Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Ratio & Proportion (Exercise 9.1)
Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 Exercise 9.1 Page No: 9.6 1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y Solution: Given x: y = 3: 5 We can write above equation as x/y = 3/5 5x = 3y x = 3y/5 By substituting the value of x in given equation 3x + 4y: 8x + 5y we get, 3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y = (9y + 20y)/5: (24y + 25y)/5 = 29y/5: 49y/5 = 29y: 49y = 29: 49 2. If x: y = 8: 9, find the ratio (7x - 4y): 3x + 2y. Solution: Given x: y = 8: 9 We can write above equation as x/y = 8/9 9x = 8y x = 8y/9 By substituting the value of x in the given equation (7x - 4y): 3x + 2y we get, (7x - 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y = (56y – 36y)/9: (24y + 18y)/9 = 20y/9: 42y/9 = 20y: 42y = 20: 42 = 10: 21 3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. Solution: Given two numbers are in the ratio 6: 13 Page 2 Exercise 9.1 Page No: 9.6 1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y Solution: Given x: y = 3: 5 We can write above equation as x/y = 3/5 5x = 3y x = 3y/5 By substituting the value of x in given equation 3x + 4y: 8x + 5y we get, 3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y = (9y + 20y)/5: (24y + 25y)/5 = 29y/5: 49y/5 = 29y: 49y = 29: 49 2. If x: y = 8: 9, find the ratio (7x - 4y): 3x + 2y. Solution: Given x: y = 8: 9 We can write above equation as x/y = 8/9 9x = 8y x = 8y/9 By substituting the value of x in the given equation (7x - 4y): 3x + 2y we get, (7x - 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y = (56y – 36y)/9: (24y + 18y)/9 = 20y/9: 42y/9 = 20y: 42y = 20: 42 = 10: 21 3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. Solution: Given two numbers are in the ratio 6: 13 Let the required number be 6x and 13x The LCM of 6x and 13x is 78x = 78x = 312 x = (312/78) x = 4 Thus the numbers are 6x = 6 (4) = 24 13x = 13 (4) = 52 4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers. Solution: Let the required numbers be 3x and 5x Given that if 8 is added to each other then ratio becomes 2: 3 That is 3x + 8: 5x + 8 = 2: 3 (3x + 8)/ (5x + 8) = 2/3 3 (3x + 8) = 2 (5x + 8) 9x + 24 = 10x + 16 By transposing 24 – 16 = 10x – 9x x = 8 Thus the numbers are 3x = 3 (8) = 24 And 5x = 5 (8) = 40 5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 Solution: Let the number to be added is x Then (7 + x) / (13 + x) = (2/3) (7 + x) 3 = 2 (13 + x) 21 + 3x = 26 + 2x 3x – 2x = 26 – 21 x = 5 Hence the required number is 5 6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers Page 3 Exercise 9.1 Page No: 9.6 1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y Solution: Given x: y = 3: 5 We can write above equation as x/y = 3/5 5x = 3y x = 3y/5 By substituting the value of x in given equation 3x + 4y: 8x + 5y we get, 3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y = (9y + 20y)/5: (24y + 25y)/5 = 29y/5: 49y/5 = 29y: 49y = 29: 49 2. If x: y = 8: 9, find the ratio (7x - 4y): 3x + 2y. Solution: Given x: y = 8: 9 We can write above equation as x/y = 8/9 9x = 8y x = 8y/9 By substituting the value of x in the given equation (7x - 4y): 3x + 2y we get, (7x - 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y = (56y – 36y)/9: (24y + 18y)/9 = 20y/9: 42y/9 = 20y: 42y = 20: 42 = 10: 21 3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. Solution: Given two numbers are in the ratio 6: 13 Let the required number be 6x and 13x The LCM of 6x and 13x is 78x = 78x = 312 x = (312/78) x = 4 Thus the numbers are 6x = 6 (4) = 24 13x = 13 (4) = 52 4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers. Solution: Let the required numbers be 3x and 5x Given that if 8 is added to each other then ratio becomes 2: 3 That is 3x + 8: 5x + 8 = 2: 3 (3x + 8)/ (5x + 8) = 2/3 3 (3x + 8) = 2 (5x + 8) 9x + 24 = 10x + 16 By transposing 24 – 16 = 10x – 9x x = 8 Thus the numbers are 3x = 3 (8) = 24 And 5x = 5 (8) = 40 5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 Solution: Let the number to be added is x Then (7 + x) / (13 + x) = (2/3) (7 + x) 3 = 2 (13 + x) 21 + 3x = 26 + 2x 3x – 2x = 26 – 21 x = 5 Hence the required number is 5 6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers Solution: Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800 Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10 First number = (2/10) × 800 = 2 × 80 = 160 Second number = (3/10) × 800 = 3 × 80 = 240 Third number = (5/10) × 800 = 5 × 80 = 400 The three numbers are 160, 240 and 400 7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages. Solution: Let present ages of two persons be 5x and 7x Given ages of two persons are in the ratio 5: 7 And also given that 18 years ago their ages were in the ratio 8: 13 Therefore (5x – 18)/ (7x – 18) = (8/13) 13 (5x – 18) = 8 (7x – 18) 65x – 234 = 56x – 144 65x – 56x = 234 – 144 9x = 90 x = 90/9 x = 10 Thus the ages are 5x = 5 (10) = 50 years And 7x = 7 (10) = 70 years 8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers. Solution: Let the required numbers be 7x and 11x If 7 is added to each of them then Page 4 Exercise 9.1 Page No: 9.6 1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y Solution: Given x: y = 3: 5 We can write above equation as x/y = 3/5 5x = 3y x = 3y/5 By substituting the value of x in given equation 3x + 4y: 8x + 5y we get, 3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y = (9y + 20y)/5: (24y + 25y)/5 = 29y/5: 49y/5 = 29y: 49y = 29: 49 2. If x: y = 8: 9, find the ratio (7x - 4y): 3x + 2y. Solution: Given x: y = 8: 9 We can write above equation as x/y = 8/9 9x = 8y x = 8y/9 By substituting the value of x in the given equation (7x - 4y): 3x + 2y we get, (7x - 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y = (56y – 36y)/9: (24y + 18y)/9 = 20y/9: 42y/9 = 20y: 42y = 20: 42 = 10: 21 3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. Solution: Given two numbers are in the ratio 6: 13 Let the required number be 6x and 13x The LCM of 6x and 13x is 78x = 78x = 312 x = (312/78) x = 4 Thus the numbers are 6x = 6 (4) = 24 13x = 13 (4) = 52 4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers. Solution: Let the required numbers be 3x and 5x Given that if 8 is added to each other then ratio becomes 2: 3 That is 3x + 8: 5x + 8 = 2: 3 (3x + 8)/ (5x + 8) = 2/3 3 (3x + 8) = 2 (5x + 8) 9x + 24 = 10x + 16 By transposing 24 – 16 = 10x – 9x x = 8 Thus the numbers are 3x = 3 (8) = 24 And 5x = 5 (8) = 40 5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 Solution: Let the number to be added is x Then (7 + x) / (13 + x) = (2/3) (7 + x) 3 = 2 (13 + x) 21 + 3x = 26 + 2x 3x – 2x = 26 – 21 x = 5 Hence the required number is 5 6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers Solution: Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800 Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10 First number = (2/10) × 800 = 2 × 80 = 160 Second number = (3/10) × 800 = 3 × 80 = 240 Third number = (5/10) × 800 = 5 × 80 = 400 The three numbers are 160, 240 and 400 7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages. Solution: Let present ages of two persons be 5x and 7x Given ages of two persons are in the ratio 5: 7 And also given that 18 years ago their ages were in the ratio 8: 13 Therefore (5x – 18)/ (7x – 18) = (8/13) 13 (5x – 18) = 8 (7x – 18) 65x – 234 = 56x – 144 65x – 56x = 234 – 144 9x = 90 x = 90/9 x = 10 Thus the ages are 5x = 5 (10) = 50 years And 7x = 7 (10) = 70 years 8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers. Solution: Let the required numbers be 7x and 11x If 7 is added to each of them then (7x + 7)/ (11x + 7) = (2/3) 3 (7x + 7) = 2 (11x + 7) 21x + 21 = 22x + 14 22x – 21x = 21 – 14 x = 21 – 14 = 7 Thus the numbers are 7x = 7 (7) =49 And 11x = 11 (7) = 77 9. Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers. Solution: Given two numbers are in the ratio 2: 7 And their sum = 810 Sum of terms in the ratio = 2 + 7 = 9 First number = (2/9) × 810 = 2 × 90 = 180 Second number = (7/9) × 810 = 7 × 90 = 630 10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3. Solution: Given total amount to be divided = 1350 Sum of the terms of the ratio = 2 + 3 = 5 Ravish share of money = (2/5) × 1350 = 2 × 270 = Rs. 540 And Shikha’s share of money = (3/5) × 1350 = 3 × 270 = Rs. 810 11. Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5. Solution: Page 5 Exercise 9.1 Page No: 9.6 1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y Solution: Given x: y = 3: 5 We can write above equation as x/y = 3/5 5x = 3y x = 3y/5 By substituting the value of x in given equation 3x + 4y: 8x + 5y we get, 3x + 4y: 8x + 5y = 3 (3y/5) + 4y: 8 (3y/5) + 5y = (9y + 20y)/5: (24y + 25y)/5 = 29y/5: 49y/5 = 29y: 49y = 29: 49 2. If x: y = 8: 9, find the ratio (7x - 4y): 3x + 2y. Solution: Given x: y = 8: 9 We can write above equation as x/y = 8/9 9x = 8y x = 8y/9 By substituting the value of x in the given equation (7x - 4y): 3x + 2y we get, (7x - 4y): 3x + 2y = 7 (8y/9) – 4y: 3 (8y/9) + 2y = (56y – 36y)/9: (24y + 18y)/9 = 20y/9: 42y/9 = 20y: 42y = 20: 42 = 10: 21 3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers. Solution: Given two numbers are in the ratio 6: 13 Let the required number be 6x and 13x The LCM of 6x and 13x is 78x = 78x = 312 x = (312/78) x = 4 Thus the numbers are 6x = 6 (4) = 24 13x = 13 (4) = 52 4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers. Solution: Let the required numbers be 3x and 5x Given that if 8 is added to each other then ratio becomes 2: 3 That is 3x + 8: 5x + 8 = 2: 3 (3x + 8)/ (5x + 8) = 2/3 3 (3x + 8) = 2 (5x + 8) 9x + 24 = 10x + 16 By transposing 24 – 16 = 10x – 9x x = 8 Thus the numbers are 3x = 3 (8) = 24 And 5x = 5 (8) = 40 5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3 Solution: Let the number to be added is x Then (7 + x) / (13 + x) = (2/3) (7 + x) 3 = 2 (13 + x) 21 + 3x = 26 + 2x 3x – 2x = 26 – 21 x = 5 Hence the required number is 5 6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers Solution: Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800 Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10 First number = (2/10) × 800 = 2 × 80 = 160 Second number = (3/10) × 800 = 3 × 80 = 240 Third number = (5/10) × 800 = 5 × 80 = 400 The three numbers are 160, 240 and 400 7. The ages of two persons are in the ratio 5: 7. Eighteen years ago their ages were in the ratio 8: 13. Find their present ages. Solution: Let present ages of two persons be 5x and 7x Given ages of two persons are in the ratio 5: 7 And also given that 18 years ago their ages were in the ratio 8: 13 Therefore (5x – 18)/ (7x – 18) = (8/13) 13 (5x – 18) = 8 (7x – 18) 65x – 234 = 56x – 144 65x – 56x = 234 – 144 9x = 90 x = 90/9 x = 10 Thus the ages are 5x = 5 (10) = 50 years And 7x = 7 (10) = 70 years 8. Two numbers are in the ratio 7: 11. If 7 is added to each of the numbers, the ratio becomes 2: 3. Find the numbers. Solution: Let the required numbers be 7x and 11x If 7 is added to each of them then (7x + 7)/ (11x + 7) = (2/3) 3 (7x + 7) = 2 (11x + 7) 21x + 21 = 22x + 14 22x – 21x = 21 – 14 x = 21 – 14 = 7 Thus the numbers are 7x = 7 (7) =49 And 11x = 11 (7) = 77 9. Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers. Solution: Given two numbers are in the ratio 2: 7 And their sum = 810 Sum of terms in the ratio = 2 + 7 = 9 First number = (2/9) × 810 = 2 × 90 = 180 Second number = (7/9) × 810 = 7 × 90 = 630 10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3. Solution: Given total amount to be divided = 1350 Sum of the terms of the ratio = 2 + 3 = 5 Ravish share of money = (2/5) × 1350 = 2 × 270 = Rs. 540 And Shikha’s share of money = (3/5) × 1350 = 3 × 270 = Rs. 810 11. Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5. Solution: Given total amount to be divided = 2000 Sum of the terms of the ratio = 2 + 3 + 5 = 10 P’s share of money = (2/10) × 2000 = 2 × 200 = Rs. 400 And Q’s share of money = (3/10) × 2000 = 3 × 200 = Rs. 600 And R’s share of money = (5/10) × 2000 = 5 × 200 = Rs. 1000 12. The boys and the girls in a school are in the ratio 7:4. If total strength of the school be 550, find the number of boys and girls. Solution: Given that boys and the girls in a school are in the ratio 7:4 Sum of the terms of the ratio = 7 + 4 = 11 Total strength = 550 Boys strength = (7/11) × 550 = 7 × 50 = 350 Girls strength = (4/11) × 550 = 4 × 50 = 200 13. The ratio of monthly income to the savings of a family is 7: 2. If the savings be of Rs. 500, find the income and expenditure. Solution: Given that the ratio of income and savings is 7: 2 Let the savings be 2x 2x = 500 So, x = 250 Therefore, Income = 7x Income = 7 × 250 = 1750Read More
|
76 videos|344 docs|39 tests
|
FAQs on Ratio & Proportion (Exercise 9.1) RD Sharma Solutions - Mathematics (Maths) Class 7
| 1. What is the definition of ratio in mathematics? |  |
Ans. In mathematics, a ratio is a comparison between two or more quantities. It represents the relationship between the sizes of different quantities and is expressed by dividing one quantity by another.
| 2. How do you simplify a ratio? | |
Ans. To simplify a ratio, you need to divide both the numerator and denominator of the ratio by their greatest common divisor (GCD). This will result in a simplified ratio with the smallest possible whole numbers.
| 3. What is the concept of proportion? | |
Ans. Proportion is a mathematical concept that relates two ratios or fractions to each other. It states that the two ratios are equal and can be represented by an equation. In a proportion, the product of the means is equal to the product of the extremes.
| 4. How do you solve problems involving ratio and proportion? | |
Ans. To solve problems involving ratio and proportion, you can use the cross-multiplication method. In this method, you set up a proportion equation with two ratios, cross-multiply the terms, and solve for the unknown variable.
| 5. Can ratios and proportions be used in real-life situations? | |
Ans. Yes, ratios and proportions are used in various real-life situations. They are commonly used in cooking recipes, financial calculations, construction projects, and even in understanding and analyzing data. Ratios and proportions help in comparing and scaling quantities accurately.
Related Exams
About this Document
|
4.83/5 Rating |
|
Dec 26, 2024 Last updated |
Document Description: RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) for Class 7 2024 is part of Mathematics (Maths) Class 7 preparation.
The notes and questions for RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) have been prepared according to the Class 7 exam syllabus. Information about RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) covers topics
like and RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Example, for Class 7 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for RD Sharma Solutions: Ratio & Proportion (Exercise 9.1).
Introduction of RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) in English is available as part of our Mathematics (Maths) Class 7
for Class 7 & RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) in Hindi for Mathematics (Maths) Class 7 course.
Download more important topics related with notes, lectures and mock test series for Class 7
Exam by signing up for free. Class 7: Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7
Description
Full syllabus notes, lecture & questions for Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7 - Class 7 | Plus excerises question with solution to help you revise complete syllabus for Mathematics (Maths) Class 7 | Best notes, free PDF download
Information about RD Sharma Solutions: Ratio & Proportion (Exercise 9.1)
In this doc you can find the meaning of RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) defined & explained in the simplest way possible. Besides explaining types of
RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) theory, EduRev gives you an ample number of questions to practice RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) tests, examples and also practice Class 7
tests
|
Explore Courses for Class 7 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,video lectures
,ppt
,Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,Summary
,past year papers
,Free
,mock tests for examination
,Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,Objective type Questions
,Sample Paper
,Important questions
,MCQs
,Exam
,Ratio & Proportion (Exercise 9.1) RD Sharma Solutions | Mathematics (Maths) Class 7
,practice quizzes
,Extra Questions
,Previous Year Questions with Solutions
,Semester Notes
,shortcuts and tricks
,study material
;
Additional Information about RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) for Class 7 Preparation
RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Free PDF Download
The RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) is an invaluable resource that delves deep into the core of the Class 7 exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) now and kickstart your journey towards success in the Class 7 exam.
Importance of RD Sharma Solutions: Ratio & Proportion (Exercise 9.1)
The importance of RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) cannot be overstated, especially for Class 7 aspirants.
This document holds the key to success in the Class 7 exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Notes
RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to RD Sharma Solutions: Ratio & Proportion (Exercise 9.1).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Notes on EduRev are your ultimate resource for success.
RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Class 7 Questions
The "RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) Class 7 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 7 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) on the App
Students of Class 7 can study RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the RD Sharma Solutions: Ratio & Proportion (Exercise 9.1),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of RD Sharma Solutions: Ratio & Proportion (Exercise 9.1) is prepared as per the latest Class 7 syllabus.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup