CBSE Class 8 Maths Chapter 13: Direct and Inverse Proportions NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 8 PDF

This chapter delves into the concepts of direct and inverse proportions, crucial for understanding relationships between quantities in mathematics. The NCERT Solutions for Class 8 Maths Chapter 13 provide step-by-step explanations for various problems. Students will learn to identify whether two quantities are directly or inversely proportional and solve problems involving these relationships. The solutions cover practical scenarios like calculating parking charges and mixing paint pigments. By working through these exercises, students will develop a strong foundation in proportion concepts, enhancing their problem-solving skills and preparing them for exams. These solutions are designed to clarify complex ideas and reinforce learning through practice.

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Quick info

Board	CBSE
Class	Class 8
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 13

Chapter summary

Chapter 13 of the CBSE Class 8 Maths syllabus focuses on Direct and Inverse Proportions. This chapter's NCERT Solutions explain how quantities change together. It covers identifying direct proportionality, where an increase in one quantity leads to a proportional increase in another, and inverse proportionality, where an increase in one leads to a proportional decrease in another. The solutions provide clear methods to solve problems related to these concepts, using examples like car parking charges and paint mixtures.

Learning outcomes

Understand the concept of direct proportion.
Identify situations where quantities are in direct proportion.
Solve problems involving direct proportion.
Understand the concept of inverse proportion.
Identify situations where quantities are in inverse proportion.
Solve problems involving inverse proportion.

Topics covered

Paper topics

Direct Proportion
Inverse Proportion
Ratio and Proportion
Constant of Proportionality
Identifying Proportional Relationships
Solving Proportionality Problems
Real-world Applications of Proportion

Important topics

Understanding Direct Proportion
Understanding Inverse Proportion
Calculating Constant of Proportionality
Solving Word Problems involving Proportions
Identifying Proportional Scenarios

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Questions and Solutions

Question 1

Following are the car parking charges near a railway station for different durations:

4 hours: ₹60

8 hours: ₹100

12 hours: ₹140

24 hours: ₹180

Check if the parking charges are in direct proportion to the parking time.

Solution:

To check if the parking charges are in direct proportion to the parking time, we need to see if the ratio of charges to time is constant for all given durations.

Let's calculate the charges per hour for each duration:

For 4 hours, the charge is ₹60. The rate is $\frac{60}{4} = ₹15$ per hour.
For 8 hours, the charge is ₹100. The rate is $\frac{100}{8} = ₹12.50$ per hour.
For 12 hours, the charge is ₹140. The rate is $\frac{140}{12} \approx ₹11.67$ per hour.
For 24 hours, the charge is ₹180. The rate is $\frac{180}{24} = ₹7.50$ per hour.

Since the charges per hour ( $₹15, ₹12.50, ₹11.67, ₹7.50$ ) are not the same for all durations, the parking charges are not in direct proportion to the parking time.

Question 2

A mixture of paint is prepared by mixing 1 part of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added for the given parts of red pigment, assuming the ratio remains constant:

Parts of red pigment: 1, 4, 7, 12, 20

Parts of base: 8, ____, ____, ____, ____

Solution:

This is a problem of direct proportion, where the ratio of parts of red pigment to parts of base is constant.

Let the parts of red pigment be $a$ and the parts of base be $b$ . The constant ratio $k$ is given by $k = \frac{a}{b}$ .

From the first case, we have $a_1 = 1$ and $b_1 = 8$ . So, the constant ratio is $k = \frac{1}{8}$ .

Now, we can find the parts of base needed for other amounts of red pigment:

When parts of red pigment $a_2 = 4$ : $\frac{4}{b_2} = \frac{1}{8} \Rightarrow b_2 = 4 \times 8 = 32$ parts of base.
When parts of red pigment $a_3 = 7$ : $\frac{7}{b_3} = \frac{1}{8} \Rightarrow b_3 = 7 \times 8 = 56$ parts of base.
When parts of red pigment $a_4 = 12$ : $\frac{12}{b_4} = \frac{1}{8} \Rightarrow b_4 = 12 \times 8 = 96$ parts of base.
When parts of red pigment $a_5 = 20$ : $\frac{20}{b_5} = \frac{1}{8} \Rightarrow b_5 = 20 \times 8 = 160$ parts of base.

So, the completed table is:

Parts of red pigment: 1, 4, 7, 12, 20

Parts of base: 8, 32, 56, 96, 160

Question 3

In the paint mixture scenario from Question 2, if 1 part of red pigment requires 75 mL of base, how much red pigment should be mixed with 1800 mL of base to maintain the same proportion?

Solution:

This problem involves direct proportion. We are given that 1 part of red pigment requires 75 mL of base. We need to find the amount of red pigment ( $x$ parts) that should be mixed with 1800 mL of base.

We can set up a proportion:

Parts of red pigment : Parts of base

1 : 75

x : 1800

Since these quantities are in direct proportion, the ratio is constant:

$\frac{1}{75} = \frac{x}{1800}$

To solve for $x$ , we cross-multiply:

$75 \times x = 1 \times 1800$

$75x = 1800$

Now, divide both sides by 75:

$x = \frac{1800}{75}$

$x = 24$

Therefore, 24 parts of red pigment should be mixed with 1800 mL of base.

Common mistakes

Confusing direct proportion with inverse proportion.
Errors in calculating the constant of proportionality.
Incorrectly setting up the proportion equation.
Calculation errors when solving for an unknown quantity.

Revision tips

Clearly distinguish between direct and inverse proportion scenarios.
Practice calculating the constant of proportionality for each type.
Work through all examples and exercises to build confidence.
Relate the concepts to real-world examples to aid understanding.

Practice MCQs

Q1. If two quantities x and y are in direct proportion, what happens when x increases?

Q2. In a direct proportion, the ratio of corresponding quantities is:

Q3. If 5 kg of sugar costs ₹150, what is the cost of 1 kg of sugar?

Q4. A mixture requires 1 part red pigment and 8 parts base. If we have 4 parts of red pigment, how much base is needed for direct proportion?

Q5. Which of the following is NOT an example of direct proportion?

Frequently asked questions

What is direct proportion?

Direct proportion means that as one quantity increases, the other quantity increases by the same factor, and vice versa. Their ratio remains constant.

What is inverse proportion?

Inverse proportion means that as one quantity increases, the other quantity decreases by the same factor, and vice versa. Their product remains constant.

How do these NCERT Solutions help Class 8 students?

These solutions provide clear, step-by-step explanations for problems in Chapter 13, helping students understand the concepts of direct and inverse proportions and how to apply them.

Are the questions in the solutions the same as the NCERT textbook?

Yes, the questions are kept the same as in the NCERT textbook, ensuring students can follow along and practice the exact problems.

What kind of real-world examples are covered?

The solutions cover practical examples such as car parking charges and mixing paint pigments, illustrating how proportions are used in everyday life.

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