What must added to each terms of the ratio 7 13 so that the ratio becomes 2 3?
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RD Sharma Solutions for Class 7 Maths Chapter 9 Ratio and Proportion is the best study material for students who are wishing to secure excellence in board exams. The PDF of this chapter are available here, and students can download for free. RD Sharma Solutions are useful for students as it helps them in scoring high marks in the examination. These solutions are formulated by subject matter experts of BYJU’S with utmost care. Show
Chapter 9, Ratio and Proportion, has three exercises. RD Sharma Solutions for Class 7 given here contains the solutions to all the questions present in these exercises. Let us have a look at some of the concepts that are being discussed in this chapter.
Chapter 9 Ratio and Proportion
Exercise
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Access answers to Maths RD Sharma Solutions For Class 7 Chapter 9 – Ratio and ProportionExercise 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: 42y/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 = 1750 Expenditure = Income – savings = 1750 – 500 = Rs.1250 14. The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides. Solution: Given sides of a triangle are in the ratio 1: 2: 3 Perimeter = 36cm Sum of the terms of the ratio = 1 + 2 + 3 = 6 First side = (1/6) × 36 = 6cm Second side = (2/6) × 36 = 2 × 6 = 12cm Third side = (3/6) × 36 = 6 × 3 = 18cm 15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get? Solution: Given total amount to be divided = 5500 Sum of the terms of the ratio = 2 + 3 = 5 Raman’s share of money = (2/5) × 5500 = 2 × 1100 = Rs. 2200 And Aman’s share of money = (3/5) × 5500 = 3 × 1100 = Rs. 3300 16. The ratio of zinc and copper in an alloy is 7: 9. It the weight of the copper in the alloy is 11.7 kg, find the weight of the zinc in the alloy. Solution: Given that ratio of zinc and copper in an alloy is 7: 9 Let their ratio = 7x: 9x Weight of copper = 11.7kg 9x = 11.7 x = 11.7/9 x = 1.3 Weight of the zinc in the alloy = 1.3 × 7 = 9.10kg 17. In the ratio 7: 8. If the consequent is 40, what a the antecedent Solution: Given ratio = 7: 8 Let the ratio of consequent and antecedent 7x: 8x Consequent = 40 8x = 40 x = 40/8 x = 5 Antecedent = 7x = 7 × 5 = 35 18. Divide Rs 351 into two parts such that one may be to the other as 2: 7. Solution: Given total amount is to be divided = 351 Ratio 2: 7 The sum of terms = 2 + 7 = 9 First ratio of amount = (2/9) × 351 = 2 × 39 = Rs. 78 Second ratio of amount = (7/9) × 351 = 7 × 39 = Rs. 273 19. Find the ratio of the price of pencil to that of ball pen, if pencil cost Rs.16 per score and ball pen cost Rs.8.40 per dozen. Solution: One score contains 20 pencils And cost per score = 16 Therefore pencil cost = 16/20 = Rs. 0.80 Cost of one dozen ball pen = 8.40 1 dozen = 12 Therefore cost of pen = 8.40/12 = Rs 0.70 Ratio of the price of pencil to that of ball pen = 0.80/0.70 = 8/7 = 8: 7 20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass? Solution: Given, total number of students = 42 One out of 6 student fails x out of 42 students 1/6 = x/42 x = 42/6 x = 7 Number of students who fail = 7 students No of students who pass =Total students – Number of students who fail = 42 – 7 = 35 students. Exercise 9.2 Page No: 9.10 1. Which ratio is larger in the following pairs? (i) 3: 4 or 9: 16 (ii) 15: 16 or 24: 25 (iii) 4: 7 or 5: 8 (iv) 9: 20 or 8: 13 (v) 1: 2 or 13: 27 Solution: (i) Given 3: 4 or 9: 16 LCM for 4 and 16 is 16 3: 4 can be written as = 3/4 3/4 × (4/4) = 12/16 And we have 9/16 Clearly 12 > 9 Therefore 3: 4 > 9: 16 (ii) Given 15: 16 or 24: 25 LCM for 16 and 25 is 400 15: 16 can be written as = 15/16 15/16 × (25/25) = 375/400 And we have 24/25 24/25 × (16/16) = 384/400 Clearly 384 > 375 Therefore 15: 16 < 24: 25 (iii) Given 4: 7 or 5: 8 LCM for 7 and 8 is 56 4: 7 can be written as = 4/7 4/7 × (8/8) = 32/56 And we have 5/8 5/8 × (7/7) = 35/56 Clearly 35 > 32 Therefore 4: 7 < 5: 8 (iv) Given 9: 20 or 8: 13 LCM for 20 and 13 is 260 9: 20 can be written as = 9/20 9/20 × (13/13) = 117/260 And we have 8/13 8/13 × (20/20) = 160/260 Clearly 160 > 117 Therefore 9: 20 < 8: 13 (v) Given 1: 2 or 13: 27 LCM for 2 and 27 is 54 1: 2 can be written as = 1/2 1/2 × (27/27) = 27/54 And we have 13/27 13/27 × (2/2) = 26/54 Clearly 27 > 26 Therefore 1: 2 > 13: 27 2. Give the equivalent ratios of 6: 8. Solution: Given 6: 8 By multiplying both numerator and denominator by 2 we equivalent ratios 6/8 × (2/2) = 12/16 And also by dividing both numerator and denominator by 2 we equivalent ratios (6/2)/ (8/2) = 3/4 Two equivalent ratios are 3: 4 = 12: 16 3. Fill in the following blanks: 12/20 = …. /5 = 9/…. Solution: 12/20 = 3/5 = 9/15 Explanation: Consider 12/20 = …. /5 Let unknown value be x Therefore 12/20 = x/5 On cross multiplying x = 60/20 x = 3 Consider 12/20 = 9/…. Let the unknown value be y Therefore 12/20 = 9/y On cross multiplying we get y = 180/12 y = 15 Exercise 9.3 Page No: 9.13 1. Find which of the following are in proportion? (i) 33, 44, 66, 88 (ii) 46, 69, 69, 46 (iii) 72, 84, 186, 217 Solution: (i) Given 33, 44, 66, 88 Product of extremes = 33 × 88 = 2904 Product of means = 44 × 66 = 2904 Therefore product of extremes = product of means Hence given numbers are in proportion. (ii) Given 46, 69, 69, 46 Product of extremes = 46 × 46 = 2116 Product of means = 69 × 69 = 4761 Therefore product of extremes is not equal to product of means Hence given numbers are not in proportion. (iii) Given 72, 84, 186, 217 Product of extremes = 72 × 217 = 15624 Product of means = 84 × 186 = 15624 Therefore product of extremes = product of means Hence given numbers are in proportion. 2. Find x in the following proportions: (i) 16: 18 = x: 96 (ii) x: 92 = 87: 116 Solution: (i) Given 16: 18 = x: 96 In proportion we know that product of extremes = product of means 16/18 = x/96 On cross multiplying x = (16 × 96)/ 18 x = 1536/18 Dividing both numerator and denominator by 6 x = 256/3 (ii) Given x: 92 = 87: 116 In proportion we know that product of extremes = product of means x/ 92 = 87/116 On cross multiplying x = (87 × 92)/ 116 x = 69 3. The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400. Solution: Given that income = 1400 Given the ratio of income and expenditure = 7: 6 7x = 1400 Therefore x = 200 Expenditure = 6x = 6 × 200 = Rs.1200 Savings = Income – Expenditure = 1400 -1200 = Rs.200 4. The scale of a map is 1: 4000000. What is the actual distance between the two towns if they are 5cm apart on the map? Solution: Given that the scale of map = 1: 4000000 Let us assume the actual distance between towns is x cm 1: 4000000 =5: x x = 5 × 4000000 x = 20000000 cm We know that 1km = 1000 m 1m = 100 cm Therefore x = 200 km 5. The ratio of income of a person to his savings is 10: 1. If his savings for one year is Rs.6000, what is his income per month? Solution: Given that the ratio of income of a person to his savings is 10: 1 Savings per year = 6000 Savings per month = 6000/12 = Rs.500 Then let income per month be x x: 500 = 10:1 x = 500 × 10 x = 5000 Income per month is Rs. 5000 6. An electric pole casts a shadow of length 20 meters at a time when a tree 6 meters high casts a shadow of length 8 meters. Find the height of the pole. What same number must be added to each term of the ratio 7 13?x=5. Was this answer helpful?
What must be added to each term of the ratio 2 is to 3?X = 2. So the ratio of 2:3 when added 2 to each term is 2*2:3+2.
What should be added to each term of the ratio 7 12 so that the ratio becomes 4 5?So, 13 must be added to each term of the ratio 7 : 12, so that the ratio becomes 4 : 5.
What number must be added to each term of the ratio 7 ratio 5 so that the ratio becomes 5 ratio 4?5 is the Answer.
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