{"id":546697,"date":"2021-10-07T10:13:24","date_gmt":"2021-10-07T10:13:24","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546697"},"modified":"2021-10-09T06:35:41","modified_gmt":"2021-10-09T06:35:41","slug":"rd-sharma-solutions-for-class-7-maths-chapter-9-ratio-and-proportion","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-9-ratio-and-proportion\/","title":{"rendered":"RD Sharma Solutions for Class 7 Maths Chapter 9\u2013Ratio And Proportion"},"content":{"rendered":"\n

Class 7: Maths Chapter 9 solutions. Complete Class 7 Maths Chapter 9 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 9\u2013Ratio And Proportion<\/h2>\n\n\n\n

RD Sharma 7th Maths Chapter 9, Class 7 Maths Chapter 9 solutions<\/p>\n\n\n\n

Exercise 9.1 Page No: 9.6<\/p>\n\n\n\n

1. If x: y = 3: 5, find the ratio 3x + 4y: 8x + 5y<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x: y = 3: 5<\/p>\n\n\n\n

We can write above equation as<\/p>\n\n\n\n

x\/y = 3\/5<\/p>\n\n\n\n

5x = 3y<\/p>\n\n\n\n

x = 3y\/5<\/p>\n\n\n\n

By substituting the value of x in given equation 3x + 4y: 8x + 5y we get,<\/p>\n\n\n\n

3x + 4y: 8x + 5y = 3 (3y\/5) + 4y: 8 (3y\/5) + 5y<\/p>\n\n\n\n

= (9y + 20y)\/5: (24y + 25y)\/5<\/p>\n\n\n\n

= 29y\/5: 49y\/5<\/p>\n\n\n\n

= 29y: 49y<\/p>\n\n\n\n

= 29: 49<\/p>\n\n\n\n

2. If x: y = 8: 9, find the ratio (7x \u2013 4y): 3x + 2y.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given x: y = 8: 9<\/p>\n\n\n\n

We can write above equation as<\/p>\n\n\n\n

x\/y = 8\/9<\/p>\n\n\n\n

9x = 8y<\/p>\n\n\n\n

x = 8y\/9<\/p>\n\n\n\n

By substituting the value of x in the given equation (7x \u2013 4y): 3x + 2y we get,<\/p>\n\n\n\n

(7x \u2013 4y): 3x + 2y = 7 (8y\/9) \u2013 4y: 3 (8y\/9) + 2y<\/p>\n\n\n\n

= (56y \u2013 36y)\/9: 42y\/9<\/p>\n\n\n\n

= 20y\/9: 42y\/9<\/p>\n\n\n\n

= 20y: 42y<\/p>\n\n\n\n

= 20: 42<\/p>\n\n\n\n

= 10: 21<\/p>\n\n\n\n

3. If two numbers are in the ratio 6: 13 and their L.C.M is 312, find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given two numbers are in the ratio 6: 13<\/p>\n\n\n\n

Let the required number be 6x and 13x<\/p>\n\n\n\n

The LCM of 6x and 13x is 78x<\/p>\n\n\n\n

= 78x = 312<\/p>\n\n\n\n

x = (312\/78)<\/p>\n\n\n\n

x = 4<\/p>\n\n\n\n

Thus the numbers are 6x = 6 (4) = 24<\/p>\n\n\n\n

13x = 13 (4) = 52<\/p>\n\n\n\n

4. Two numbers are in the ratio 3: 5. If 8 is added to each number, the ratio becomes 2:3. Find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the required numbers be 3x and 5x<\/p>\n\n\n\n

Given that if 8 is added to each other then ratio becomes 2: 3<\/p>\n\n\n\n

That is 3x + 8: 5x + 8 = 2: 3<\/p>\n\n\n\n

(3x + 8)\/ (5x + 8) = 2\/3<\/p>\n\n\n\n

3 (3x + 8) = 2 (5x + 8)<\/p>\n\n\n\n

9x + 24 = 10x + 16<\/p>\n\n\n\n

By transposing<\/p>\n\n\n\n

24 \u2013 16 = 10x \u2013 9x<\/p>\n\n\n\n

x = 8<\/p>\n\n\n\n

Thus the numbers are 3x = 3 (8) = 24<\/p>\n\n\n\n

And 5x = 5 (8) = 40<\/p>\n\n\n\n

5. What should be added to each term of the ratio 7: 13 so that the ratio becomes 2: 3<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the number to be added is x<\/p>\n\n\n\n

Then (7 + x)\/ (13 + x) = (2\/3)<\/p>\n\n\n\n

(7 + x) 3 = 2 (13 + x)<\/p>\n\n\n\n

21 + 3x = 26 + 2x<\/p>\n\n\n\n

3x \u2013 2x = 26 \u2013 21<\/p>\n\n\n\n

x = 5<\/p>\n\n\n\n

Hence the required number is 5<\/p>\n\n\n\n

6. Three numbers are in the ratio 2: 3: 5 and the sum of these numbers is 800. Find the numbers<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that three numbers are in the ratio 2: 3: 5 and sum of them is 800<\/p>\n\n\n\n

Therefore sum of the terms of the ratio = 2 + 3 + 5 = 10<\/p>\n\n\n\n

First number = (2\/10) \u00d7 800<\/p>\n\n\n\n

= 2 \u00d7 80<\/p>\n\n\n\n

= 160<\/p>\n\n\n\n

Second number = (3\/10) \u00d7 800<\/p>\n\n\n\n

= 3 \u00d7 80<\/p>\n\n\n\n

= 240<\/p>\n\n\n\n

Third number = (5\/10) \u00d7 800<\/p>\n\n\n\n

= 5 \u00d7 80<\/p>\n\n\n\n

= 400<\/p>\n\n\n\n

The three numbers are 160, 240 and 400<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let present ages of two persons be 5x and 7x<\/p>\n\n\n\n

Given ages of two persons are in the ratio 5: 7<\/p>\n\n\n\n

And also given that 18 years ago their ages were in the ratio 8: 13<\/p>\n\n\n\n

Therefore (5x \u2013 18)\/ (7x \u2013 18) = (8\/13)<\/p>\n\n\n\n

13 (5x \u2013 18) = 8 (7x \u2013 18)<\/p>\n\n\n\n

65x \u2013 234 = 56x \u2013 144<\/p>\n\n\n\n

65x \u2013 56x = 234 \u2013 144<\/p>\n\n\n\n

9x = 90<\/p>\n\n\n\n

x = 90\/9<\/p>\n\n\n\n

x = 10<\/p>\n\n\n\n

Thus the ages are 5x = 5 (10) = 50 years<\/p>\n\n\n\n

And 7x = 7 (10) = 70 years<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the required numbers be 7x and 11x<\/p>\n\n\n\n

If 7 is added to each of them then<\/p>\n\n\n\n

(7x + 7)\/ (11x + 7) = (2\/3)<\/p>\n\n\n\n

3 (7x + 7) = 2 (11x + 7)<\/p>\n\n\n\n

21x + 21 = 22x + 14<\/p>\n\n\n\n

22x \u2013 21x = 21 \u2013 14<\/p>\n\n\n\n

x = 21 \u2013 14 = 7<\/p>\n\n\n\n

Thus the numbers are 7x = 7 (7) =49<\/p>\n\n\n\n

And 11x = 11 (7) = 77<\/p>\n\n\n\n

9. Two numbers are in the ratio 2: 7. 11 the sum of the numbers is 810. Find the numbers.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given two numbers are in the ratio 2: 7<\/p>\n\n\n\n

And their sum = 810<\/p>\n\n\n\n

Sum of terms in the ratio = 2 + 7 = 9<\/p>\n\n\n\n

First number = (2\/9) \u00d7 810<\/p>\n\n\n\n

= 2 \u00d7 90<\/p>\n\n\n\n

= 180<\/p>\n\n\n\n

Second number = (7\/9) \u00d7 810<\/p>\n\n\n\n

= 7 \u00d7 90<\/p>\n\n\n\n

= 630<\/p>\n\n\n\n

10. Divide Rs 1350 between Ravish and Shikha in the ratio 2: 3.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given total amount to be divided = 1350<\/p>\n\n\n\n

Sum of the terms of the ratio = 2 + 3 = 5<\/p>\n\n\n\n

Ravish share of money = (2\/5) \u00d7 1350<\/p>\n\n\n\n

= 2 \u00d7 270<\/p>\n\n\n\n

= Rs. 540<\/p>\n\n\n\n

And Shikha\u2019s share of money = (3\/5) \u00d7 1350<\/p>\n\n\n\n

= 3 \u00d7 270<\/p>\n\n\n\n

= Rs. 810<\/p>\n\n\n\n

11. Divide Rs 2000 among P, Q, R in the ratio 2: 3: 5.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given total amount to be divided = 2000<\/p>\n\n\n\n

Sum of the terms of the ratio = 2 + 3 + 5 = 10<\/p>\n\n\n\n

P\u2019s share of money = (2\/10) \u00d7 2000<\/p>\n\n\n\n

= 2 \u00d7 200<\/p>\n\n\n\n

= Rs. 400<\/p>\n\n\n\n

And Q\u2019s share of money = (3\/10) \u00d7 2000<\/p>\n\n\n\n

= 3 \u00d7 200<\/p>\n\n\n\n

= Rs. 600<\/p>\n\n\n\n

And R\u2019s share of money = (5\/10) \u00d7 2000<\/p>\n\n\n\n

= 5 \u00d7 200<\/p>\n\n\n\n

= Rs. 1000<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that boys and the girls in a school are in the ratio 7:4<\/p>\n\n\n\n

Sum of the terms of the ratio = 7 + 4 = 11<\/p>\n\n\n\n

Total strength = 550<\/p>\n\n\n\n

Boys strength = (7\/11) \u00d7 550<\/p>\n\n\n\n

= 7 \u00d7 50<\/p>\n\n\n\n

= 350<\/p>\n\n\n\n

Girls strength = (4\/11) \u00d7 550<\/p>\n\n\n\n

= 4 \u00d7 50<\/p>\n\n\n\n

= 200<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that the ratio of income and savings is 7: 2<\/p>\n\n\n\n

Let the savings be 2x<\/p>\n\n\n\n

2x = 500<\/p>\n\n\n\n

So, x = 250<\/p>\n\n\n\n

Therefore,<\/p>\n\n\n\n

Income = 7x<\/p>\n\n\n\n

Income = 7 \u00d7 250 = 1750<\/p>\n\n\n\n

Expenditure = Income \u2013 savings<\/p>\n\n\n\n

= 1750 \u2013 500<\/p>\n\n\n\n

= Rs.1250<\/p>\n\n\n\n

14. The sides of a triangle are in the ratio 1: 2: 3. If the perimeter is 36 cm, find its sides.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given sides of a triangle are in the ratio 1: 2: 3<\/p>\n\n\n\n

Perimeter = 36cm<\/p>\n\n\n\n

Sum of the terms of the ratio = 1 + 2 + 3 = 6<\/p>\n\n\n\n

First side = (1\/6) \u00d7 36<\/p>\n\n\n\n

= 6cm<\/p>\n\n\n\n

Second side = (2\/6) \u00d7 36<\/p>\n\n\n\n

= 2 \u00d7 6<\/p>\n\n\n\n

= 12cm<\/p>\n\n\n\n

Third side = (3\/6) \u00d7 36<\/p>\n\n\n\n

= 6 \u00d7 3<\/p>\n\n\n\n

= 18cm<\/p>\n\n\n\n

15. A sum of Rs 5500 is to be divided between Raman and Amen in the rate 2: 3. How much will each get?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given total amount to be divided = 5500<\/p>\n\n\n\n

Sum of the terms of the ratio = 2 + 3 = 5<\/p>\n\n\n\n

Raman\u2019s share of money = (2\/5) \u00d7 5500<\/p>\n\n\n\n

= 2 \u00d7 1100<\/p>\n\n\n\n

= Rs. 2200<\/p>\n\n\n\n

And Aman\u2019s share of money = (3\/5) \u00d7 5500<\/p>\n\n\n\n

= 3 \u00d7 1100<\/p>\n\n\n\n

= Rs. 3300<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that ratio of zinc and copper in an alloy is 7: 9<\/p>\n\n\n\n

Let their ratio = 7x: 9x<\/p>\n\n\n\n

Weight of copper = 11.7kg<\/p>\n\n\n\n

9x = 11.7<\/p>\n\n\n\n

x = 11.7\/9<\/p>\n\n\n\n

x = 1.3<\/p>\n\n\n\n

Weight of the zinc in the alloy = 1.3 \u00d7 7<\/p>\n\n\n\n

= 9.10kg<\/p>\n\n\n\n

17. In the ratio 7: 8. If the consequent is 40, what a the antecedent<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given ratio = 7: 8<\/p>\n\n\n\n

Let the ratio of consequent and antecedent 7x: 8x<\/p>\n\n\n\n

Consequent = 40<\/p>\n\n\n\n

8x = 40<\/p>\n\n\n\n

x = 40\/8<\/p>\n\n\n\n

x = 5<\/p>\n\n\n\n

Antecedent = 7x = 7 \u00d7 5 = 35<\/p>\n\n\n\n

18. Divide Rs 351 into two parts such that one may be to the other as 2: 7.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given total amount is to be divided = 351<\/p>\n\n\n\n

Ratio 2: 7<\/p>\n\n\n\n

The sum of terms = 2 + 7<\/p>\n\n\n\n

= 9<\/p>\n\n\n\n

First ratio of amount = (2\/9) \u00d7 351<\/p>\n\n\n\n

= 2 \u00d7 39<\/p>\n\n\n\n

= Rs. 78<\/p>\n\n\n\n

Second ratio of amount = (7\/9) \u00d7 351<\/p>\n\n\n\n

= 7 \u00d7 39<\/p>\n\n\n\n

= Rs. 273<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

One score contains 20 pencils<\/p>\n\n\n\n

And cost per score = 16<\/p>\n\n\n\n

Therefore pencil cost = 16\/20<\/p>\n\n\n\n

= Rs. 0.80<\/p>\n\n\n\n

Cost of one dozen ball pen = 8.40<\/p>\n\n\n\n

1 dozen = 12<\/p>\n\n\n\n

Therefore cost of pen = 8.40\/12<\/p>\n\n\n\n

= Rs 0.70<\/p>\n\n\n\n

Ratio of the price of pencil to that of ball pen = 0.80\/0.70<\/p>\n\n\n\n

= 8\/7<\/p>\n\n\n\n

= 8: 7<\/p>\n\n\n\n

20. In a class, one out of every six students fails. If there are 42 students in the class, how many pass?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given, total number of students = 42<\/p>\n\n\n\n

One out of 6 student fails<\/p>\n\n\n\n

x out of 42 students<\/p>\n\n\n\n

1\/6 = x\/42<\/p>\n\n\n\n

x = 42\/6<\/p>\n\n\n\n

x = 7<\/p>\n\n\n\n

Number of students who fail = 7 students<\/p>\n\n\n\n

No of students who pass =Total students \u2013 Number of students who fail<\/p>\n\n\n\n

= 42 \u2013 7<\/p>\n\n\n\n

= 35 students.<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.2 Page No: 9.10<\/p>\n\n\n\n

1. Which ratio is larger in the following pairs?<\/strong><\/p>\n\n\n\n

(i) 3: 4 or 9: 16<\/strong><\/p>\n\n\n\n

(ii) 15: 16 or 24: 25<\/strong><\/p>\n\n\n\n

(iii) 4: 7 or 5: 8<\/strong><\/p>\n\n\n\n

(iv) 9: 20 or 8: 13<\/strong><\/p>\n\n\n\n

(v) 1: 2 or 13: 27<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 3: 4 or 9: 16<\/p>\n\n\n\n

LCM for 4 and 16 is 16<\/p>\n\n\n\n

3: 4 can be written as = 3\/4<\/p>\n\n\n\n

3\/4 \u00d7 (4\/4) = 12\/16<\/p>\n\n\n\n

And we have 9\/16<\/p>\n\n\n\n

Clearly 12 > 9<\/p>\n\n\n\n

Therefore 3: 4 > 9: 16<\/p>\n\n\n\n

(ii) Given 15: 16 or 24: 25<\/p>\n\n\n\n

LCM for 16 and 25 is 400<\/p>\n\n\n\n

15: 16 can be written as = 15\/16<\/p>\n\n\n\n

15\/16 \u00d7 (25\/25) = 375\/400<\/p>\n\n\n\n

And we have 24\/25<\/p>\n\n\n\n

24\/25 \u00d7 (16\/16) = 384\/400<\/p>\n\n\n\n

Clearly 384 > 375<\/p>\n\n\n\n

Therefore 15: 16 < 24: 25<\/p>\n\n\n\n

(iii) Given 4: 7 or 5: 8<\/p>\n\n\n\n

LCM for 7 and 8 is 56<\/p>\n\n\n\n

4: 7 can be written as = 4\/7<\/p>\n\n\n\n

4\/7 \u00d7 (8\/8) = 32\/56<\/p>\n\n\n\n

And we have 5\/8<\/p>\n\n\n\n

5\/8 \u00d7 (7\/7) = 35\/56<\/p>\n\n\n\n

Clearly 35 > 32<\/p>\n\n\n\n

Therefore 4: 7 < 5: 8<\/p>\n\n\n\n

(iv) Given 9: 20 or 8: 13<\/p>\n\n\n\n

LCM for 20 and 13 is 260<\/p>\n\n\n\n

9: 20 can be written as = 9\/20<\/p>\n\n\n\n

9\/20 \u00d7 (13\/13) = 117\/260<\/p>\n\n\n\n

And we have 8\/13<\/p>\n\n\n\n

8\/13 \u00d7 (20\/20) = 160\/260<\/p>\n\n\n\n

Clearly 160 > 117<\/p>\n\n\n\n

Therefore 9: 20 < 8: 13<\/p>\n\n\n\n

(v) Given 1: 2 or 13: 27<\/p>\n\n\n\n

LCM for 2 and 27 is 54<\/p>\n\n\n\n

1: 2 can be written as = 1\/2<\/p>\n\n\n\n

1\/2 \u00d7 (27\/27) = 27\/54<\/p>\n\n\n\n

And we have 13\/27<\/p>\n\n\n\n

13\/27 \u00d7 (2\/2) = 26\/54<\/p>\n\n\n\n

Clearly 27 > 26<\/p>\n\n\n\n

Therefore 1: 2 > 13: 27<\/p>\n\n\n\n

2. Give the equivalent ratios of 6: 8.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given 6: 8<\/p>\n\n\n\n

By multiplying both numerator and denominator by 2 we equivalent ratios<\/p>\n\n\n\n

6\/8 \u00d7 (2\/2) = 12\/16<\/p>\n\n\n\n

And also by dividing both numerator and denominator by 2 we equivalent ratios<\/p>\n\n\n\n

(6\/2)\/ (8\/2) = 3\/4<\/p>\n\n\n\n

Two equivalent ratios are 3: 4 = 12: 16<\/p>\n\n\n\n

3. Fill in the following blanks:<\/strong><\/p>\n\n\n\n

12\/20 = \u2026. \/5 = 9\/\u2026.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

12\/20 = 3\/5 = 9\/15<\/p>\n\n\n\n

Explanation:<\/strong><\/p>\n\n\n\n

Consider 12\/20 = \u2026. \/5<\/p>\n\n\n\n

Let unknown value be x<\/p>\n\n\n\n

Therefore 12\/20 = x\/5<\/p>\n\n\n\n

On cross multiplying<\/p>\n\n\n\n

x = 60\/20<\/p>\n\n\n\n

x = 3<\/p>\n\n\n\n

Consider 12\/20 = 9\/\u2026.<\/p>\n\n\n\n

Let the unknown value be y<\/p>\n\n\n\n

Therefore 12\/20 = 9\/y<\/p>\n\n\n\n

On cross multiplying we get<\/p>\n\n\n\n

y = 180\/12<\/p>\n\n\n\n

y = 15<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.3 Page No: 9.13<\/p>\n\n\n\n

1. Find which of the following are in proportion?<\/strong><\/p>\n\n\n\n

(i) 33, 44, 66, 88<\/strong><\/p>\n\n\n\n

(ii) 46, 69, 69, 46<\/strong><\/p>\n\n\n\n

(iii) 72, 84, 186, 217<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 33, 44, 66, 88<\/p>\n\n\n\n

Product of extremes = 33 \u00d7 88 = 2904<\/p>\n\n\n\n

Product of means = 44 \u00d7 66 = 2904<\/p>\n\n\n\n

Therefore product of extremes = product of means<\/p>\n\n\n\n

Hence given numbers are in proportion.<\/p>\n\n\n\n

(ii) Given 46, 69, 69, 46<\/p>\n\n\n\n

Product of extremes = 46 \u00d7 46 = 2116<\/p>\n\n\n\n

Product of means = 69 \u00d7 69 = 4761<\/p>\n\n\n\n

Therefore product of extremes is not equal to product of means<\/p>\n\n\n\n

Hence given numbers are not in proportion.<\/p>\n\n\n\n

(iii) Given 72, 84, 186, 217<\/p>\n\n\n\n

Product of extremes = 72 \u00d7 217 = 15624<\/p>\n\n\n\n

Product of means = 84 \u00d7 186 = 15624<\/p>\n\n\n\n

Therefore product of extremes = product of means<\/p>\n\n\n\n

Hence given numbers are in proportion.<\/p>\n\n\n\n

2. Find x in the following proportions:<\/strong><\/p>\n\n\n\n

(i) 16: 18 = x: 96<\/strong><\/p>\n\n\n\n

(ii) x: 92 = 87: 116<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

(i) Given 16: 18 = x: 96<\/p>\n\n\n\n

In proportion we know that product of extremes = product of means<\/p>\n\n\n\n

16\/18 = x\/96<\/p>\n\n\n\n

On cross multiplying<\/p>\n\n\n\n

x = (16 \u00d7 96)\/ 18<\/p>\n\n\n\n

x = 1536\/18<\/p>\n\n\n\n

Dividing both numerator and denominator by 6<\/p>\n\n\n\n

x = 256\/3<\/p>\n\n\n\n

(ii) Given x: 92 = 87: 116<\/p>\n\n\n\n

In proportion we know that product of extremes = product of means<\/p>\n\n\n\n

x\/ 92 = 87\/116<\/p>\n\n\n\n

On cross multiplying<\/p>\n\n\n\n

x = (87 \u00d7 92)\/ 116<\/p>\n\n\n\n

x = 69<\/p>\n\n\n\n

3. The ratio of income to the expenditure of a family is 7: 6. Find the savings if the income is Rs.1400.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that income = 1400<\/p>\n\n\n\n

Given the ratio of income and expenditure = 7: 6<\/p>\n\n\n\n

7x = 1400<\/p>\n\n\n\n

Therefore x = 200<\/p>\n\n\n\n

Expenditure = 6x = 6 \u00d7 200 = Rs.1200<\/p>\n\n\n\n

Savings = Income \u2013 Expenditure<\/p>\n\n\n\n

= 1400 -1200<\/p>\n\n\n\n

= Rs.200<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that the scale of map = 1: 4000000<\/p>\n\n\n\n

Let us assume the actual distance between towns is x cm<\/p>\n\n\n\n

1: 4000000 =5: x<\/p>\n\n\n\n

x = 5 \u00d7 4000000<\/p>\n\n\n\n

x = 20000000 cm<\/p>\n\n\n\n

We know that 1km = 1000 m<\/p>\n\n\n\n

1m = 100 cm<\/p>\n\n\n\n

Therefore<\/p>\n\n\n\n

x = 200 km<\/p>\n\n\n\n

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?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that the ratio of income of a person to his savings is 10: 1<\/p>\n\n\n\n

Savings per year = 6000<\/p>\n\n\n\n

Savings per month = 6000\/12<\/p>\n\n\n\n

= Rs.500<\/p>\n\n\n\n

Then let income per month be x<\/p>\n\n\n\n

x: 500 = 10:1<\/p>\n\n\n\n

x = 500 \u00d7 10<\/p>\n\n\n\n

x = 5000<\/p>\n\n\n\n

Income per month is Rs. 5000<\/p>\n\n\n\n

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.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given that length electric pole shadow is 20m<\/p>\n\n\n\n

Height of the tree: Length of the shadow of tree<\/p>\n\n\n\n

Height of the pole: Length of the shadow of pole<\/p>\n\n\n\n

x: 20 = 6: 8<\/p>\n\n\n\n

x = 120\/8<\/p>\n\n\n\n

x = 15<\/p>\n\n\n\n

Therefore height of the pole is 15 meters<\/p>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 7 Maths Chapter 9\u2013Ratio And Proportion<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 7 Maths Chapter 9\u2013Ratio And Proportion PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 7 Maths :<\/strong><\/h2>\n\n\n\n