiply each value by its corresponding weight and sum it up. Then divide it by the sum of the weights.

The formula of weighted average can be written as:

$\bar x = \frac{w_1x_1+w_2x_2+w_3x_3+.....+w_nx_n}{w_1+w_2+w_3+.....+w_n}$,

where, $w_1,w_2,w_3,.....,w_n$ are corresponding weights of $x_1,x_2,x_3,....,x_n$, respectively.

In summation, formula can be written as $\frac{\displaystyle\sum _{i=1} ^{n}w_ix_i}{\displaystyle\sum _{i=1} ^{n}w_i}$,
where $i$ = 1 , 2, 3, . . . . ., n

In a simple way, the formula is, weighted average = $\frac{\text{Sum of weighted terms}}{\text{Total number of terms}}$

Example:

The average of marks secured by 50 students in Division A of class X is 61, 25 students in Division B is 57 and that of 50 students in Division C is 55. What will be the average marks of the students of the three divisions of Class X?

Sol: Total marks secured by 50 students in Division A of Class X = 50 × 61 = 3050

Total marks secured by 25 students in Division B of Class X = 25 × 57 = 1425

Total marks secured by 50 students in Division C of Class X = 50 × 55 = 2750

Total marks secured by all the students of Class X = 3050+1425+2750=7225

The average of the marks of three divisions of Class X $= \frac{\text{Total marks secured by all the students}}{\text{Number of Students}}=\frac{7225}{50+25+50}=\frac{7225}{125} = 57.8$

Hence, the average mark of the students of the three divisions of Class X is 57.8.

Application of weighted Average

There are wide ranges of use of weighted averages in Mathematics as well as real-life scenarios.

• Weighted averages can be used to calculate the average marks of a school or a class when different sets of students get different average marks.

• Weighted averages are used for finance or investment purposes when different amounts of money are invested for different time periods.

• It is also used in statistics and data analysis.

• In pharmacology, weighted averages can help determine appropriate drug dosages based on patient-specific factors such as weight or age.

• Weighted averages can be used to determine overall product quality scores by considering different quality metrics with different weights.

Tips and Tricks

• Average = $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$

• The average of the first $n$ natural numbers is $\frac{(n+1)}{2}$.

• The average of the first $n$ natural numbers squares is $\frac{(n+1)(2n+1)}{6}$.

• The average of the first $n$ natural numbers cubes is $\frac{n(n+1)^2}{4}$.

• If K goes P from Q with a speed $x$ km/hr and returns from Q to P with a speed $y$ km/hr, then Average speed = $\frac{2xy}{x+y}$

• weighted average = $\frac{\text{Sum of weighted terms}}{\text{Total number of terms}}$

Practice Questions

Q1. In an examination, the average marks obtained by John in English, Maths, Hindi and Drawing were 50. His average marks in Maths, Science, Social Studies and Craft were 70. If the average marks in all seven subjects are 58, his score in Maths is:

1. 50

2. 52

3. 60

4. 74

Hint: Use the formula: Sum of marks = average marks × number of subjects

Answer:

Given: The average marks of John in English, Maths, Hindi and Drawing = 50

$\therefore$ The sum of marks in English, Maths, Hindi and Drawing = 50 × 4 = 200

The average marks in Maths, Science, Social Studies and Craft = 70

$\therefore$ The sum of marks in Maths, Science, Social Studies and Craft = 70 × 4 = 280

The average marks in all seven subjects = 58

$\therefore$ The sum of marks in all seven subjects = 58 × 7 = 406

Marks scored in Maths = (200 + 280) – 406 = 74

Hence, the correct answer is 74.

Q2. The average mark obtained by a class of 60 students is 65. The average marks of half of the students were found to be 85. The average marks of the remaining students are:

1. 35

2. 45

3. 55

4. 65

Hint: Use this formula: Total marks = average marks × number of students

Answer:

Given that the average mark of 60 students is 65.

Total marks = (60 × 65) = 3900

The average mark of half of the students = 85

Therefore, total marks of 30 students = 30 × 85 = 2550

Total marks of the remaining students = 3900 – 2550 = 1350

So, the Average mark of the remaining 30 students = $\frac{1350}{30}$ = 45

Hence, the correct answer is 45.

Q3. In the first 30 overs of a cricket match, the run rate was 5.2 runs/over. What is the required run rate in the remaining 20 overs to reach the target of 280 runs?

1. 6.8

2. 7.4

3. 6.2

4. 5.8

Hint: Use the formula given below to get the unknown value,

The required run rate = $\frac{\text{The remaining runs}}{\text{The remaining overs}}$

Answer:

Given:

The run rate in the first 30 overs was 5.2 runs per over.

The total runs in the first 30 overs = 30 × 5.2 = 156

The target score is 280.

The remaining runs = 280 – 156 = 124

The remaining overs are 20.

$\therefore$ The required run rate = $\frac{\text{The remaining runs}}{\text{The remaining overs}}$=$\frac{124}{20}$ = 6.2

Hence, the correct answer is 6.2.

Q4. Three Science classes A, B, and C take a Life Science test. The average score for class A is 83. The average score for class B is 76. The average score in class C is 85. The average score of classes A and B is 79, and the average score of classes B and C is 81. Then the average score of classes A, B, and C is:

1. 81.5

2. 81

3. 80.5

4. 80

Hint: Let the number of students in classes A, B, and C be $x$, $y$, and $z$ and find the ratio of $x, y$, and $z$. Then use the formula given below to get the desired value.

Average = $\frac{\text{Total Score}}{x+y+z}$

Answer:

The average score of classes A, B, and C are 83, 76, and 85 respectively.

The average score of class A and class B = 79

The average score of class B and class C = 81

Let the number of students in classes A, B, and C be $x, y$, and $z$, respectively.

So, the total score of classes A, B, and C = $83x, 76y$, and $85z$

Also, total score of class (A + B) is $= 79(x + y) = 79x + 79y$

and total score of class (B + C) is $= 81(y + z) = 81y + 81z$

Now, $83x + 76y = 79x + 79y$

⇒ $4x = 3y$

⇒ $\frac{x}{y}$ = $\frac{3}{4}$

And, $76y + 85z = 81y + 81z$

⇒ $5y = 4z$

⇒ $\frac{\text{y}}{\text{z}}$ = $\frac{4}{5}$

So, $x : y : z = 3 : 4 : 5$

$\therefore$ Required average = $\frac{83×3+76×4+85×5}{12}$ = $\frac{249+304+425}{12}$ = $\frac{978}{12}$ = 81.5

Hence, the correct answer is 81.5.

Q5. The average of 11 numbers is 36, whereas the average of 9 of them is 34. If the remaining two numbers are in the ratio of 2 : 3, find the value of the smaller number (between the remaining two numbers).

1. 45

2. 48

3. 54

4. 36

Hint: Use this formula: Sum of all observations = Average × Number of observations

Answer:

Given:

The average of 11 numbers is 36.

⇒ Total of 11 numbers = 36 × 11 = 396

The average of 9 of them is 34.

⇒ Total of 9 numbers = 34 × 9 = 306

$\therefore$ Total of the remaining two numbers = (396 – 306) = 90

The remaining two numbers are in the ratio of 2 : 3.

Therefore, the smaller number = $\frac{2}{2+3}×90$ = 36

Hence, the correct answer is 36.

Q6. The average number of runs scored by a batsman in 7 matches is 53 and in 9 other matches, his average is 33. What is the average number of runs scored by the batsman in 16 matches?

1. 41.75

2. 44.25

3. 47

4. 49.175

Hint: Use the formula below.

Average runs = $\frac{\text{Total runs}}{\text{Total matches}}$

Answer:

According to the question statement:

The sum of runs in 7 matches = 7 ⨯ 53 = 371

The sum of runs in 9 matches = 9 ⨯ 33 = 297

Total runs in 16 matches = 371 + 297 = 668

So, average runs scored in 16 matches = $\frac{668} {16}$ = 41.75

Hence, the correct answer is 41.75.

Q7. The average of some natural numbers is 15. If 30 is added to the first number and 5 is subtracted from the last number, the new average becomes 17.5. What is the total number of natural numbers?

1. 15

2. 30

3. 20

4. 10

Hint: Let the number of natural numbers = n. According to the question,

$\frac{15n+30-5}{n}$ = 17.5

Answer:

Let the number of natural numbers = n

The sum of these natural numbers = 15 × n = 15n

Since 30 is added and 5 is subtracted.

15n + 30 – 5 = 15n + 25

According to the question,

$\frac{15n+25}{n}$ = 17.5

⇒ 15n + 25 = 17.5n

⇒ 2.5n = 25

$\therefore$ n = 10

Hence, the correct answer is 10.

Q8. The average of prime numbers between 1 and 20 is:

1. 9

2. $9\frac{5}{8}$

3. $10\frac{1}{8}$

4. 8

Hint: Use this formula:

Average = $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$

Answer:

Prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17 and 19.

Average = $\frac{\text{Sum of all observations}}{{\text{Total number of observations}}}=\frac{(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)}{8}=\frac{77}{8}=9\frac{5}{8}$

Hence, the correct answer is $9\frac{5}{8}$.

Q9. In a class of 60 students, 20 are girls. The average weight of the boys in the class is 40 kg, while that of all the girls is 35 kg. What is the average weight (in kg) of the entire class (correct to two decimal places)?

1. 36.67

2. 38.33

3. 33.33

4. 40.67

Hint: Average weight of the entire class = $\frac{\text{Total weight of the boys + Total weight of the girls}}{\text{Total number of boys and girls}}$

Answer:

In a class of 60 students, 20 are girls.

Number of boys = 60 – 20 = 40

The average weight of the boys in the class is 40 kg.

Total weight of the boys in the class = 40 × 40 = 1600

The average weight of all the girls is 35 kg.

Total weight of all the girls = 35 × 20 = 700

$\therefore$ Average weight of the entire class = $\frac{700 + 1600}{60} = 38.33$ kg

Hence, the correct answer is 38.33 kg.

Q10. The average of three numbers of which the greatest is 16 is 12. If the smallest is half of the greatest, the remaining number is:

1. 12

2. 8

3. 14

4. 10

Hint: Using this formula:

Average = $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$

Answer:

The greatest number = 16

The smallest number = half of the greatest number = $\frac{16}{2} = 8$

Let the third number be $z$.

Average = $\frac{\text{Sum of all the observations}}{{\text{Total number of observations}}}$

$⇒\frac{16+8+z}{3}=12$

$⇒24+z=36$

$\therefore z =12$

Hence, the correct answer is 12.

Q11. During a school excursion, each student of the junior school was charged Rs. 325 and each student of the senior school was charged Rs. 400. If there were 80 students from junior school and the combined average amount charged per student was Rs. 352, how many students from senior school went for the excursion?

1. 55

2. 45

3. 50

4. 40

Hint: Total amount of money charged from the juniors + Total amount of money charged from the seniors = Total amount of money charged from all students

Answer:

Let the number of seniors be $x$.

The total amount of money charged from the juniors $=325 \times 80= 26000$

Total amount of money charged from all students $= 352 \times (80 + x)$

The total amount charged from the seniors $ =400 \times x$

So, $400 \times x + 26000 = 352 \times (80 + x)$

$⇒ 400x - 352x = 28160 - 26000$

$⇒ 48x = 2160$

$\therefore x = 45$

Hence, the correct answer is 45.