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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
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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.
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