To help students score well in their annual exams, subject matter experts at Infinity Learn have created detailed solutions for Chapter 10 – Direct and Inverse Variations from RD Sharma Class 8 Maths. These solutions are designed in an easy-to-understand and engaging format, making complex concepts simple and enjoyable to learn.
This chapter, aligned with the latest CBSE syllabus, focuses on the concept of variations, particularly:
- Direct Variation
- Inverse Variation
The chapter includes two key exercises, both of which are fully solved and explained step-by-step in our solution module. These RD Sharma Class 8 Maths Solutions help students strengthen their problem-solving skills and build confidence for their final exams.
Students can also download the free PDF of these solutions using the link provided, making it easy to revise anytime, anywhere.
Key Concepts Covered in Chapter 10:
- What are Variations?
- Types of Variations
- Understanding Direct Variation
- Understanding Inverse Variation
Download Free RD Sharma Solutions PDF for Class 8 Maths Chapter 10 Direct and Inverse variations
You can download the free PDF of RD Sharma Class 8 Maths Chapter 10 – Direct and Inverse Variations solutions, prepared by expert Maths teachers. These solutions cover all exercise questions from Chapter 10 and are designed to help you understand the concepts easily and score better in your exams.
At Home-Tution, we provide free RD Sharma Solutions and NCERT solutions, along with other helpful study materials for students. You can also get RD Sharma Class 8 Solutions for all chapters in one place. These chapter-wise solutions are created by experienced teachers to help you revise the full syllabus and improve your marks in a simple and effective way.
Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 10
Q. Which of the following quantities vary directly with each other?
(i) Number of articles (x) and their price (y).
(ii) Weight of articles (x) and their cost (y).
(iii) Distance x and time y, speed remaining the same.
(iv) Wages (y) and number of hours (x) of work.
(v) Speed (x) and time (y) distance covered remaining the same).
(vi) Area of a land (x) and its cost (y).
Solution:
(i) Number of Articles (x) and Their Price (y):
When the number of articles increases, the total price also increases.
This is a case of direct proportion.
(ii) Weight of Articles (x) and Their Cost (y):
As the weight of the articles increases, the cost increases too.
This is a case of direct proportion.
(iii) Distance (x) and Time (y), with Constant Speed:
If the speed stays the same, more distance means more time.
This is a case of direct proportion.
(iv) Wages (y) and Number of Working Hours (x):
The more hours a person works, the more wages they earn.
This is a case of direct proportion.
(v) Speed (x) and Time (y), with Fixed Distance:
If the distance is fixed, increasing speed decreases time.
This is not a case of direct proportion.
(vi) Area of Land (x) and Its Cost (y):
As the area of land increases, the cost also increases.
This is a case of direct proportion.
Q. In which of the following tables x and y vary directly?
(i)
| a | 7 | 9 | 13 | 21 | 25 |
| b | 21 | 27 | 39 | 63 | 75 |
(ii)
| a | 10 | 20 | 30 | 40 | 46 |
| b | 5 | 10 | 15 | 20 | 23 |
(iii)
| a | 2 | 3 | 4 | 5 | 6 |
| b | 6 | 9 | 12 | 17 | 20 |
(iv)
| a | 12 | 22 | 32 | 42 | 52 |
| b | 13 | 23 | 33 | 43 | 53 |
Solution:
(i) Case 1 – Directly Proportional
In this table, the value of ‘b’ is always three times the value of ‘a’ in every column.
Therefore, ‘a’ and ‘b’ are directly proportional.
(ii) Case 2 – Directly Proportional
Here, the value of ‘b’ is always half of ‘a’ across all columns.
So, ‘a’ and ‘b’ are directly proportional.
(iii) Case 3 – Not Directly Proportional
In this table, the value of ‘b’ is not consistently three times ‘a’ in every column.
Hence, ‘a’ and ‘b’ are not directly proportional.
(iv) Case 4 – Not Directly Proportional
Here, the ratio between ‘a’ and ‘b’ keeps changing across columns.
So, ‘a’ and ‘b’ are not directly proportional.
Q. Fill in the blanks in each of the following so as to make the statement true:
(i) Two quantities are said to vary…. with each other if they increase (decrease) together in such a way that the ratio of the corresponding values remains same.
(ii) x and y are said to vary directly with each if for some positive number k, ………= k.
(iii) if u = 3v, then u and v vary…. with each other.
Solution:
(i) Two quantities are said to vary directly when both increase or decrease together in such a way that their ratio remains constant.
That means, if one value goes up, the other also goes up in the same ratio — and vice versa.
(ii) If two variables x and y vary directly, there exists a constant positive number k such that: x/y = k or x = ky
This constant k shows that the ratio between x and y always stays the same.
(iii) For example, if u = 3v, it means u and v change in the same ratio.
Hence, u and v are in direct variation.
Complete the following tables given that x varies directly as y.
(i)
| x | 2.5 | … | … | 15 |
| y | 5 | 8 | 12 | … |
(ii)
| x | 5 | … | 10 | 35 | 25 | … |
| y | 8 | 12 | … | … | … | 32 |
(iii)
| x | 6 | 8 | 10 | … | 20 |
| y | 15 | 20 | … | 40 | … |
(iv)
| x | 4 | 9 | … | … | 3 | … |
| y | 16 | … | 48 | 36 | … | 4 |
(v)
| x | 3 | 5 | 7 | 9 |
| y | … | 20 | 28 | … |
Solution:
(i)
We know k = x/y
2.5/5 = x1/8
By cross-multiplying
8(2.5) = 5×1
20 = 5x1
x1 = 20/5
= 4
We know k = x/y
4/8 = x2/12
By cross-multiplying
12(4) = 8x2
48 = 8x2
x2 = 48/8
= 6
We know k = x/y
6/12 = 15/y1
By cross-multiplying
6y1 = 15(12)
6y1 = 180
y1 = 180/6
= 30
| x | 2.5 | 4 | 6 | 15 |
| y | 5 | 8 | 12 | 30 |
(ii)
We know k = x/y
5/8 = x1/12
By cross-multiplying
12(5) = 8x1
60 = 8x1
x1 = 60/8
= 7.5
We know k = x/y
7.5/12 = 10/y1
By cross-multiplying
7.5y1 = 10(12)
7.5y1 = 120
y1 = 120/7.5
= 16
We know k = x/y
10/16 = 35/y2
By cross-multiplying
10y2 = 35(16)
10y2 = 560
y2 = 560/10
= 56
We know k = x/y
35/56 = 25/y3
By cross-multiplying
35y3 = 56(25)
35y3 = 1400
y3 = 1400/35
= 40
We know k = x/y
25/40 = x2/32
By cross-multiplying
25(32) = 40x2
800 = 40x2
x2 = 800/40
= 20
| x | 5 | 7.5 | 10 | 35 | 25 | 20 |
| y | 8 | 12 | 16 | 56 | 40 | 32 |
(iii)
We know k = x/y
8/20 = 10/y1
By cross-multiplying
8y1 = 10(20)
8y1 = 200
y1 = 200/8
= 25
We know k = x/y
10/25 = x1/40
By cross-multiplying
10(40) = 25x1
400 = 25x1
x1 = 400/25
= 16
We know k = x/y
16/40 = 20/y2
By cross-multiplying
16y2 = 20(40)
16y2 = 800
y2 = 800/16
= 50
| x | 6 | 8 | 10 | 16 | 20 |
| y | 15 | 20 | 25 | 40 | 50 |
(iv)
We know k = x/y
4/16 = 9/y1
By cross-multiplying
4y1 = 9(16)
= 144
y1 = 144/4
= 36
We know k = x/y
9/36 = x1/48
By cross-multiplying
9(48) = 36x1
432 = 36x1
x1 = 432/36
= 12
We know k = x/y
12/48 = x2/36
By cross-multiplying
12(36) = 48x2
432 = 48x2
x2 = 432/48
= 9
We know k = x/y
9/36 = 3/y2
By cross-multiplying
9y2 = 3(36)
= 108
y2 = 108/9
= 12
We know k = x/y
3/12 = x3/4
By cross-multiplying
3(4) = 12x3
12 = 12x3
x3 = 12/12
= 1
| x | 4 | 9 | 12 | 9 | 3 | 1 |
| y | 16 | 36 | 48 | 36 | 12 | 4 |
(v)
We know k = x/y
3/y1 = 5/20
By cross-multiplying
3(20) = 5y1
60 = 5y1
y1 = 60/5
= 12
We know k = x/y
7/28 = 9/y2
By cross-multiplying
7y2 = 9(28)
= 252
y2 = 252/7
= 36
| x | 3 | 5 | 7 | 9 |
| y | 12 | 20 | 28 | 36 |
Also Read:
- RD Sharma Solutions for Class 8 Maths Chapter 1
- RD Sharma Solutions for Class 8 Maths Chapter 2
- RD Sharma Solutions for Class 8 Maths Chapter 3
- RD Sharma Solutions for Class 8 Maths Chapter 4
- RD Sharma Solutions for Class 8 Maths Chapter 5
- RD Sharma Solutions for Class 8 Maths Chapter 6
- RD Sharma Solutions for Class 8 Maths Chapter 7
- RD Sharma Solutions for Class 8 Maths Chapter 8
Importance of RD Sharma Maths Solutions Class 8 Chapter 10
- RD Sharma solutions for Chapter 10 help students clearly understand the two key types of variation direct and inverse. These are foundational concepts in algebra and real-life problem-solving, and mastering them ensures better understanding of future mathematical topics like linear equations, ratios, and proportions.
- The solutions are strictly based on the latest CBSE syllabus Class 10, ensuring that students are preparing exactly what’s required for school exams. The exercises and problems match the NCERT Solutions Class 10 pattern and promote structured learning.
- Each problem in Chapter 10 is solved in a step-by-step manner, making it easy for students to follow the logic behind each step. This encourages logical thinking and analytical reasoning, which are essential for competitive exams.
- Chapter 10 includes two key exercises, one focusing on direct variation and the other on inverse variation. The RD Sharma solutions provide accurate and detailed answers for every question, helping students practice effectively.
- By regularly practicing these solutions, students develop the problem-solving speed and accuracy needed in their annual exams. The variety of problems boosts confidence and reduces exam fear.
- Students who rely on self-study or need help with their homework will find these solutions incredibly useful. The answers are explained in simple language, making it easy to learn independently without external help.
- Many educational platforms like Home-Tution offer free PDF downloads of RD Sharma Solutions. This makes them easily accessible for quick reference during revision or doubt-clearing sessions.
Frequently Asked Questions
Ans. Chapter 10 explains both direct variation (where y increases as x increases) and inverse variation (where y decreases as x increases), covering related formulas and real-life examples.
Ans. The solutions offer clear, step-by-step explanations, helping students grasp variation concepts, improve accuracy, and prepare effectively for exams.
Ans. You can download the free PDF of Chapter 10 solutions from platforms like Home-Tution.
Ans. Chapter 10 consists of Exercise 10.1 (focused on direct variation) and Exercise 10.2 (focused on inverse variation), with full solutions available online.
Ans. Practicing these exercises helps fortify conceptual understanding, enhance problem-solving speed, and align with the CBSE syllabus boosting academic performance.