{"id":547292,"date":"2021-10-09T10:51:01","date_gmt":"2021-10-09T10:51:01","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=547292"},"modified":"2021-10-11T07:46:23","modified_gmt":"2021-10-11T07:46:23","slug":"rd-sharma-solutions-for-class-6-maths-chapter-9-ratio-proportion-and-unitary-method","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-6-maths-chapter-9-ratio-proportion-and-unitary-method\/","title":{"rendered":"RD Sharma Solutions for Class 6 Maths Chapter 9\u2013Ratio, Proportion and Unitary Method"},"content":{"rendered":"\n

Class 6: Maths Chapter 9 solutions. Complete Class 6 Maths Chapter 9 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 9\u2013Ratio, Proportion and Unitary Method<\/h2>\n\n\n\n

RD Sharma 6th Maths Chapter 9, Class 6 Maths Chapter 9 solutions<\/p>\n\n\n\n

Exercise 9.1 page: 9.5<\/h4>\n\n\n\n

1. Express each of the following in the language of ratios:<\/strong><\/p>\n\n\n\n

(i) In a class, the number of girls in the merit list of the board examination is two times that of boys.<\/strong><\/p>\n\n\n\n

(ii) The number of students passing mathematics test is 2\/3 of the number that appeared.<\/strong><\/p>\n\n\n\n

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

(i) Ratio of the number of girls to that of boys in the merit list is 2: 1.<\/p>\n\n\n\n

(ii) Ratio of the number of students passing a mathematics test to that of total students appearing in the test is 2: 3.<\/p>\n\n\n\n

2. Express the following ratios in language of daily life:<\/strong><\/p>\n\n\n\n

(i) The ratio of the number of bad pencils to that of good pencils produced in a factory is 1: 9.<\/strong><\/p>\n\n\n\n

(ii) In India, the ratio of the number of villages to that of cities is about 2000: 1.<\/strong><\/p>\n\n\n\n

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

(i) The number of bad pencils produced in a factory is 1\/9 of the number of good pencils produced in the factory.<\/p>\n\n\n\n

(ii) The number of villages is 2000 times that of cities in India.<\/p>\n\n\n\n

3. Express each of the following ratios in its simplest form:<\/strong><\/p>\n\n\n\n

(i) 60: 72<\/strong><\/p>\n\n\n\n

(ii) 324: 144<\/strong><\/p>\n\n\n\n

(iii) 85: 391<\/strong><\/p>\n\n\n\n

(iv) 186: 403<\/strong><\/p>\n\n\n\n

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

(i) 60: 72<\/p>\n\n\n\n

It can be written as 60\/72<\/p>\n\n\n\n

We know that the HCF of 60 and 72 is 12<\/p>\n\n\n\n

By dividing the term by 12 we get<\/p>\n\n\n\n

(60\/72) \u00d7 (12\/12) = 5\/6<\/p>\n\n\n\n

So we get 60: 72 = 5: 6<\/p>\n\n\n\n

(ii) 324: 144<\/p>\n\n\n\n

It can be written as 324\/144<\/p>\n\n\n\n

We know that the HCF of 324 and 144 is 36<\/p>\n\n\n\n

By dividing the term by 36 we get<\/p>\n\n\n\n

(324\/144) \u00d7 (36\/36) = 9\/4<\/p>\n\n\n\n

So we get 324: 144 = 9: 4<\/p>\n\n\n\n

(iii) 85: 391<\/p>\n\n\n\n

It can be written as 85\/391<\/p>\n\n\n\n

We know that the HCF of 85 and 391 is 17<\/p>\n\n\n\n

By dividing the term by 17 we get<\/p>\n\n\n\n

(85\/391) \u00d7 (17\/17) = 5\/23<\/p>\n\n\n\n

So we get 85: 391 = 5: 23<\/p>\n\n\n\n

(iv) 186: 403<\/p>\n\n\n\n

It can be written as 186\/403<\/p>\n\n\n\n

We know that the HCF of 186 and 403 is 31<\/p>\n\n\n\n

By dividing the term by 31 we get<\/p>\n\n\n\n

(186\/403) \u00d7 (31\/31) = 6\/13<\/p>\n\n\n\n

So we get 186: 403 = 6: 13<\/p>\n\n\n\n

4. Find the ratio of the following in the simplest form:<\/strong><\/p>\n\n\n\n

(i) 75 paise to Rs 3<\/strong><\/p>\n\n\n\n

(ii) 35 minutes to 45 minutes<\/strong><\/p>\n\n\n\n

(iii) 8 kg to 400 gm<\/strong><\/p>\n\n\n\n

(iv) 48 minutes to 1 hour<\/strong><\/p>\n\n\n\n

(v) 2 metres to 35 cm<\/strong><\/p>\n\n\n\n

(vi) 35 minutes to 45 seconds<\/strong><\/p>\n\n\n\n

(vii) 2 dozen to 3 scores<\/strong><\/p>\n\n\n\n

(viii) 3 weeks to 3 days<\/strong><\/p>\n\n\n\n

(ix) 48 min to 2 hours 40 min<\/strong><\/p>\n\n\n\n

(x) 3 m 5 cm to 35 cm<\/strong><\/p>\n\n\n\n

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

(i) 75 paise to Rs 3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

75 paise to Rs 3 = 75 paise: Rs 3<\/p>\n\n\n\n

We know that 1 Rs = 100 paise<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

75 paise to Rs 3 = 75 paise: 300 paise<\/p>\n\n\n\n

Dividing the two terms by HCF 75<\/p>\n\n\n\n

75 paise to Rs 3 = 1: 4<\/p>\n\n\n\n

(ii) 35 minutes to 45 minutes<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

35 minutes to 45 minutes = 35 minutes: 45 minutes<\/p>\n\n\n\n

Dividing the two terms by HCF 5<\/p>\n\n\n\n

35 minutes to 45 minutes = 7: 9<\/p>\n\n\n\n

(iii) 8 kg to 400 gm<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

8 kg to 400 gm = 8 kg: 400 gm<\/p>\n\n\n\n

We know that 1 kg = 1000 gm<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

8 kg to 400 gm = 8000 gm: 400 gm<\/p>\n\n\n\n

Dividing the two terms by HCF 400<\/p>\n\n\n\n

8 kg to 400 gm = 20: 1<\/p>\n\n\n\n

(iv) 48 minutes to 1 hour<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

48 minutes to 1 hour = 48 minutes: 1 hour<\/p>\n\n\n\n

We know that 1 hour = 60 minutes<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

48 minutes to 1 hour = 48 minutes: 60 minutes<\/p>\n\n\n\n

Dividing the two terms by HCF 12<\/p>\n\n\n\n

48 minutes to 1 hour = 4: 5<\/p>\n\n\n\n

(v) 2 metres to 35 cm<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

2 metres to 35 cm = 2 metres: 35 cm<\/p>\n\n\n\n

We know that 1 m = 100 cm<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

2 metres to 35 cm = 200 cm: 35 cm<\/p>\n\n\n\n

Dividing the two terms by HCF 5<\/p>\n\n\n\n

2 metres to 35 cm = 40: 7<\/p>\n\n\n\n

(vi) 35 minutes to 45 seconds<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

35 minutes to 45 seconds = 35 minutes: 45 seconds<\/p>\n\n\n\n

We know that 1 minute = 60 seconds<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

35 minutes to 45 seconds = 2100 seconds: 45 seconds<\/p>\n\n\n\n

Dividing the two terms by HCF 15<\/p>\n\n\n\n

35 minutes to 45 seconds = 140: 3<\/p>\n\n\n\n

(vii) 2 dozen to 3 scores<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

2 dozen to 3 scores = 2 dozen: 3 scores<\/p>\n\n\n\n

We know that 1 dozen = 12 score = 20<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

2 dozen to 3 scores = 24: 60<\/p>\n\n\n\n

Dividing the two terms by HCF 12<\/p>\n\n\n\n

2 dozen to 3 scores = 2: 5<\/p>\n\n\n\n

(viii) 3 weeks to 3 days<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3 weeks to 3 days = 3 weeks: 3 days<\/p>\n\n\n\n

We know that 1 week = 7 days<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

3 weeks to 3 days = 21 days: 3 days<\/p>\n\n\n\n

Dividing the two terms by HCF 3<\/p>\n\n\n\n

3 weeks to 3 days = 7: 1<\/p>\n\n\n\n

(ix) 48 min to 2 hours 40 min<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

48 min to 2 hours 40 min = 48 min: 2 hours 40 min<\/p>\n\n\n\n

We know that 1 hour = 60 minutes<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

48 min to 2 hours 40 min = 48 min: 160 min<\/p>\n\n\n\n

Dividing the two terms by HCF 16<\/p>\n\n\n\n

48 min to 2 hours 40 min = 3: 10<\/p>\n\n\n\n

(x) 3 m 5 cm to 35 cm<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3 m 5 cm to 35 cm = 3 m 5 cm: 35 cm<\/p>\n\n\n\n

We know that 1 m = 100 cm<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

3 m 5 cm to 35 cm = 305 cm: 35 cm<\/p>\n\n\n\n

Dividing the two terms by HCF 5<\/p>\n\n\n\n

3 m 5 cm to 35 cm = 61: 7<\/p>\n\n\n\n

5. Find the ratio of<\/strong><\/p>\n\n\n\n

(i) 3.2 metres to 56 metres<\/strong><\/p>\n\n\n\n

(ii) 10 metres to 25 cm<\/strong><\/p>\n\n\n\n

(iii) 25 paise to Rs 60<\/strong><\/p>\n\n\n\n

(iv) 10 litres to 0.25 litre<\/strong><\/p>\n\n\n\n

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

(i) 3.2 metres to 56 metres<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3.2 metres to 56 metres = 3.2 metres: 56 metres<\/p>\n\n\n\n

Dividing the two terms by HCF 1.6<\/p>\n\n\n\n

3.2 metres to 56 metres = 2: 35<\/p>\n\n\n\n

(ii) 10 metres to 25 cm<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

10 metres to 25 cm = 10 m: 25 cm<\/p>\n\n\n\n

We know that 1 m = 100 cm<\/p>\n\n\n\n

10 metres to 25 cm = 1000 cm: 25 cm<\/p>\n\n\n\n

Dividing the two terms by HCF 25<\/p>\n\n\n\n

10 metres to 25 cm = 40: 1<\/p>\n\n\n\n

(iii) 25 paise to Rs 60<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

25 paise to Rs 60 = 25 paise: Rs 60<\/p>\n\n\n\n

We know that 1 Rs = 100 paise<\/p>\n\n\n\n

25 paise to Rs 60 = 25 paise: 6000 paise<\/p>\n\n\n\n

Dividing the two terms by HCF 25<\/p>\n\n\n\n

25 paise to Rs 60 = 1: 240<\/p>\n\n\n\n

(iv) 10 litres to 0.25 litre<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

10 litres to 0.25 litre = 10 litres: 0.25 litre<\/p>\n\n\n\n

Dividing the two terms by HCF 0.25<\/p>\n\n\n\n

10 litres to 0.25 litre = 40: 1<\/p>\n\n\n\n

6. The number of boys and girls in a school are 1168 and 1095 respectively. Express the ratio of the number of boys to that of the girls in the simplest form.<\/strong><\/p>\n\n\n\n

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

No. of boys = 1168<\/p>\n\n\n\n

No. of girls = 1095<\/p>\n\n\n\n

So the ratio of the number of boys to that of the girls = 1168: 1095<\/p>\n\n\n\n

Dividing the two terms by HCF 73<\/p>\n\n\n\n

Ratio of number of boys to that of the girls = 16: 15<\/p>\n\n\n\n

Hence, the ratio of the number of boys to that of girls in simplest form is 16: 15.<\/p>\n\n\n\n

7. Avinash works as a lecturer and earns Rs 12000 per month. His wife who is a doctor earns Rs 15000 per month. Find the following ratios:<\/strong><\/p>\n\n\n\n

(i) Avinash\u2019s income to the income of his wife.<\/strong><\/p>\n\n\n\n

(ii) Avinash\u2019s income to their total income.<\/strong><\/p>\n\n\n\n

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

Avinash salary earned per month = Rs 12000<\/p>\n\n\n\n

Avinash wife salary per month = Rs 15000<\/p>\n\n\n\n

(i) Avinash\u2019s income to the income of his wife = 12000\/15000 = 4: 5<\/p>\n\n\n\n

(ii) Avinash\u2019s income to their total income = 12000\/ (12000 + 15000) = 4: 9<\/p>\n\n\n\n

8. Of the 72 persons working in an office, 28 are men and the remaining are women. Find the ratio of the number of:<\/strong><\/p>\n\n\n\n

(i) men to that of women,<\/strong><\/p>\n\n\n\n

(ii) men to the total number of persons<\/strong><\/p>\n\n\n\n

(iii) persons to that of women.<\/strong><\/p>\n\n\n\n

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

No. of persons working in an office = 72<\/p>\n\n\n\n

No. of men = 28<\/p>\n\n\n\n

So the number of women = 72 \u2013 28 = 44<\/p>\n\n\n\n

(i) men to that of women = 28: 44<\/p>\n\n\n\n

Multiplying and dividing the equation by HCF 4<\/p>\n\n\n\n

Men to that of women = (28\/44) \u00d7 (4\/4) = 7: 11<\/p>\n\n\n\n

(ii) men to the total number of persons = 28: 72<\/p>\n\n\n\n

Multiplying and dividing the equation by HCF 4<\/p>\n\n\n\n

Men to the total number of persons = (28\/72) \u00d7 (4\/4) = 7: 18<\/p>\n\n\n\n

(iii) persons to that of women = 72: 44<\/p>\n\n\n\n

Multiplying and dividing the equation by HCF 4<\/p>\n\n\n\n

Persons to that of women = (72\/44) \u00d7 (4\/4) = 18: 11<\/p>\n\n\n\n

9. The length of a steel tape for measurements of buildings is 10 m and its width is 2.4 cm. What is the ratio of its length to width?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Length of a steel tape = 10 m<\/p>\n\n\n\n

Width of steel tape = 2.4 cm<\/p>\n\n\n\n

So the ratio of its length to width = 10 m\/ 2.4 cm<\/p>\n\n\n\n

We know that 1 m = 100 cm<\/p>\n\n\n\n

Ratio of its length to width = 1000 cm\/ 2.4 cm<\/p>\n\n\n\n

Dividing the two terms by HCF 0.8 cm<\/p>\n\n\n\n

Ratio of its length to width = 1250: 3<\/p>\n\n\n\n

Hence, the ratio of its length to width is 1250: 3.<\/p>\n\n\n\n

10. An office opens at 9 am and closes at 5 pm with a lunch interval of 30 minutes. What is the ratio of lunch interval to the total period in office?<\/strong><\/p>\n\n\n\n

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

Duration of office = 9 am to 5 pm = 8 hours<\/p>\n\n\n\n

Lunch interval = 30 minutes<\/p>\n\n\n\n

So the ratio of lunch interval to the period in office = 30 minutes\/8 hours<\/p>\n\n\n\n

We know that 1 hour = 60 minutes<\/p>\n\n\n\n

Ratio of lunch interval to the period in office = 30\/ (8 \u00d7 60) = 30\/480<\/p>\n\n\n\n

Dividing the two terms by HCF 30<\/p>\n\n\n\n

Ratio of lunch interval to the period in office = (30\/480) \u00d7 (30\/30) = 1: 16<\/p>\n\n\n\n

Hence, the ratio of lunch interval to the total period in office is 1: 16.<\/p>\n\n\n\n

11. A bullock-cart travels 24 km in 3 hours and a train travels 120 km in 2 hours. Find the ratio of their speeds.<\/strong><\/p>\n\n\n\n

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

Distance travelled by bullock-cart = 24 km in 3 hours<\/p>\n\n\n\n

Distance travelled by train = 120 km in 2 hours<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

Distance travelled by bullock-cart = 24 km\/ 3 = 8 km<\/p>\n\n\n\n

Distance travelled by train = 120 km\/2 = 60 km<\/p>\n\n\n\n

So the ratio of their speeds = 8\/60<\/p>\n\n\n\n

Dividing the two terms by HCF 4<\/p>\n\n\n\n

Ratio of their speeds = (8\/60) \u00d7 (4\/4) = 2:15<\/p>\n\n\n\n

Hence, the ratio of their speeds is 2: 15.<\/p>\n\n\n\n

12. Margarette works in a factory and earns Rs 955 per month. She saves Rs 185 per month from her earnings. Find the ratio of:<\/strong><\/p>\n\n\n\n

(i) her savings to her income<\/strong><\/p>\n\n\n\n

(ii) her income to her expenditure<\/strong><\/p>\n\n\n\n

(iii) her savings to her expenditure.<\/strong><\/p>\n\n\n\n

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

Margarette monthly income = Rs 955<\/p>\n\n\n\n

Margarette monthly savings = Rs 185<\/p>\n\n\n\n

Margarette expenditure = 955 \u2013 185 = Rs 770<\/p>\n\n\n\n

(i) her savings to her income = 185\/955<\/p>\n\n\n\n

Dividing the two terms by HCF 5<\/p>\n\n\n\n

Her savings to her income = (185\/955) \u00d7 (5\/5) = 37: 191<\/p>\n\n\n\n

(ii) her income to her expenditure = 955\/770 = 191: 154<\/p>\n\n\n\n

(iii) her savings to her expenditure = 185\/770 = 37: 154<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.2 page: 9.9<\/h4>\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) 3: 4 or 9: 16<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

3: 4 = 3\/4 and 9: 16 = 9\/16<\/p>\n\n\n\n

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

Multiplying both numerator and denominator of the term \u00be by 4 to make the denominator 16<\/p>\n\n\n\n

3\/4 = (3\/4) \u00d7 (4\/4) = 12\/16 and 9\/16<\/p>\n\n\n\n

We know that 12 > 9<\/p>\n\n\n\n

So we get 12\/16 > 9\/16<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

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

Hence, 3: 4 > 9: 16.<\/p>\n\n\n\n

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

It can be written as<\/p>\n\n\n\n

15: 16 = 15\/16 and 24: 25 = 24\/25<\/p>\n\n\n\n

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

Multiplying both the terms by relevant numbers to make denominator as 400<\/p>\n\n\n\n

15\/16 = (15\/16) \u00d7 (25\/25) = 375\/400 and 24\/25 = (24\/25) \u00d7 (16\/16) = 384\/400<\/p>\n\n\n\n

We know that 384 > 375<\/p>\n\n\n\n

So we get 384\/400 > 375\/400<\/p>\n\n\n\n

We can write it as 24\/25 > 15\/16<\/p>\n\n\n\n

Hence, 24: 25 > 15: 16.<\/p>\n\n\n\n

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

It can be written as<\/p>\n\n\n\n

4: 7 = 4\/7 and 5: 8 = 5\/8<\/p>\n\n\n\n

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

Multiplying both the terms by relevant numbers to make denominator as 56<\/p>\n\n\n\n

4\/7 = (4\/7) \u00d7 (8\/8) = 32\/56 and 5\/8 = (5\/8) \u00d7 (7\/7) = 35\/56<\/p>\n\n\n\n

We know that 35 > 32<\/p>\n\n\n\n

So we get 35\/56 > 32\/56<\/p>\n\n\n\n

We can write it as 5\/8 > 4\/7<\/p>\n\n\n\n

Hence, 5: 8 > 4: 7.<\/p>\n\n\n\n

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

It can be written as<\/p>\n\n\n\n

9: 20 = 9\/20 and 8: 13 = 8\/13<\/p>\n\n\n\n

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

Multiplying both the terms by relevant numbers to make denominator as 260<\/p>\n\n\n\n

9\/20 = (9\/20) \u00d7 (13\/13) = 117\/260 and 8\/13 = (8\/13) \u00d7 (20\/20) = 160\/260<\/p>\n\n\n\n

We know that 160 > 117<\/p>\n\n\n\n

So we get 160\/260 > 117\/260<\/p>\n\n\n\n

We can write it as 8\/13 > 9\/20<\/p>\n\n\n\n

Hence, 8: 13 > 9: 20.<\/p>\n\n\n\n

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

It can be written as<\/p>\n\n\n\n

1: 2 = 1\/2 and 13: 27 = 13\/27<\/p>\n\n\n\n

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

Multiplying both the terms by relevant numbers to make denominator as 54<\/p>\n\n\n\n

1\/2 = (1\/2) \u00d7 (27\/27) = 27\/54 and 13\/27 = (13\/27) \u00d7 (2\/2) = 26\/54<\/p>\n\n\n\n

We know that 27 > 26<\/p>\n\n\n\n

So we get 27\/54 > 26\/54<\/p>\n\n\n\n

We can write it as 1\/2 > 13\/27<\/p>\n\n\n\n

Hence, 1: 2 > 13: 27.<\/p>\n\n\n\n

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

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

The given ratio = 6: 8<\/p>\n\n\n\n

It can be written as = 6\/8<\/p>\n\n\n\n

Dividing the fraction by 2 we get<\/p>\n\n\n\n

6\/8 = (6\/8) \u00f7 (2\/2) = 3\/4<\/p>\n\n\n\n

Hence, 3: 4 is an equivalent ratio of 6: 8<\/p>\n\n\n\n

Multiply the fraction by 2 we get<\/p>\n\n\n\n

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

Hence, 12: 16 is an equivalent ratio of 6: 8<\/p>\n\n\n\n

So, 3: 4 and 12: 16 are the equivalent ratios of 6: 8.<\/p>\n\n\n\n

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

12\/20 = \u2610\/5 = 9\/\u2610<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

12\/20 = \u2610\/5 = 9\/\u2610<\/p>\n\n\n\n

We know that LCM of 20 and 5 is 20<\/p>\n\n\n\n

It can be written as 20\/4 = 5<\/p>\n\n\n\n

Dividing the fraction by 4<\/p>\n\n\n\n

12\/20 = (12\/20) \u00d7 (4\/4) = 3\/5<\/p>\n\n\n\n

So the first number is 3 and the ratio is 3\/5.<\/p>\n\n\n\n

In the same way,<\/p>\n\n\n\n

Consider 2\/3 + 3\/5 = 9\/\u2610<\/p>\n\n\n\n

We know that 9\/3 = 3<\/p>\n\n\n\n

Multiply the fraction by 3<\/p>\n\n\n\n

3\/5 = (3\/5) \u00d7 (3\/3) = 9\/15<\/p>\n\n\n\n

So the second number is 15 and the ratio is 9\/15.<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.3 page: 9.14<\/h4>\n\n\n\n

1. Which of the following statements are true?<\/strong><\/p>\n\n\n\n

(i) 16: 24 = 20: 30<\/strong><\/p>\n\n\n\n

(ii) 21: 6 = 35: 10<\/strong><\/p>\n\n\n\n

(iii) 12: 18 = 28: 12<\/strong><\/p>\n\n\n\n

(iv) 51: 58 = 85: 102<\/strong><\/p>\n\n\n\n

(v) 40 men: 200 men = Rs 5: Rs 25<\/strong><\/p>\n\n\n\n

(vi) 99 kg: 45 kg = Rs 44: Rs 20<\/strong><\/p>\n\n\n\n

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

(i) 16: 24 = 20: 30<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

16\/24 = 20\/30<\/p>\n\n\n\n

Dividing 16\/24 by 4\/4 and 20\/30 by 5\/5<\/p>\n\n\n\n

(16\/24) \u00f7 (4\/4) = (20\/30) \u00f7 (5\/5)<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

4\/6 = 4\/6<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

2\/3 = 2\/3<\/p>\n\n\n\n

Hence, 16: 24 = 20: 30 is true.<\/p>\n\n\n\n

(ii) 21: 6 = 35: 10<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

21\/6 = 35\/10<\/p>\n\n\n\n

Dividing 21\/6 by 3\/3 and 35\/10 by 5\/5<\/p>\n\n\n\n

(21\/6) \u00f7 (3\/3) = (35\/10) \u00f7 (5\/5)<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

7\/2 = 7\/2<\/p>\n\n\n\n

Hence, 21: 6 = 35: 10 is true.<\/p>\n\n\n\n

(iii) 12: 18 = 28: 12<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

12\/18 = 28\/12<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

6\/9 \u2260 14\/6<\/p>\n\n\n\n

Hence, 12: 18 = 28: 12 is false.<\/p>\n\n\n\n

(iv) 51: 58 = 85: 102<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

51\/58 = 85\/102<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

51\/58 \u2260 5\/6<\/p>\n\n\n\n

Hence, 51: 58 = 85: 102 is false.<\/p>\n\n\n\n

(v) 40 men: 200 men = Rs 5: Rs 25<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

40\/200 = 5\/25<\/p>\n\n\n\n

We get 40\/200 = 1\/5 and 5\/25 = 1\/5<\/p>\n\n\n\n

Hence, 40 men: 200 men = Rs 5: Rs 25 is true.<\/p>\n\n\n\n

(vi) 99 kg: 45 kg = Rs 44: Rs 20<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

99\/45 = 44\/20<\/p>\n\n\n\n

Dividing the fraction by 9<\/p>\n\n\n\n

(99\/45) \u00f7 (9\/9) = (44\/20) \u00f7 (9\/9)<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

11\/5 = 11\/5<\/p>\n\n\n\n

Hence, 99 kg: 45 kg = Rs 44: Rs 20 is true.<\/p>\n\n\n\n

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

(i) 8, 16, 6, 12<\/strong><\/p>\n\n\n\n

(ii) 6, 2, 4, 3<\/strong><\/p>\n\n\n\n

(iii) 150, 250, 200, 300<\/strong><\/p>\n\n\n\n

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

(i) 8, 16, 6, 12<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

8: 16 = 8\/16 = 1\/2<\/p>\n\n\n\n

6: 12 = 6\/12 = 1\/2<\/p>\n\n\n\n

So we get 8\/16 = 6\/12<\/p>\n\n\n\n

Therefore, 8, 16, 6, 12 are in proportion.<\/p>\n\n\n\n

(ii) 6, 2, 4, 3<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

6: 2 = 6\/2 = 3\/1<\/p>\n\n\n\n

4: 3 = 4\/3<\/p>\n\n\n\n

So we get 3\/1 \u2260 4\/3<\/p>\n\n\n\n

Therefore, 6, 2, 4, 3 are not in proportion.<\/p>\n\n\n\n

(iii) 150, 250, 200, 300<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

150: 250 = 150\/250 = 3\/5<\/p>\n\n\n\n

200: 300 = 200\/300 = 4\/6 = 2\/3<\/p>\n\n\n\n

So we get 3\/5 \u2260 2\/3<\/p>\n\n\n\n

Therefore, 150, 250, 200, 300 are not in proportion.<\/p>\n\n\n\n

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

(i) x: 6 = 55: 11<\/strong><\/p>\n\n\n\n

(ii) 18: x = 27: 3<\/strong><\/p>\n\n\n\n

(iii) 7: 14 = 15: x<\/strong><\/p>\n\n\n\n

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

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

(i) x: 6 = 55: 11<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

x\/6 = 55\/11<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

x\/6 = 5\/1<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = 5 (6) = 30<\/p>\n\n\n\n

(ii) 18: x = 27: 3<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

18\/x = 27\/3<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

18\/x = 9\/1<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = 18\/9 = 2<\/p>\n\n\n\n

(iii) 7: 14 = 15: x<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

7\/14 = 15\/x<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

1\/2 = 15\/x<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = 15 (2) = 30<\/p>\n\n\n\n

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

It can be written as<\/p>\n\n\n\n

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

We get<\/p>\n\n\n\n

8\/9 = x\/96<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

x = 8\/9 (96) = 256\/3<\/p>\n\n\n\n

4. Set up all proportions from the numbers 9, 150, 105, 1750.<\/strong><\/p>\n\n\n\n

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

The proportions from the numbers are<\/p>\n\n\n\n

9: 150 = 3: 50<\/p>\n\n\n\n

9: 105 = 3: 35<\/p>\n\n\n\n

9: 1750<\/p>\n\n\n\n

150: 9 = 50: 3<\/p>\n\n\n\n

150: 105 = 10: 7<\/p>\n\n\n\n

150: 1750 = 3: 35<\/p>\n\n\n\n

105: 9 = 35: 3<\/p>\n\n\n\n

105: 150 = 7: 10<\/p>\n\n\n\n

105: 1750 = 3: 50<\/p>\n\n\n\n

1750: 9<\/p>\n\n\n\n

1750: 150 = 35: 3<\/p>\n\n\n\n

1750: 105 = 50: 3<\/p>\n\n\n\n

Hence, the proportions that are formed are<\/p>\n\n\n\n

9: 150 :: 105: 1750<\/p>\n\n\n\n

150: 9 :: 1750: 105<\/p>\n\n\n\n

1750: 150 :: 105: 9<\/p>\n\n\n\n

9: 105 :: 150: 1750<\/p>\n\n\n\n

5. Find the other three proportions involving terms of each of the following:<\/strong><\/p>\n\n\n\n

(i) 45: 30 = 24: 16<\/strong><\/p>\n\n\n\n

(ii) 12: 18 = 14: 21<\/strong><\/p>\n\n\n\n

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

(i) 45: 30 = 24: 16 can be written as 3: 2 in simplest form<\/p>\n\n\n\n

So the other three proportions involving terms are<\/p>\n\n\n\n

45: 24 = 30: 16 can be written as 15: 8 in simplest form<\/p>\n\n\n\n

30: 45 = 16: 24 can be written as 2: 3 in simplest form<\/p>\n\n\n\n

16: 30 = 24: 45 can be written as 8: 15 in simplest form<\/p>\n\n\n\n

(ii) 12: 18 = 14: 21 can be written as 2: 3 in simplest form<\/p>\n\n\n\n

So the other three proportions involving terms are<\/p>\n\n\n\n

12: 14 = 18: 21 can be written as 6: 7 in simplest form<\/p>\n\n\n\n

21: 18 = 14: 12 can be written as 7: 6 in simplest form<\/p>\n\n\n\n

18: 12 = 21: 14 can be written as 3: 2 in simplest form<\/p>\n\n\n\n

6. If 4, x, 9 are in continued proportion, find the value of x.<\/strong><\/p>\n\n\n\n

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

We know that 4, x, 9 are in continued proportion<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

4: x :: x: 9<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

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

On further calculation<\/p>\n\n\n\n

x2<\/sup> = 9 (4) = 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

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

7. If in a proportion, the first, second and fourth terms are 32, 112 and 217 respectively, find the third term.<\/strong><\/p>\n\n\n\n

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

It is given that in a proportion the first, second and fourth terms are 32, 112 and 217<\/p>\n\n\n\n

Consider x as the third term<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

32: 112 :: x: 217<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

32\/112 = x\/217<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 32\/112 (217) = 62<\/p>\n\n\n\n

8. Show that the following numbers are in continued proportion:<\/strong><\/p>\n\n\n\n

(i) 36, 90, 225<\/strong><\/p>\n\n\n\n

(ii) 48, 60, 75<\/strong><\/p>\n\n\n\n

(iii) 16, 84, 441<\/strong><\/p>\n\n\n\n

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

(i) 36, 90, 225<\/p>\n\n\n\n

Consider the fraction 36\/90<\/p>\n\n\n\n

By dividing the fraction by 18<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

36\/90 = 2\/5<\/p>\n\n\n\n

Consider the fraction 90\/225<\/p>\n\n\n\n

By dividing the fraction by 45<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

90\/225 = 2\/5<\/p>\n\n\n\n

Hence, 36: 90 :: 90: 225.<\/p>\n\n\n\n

(ii) 48, 60, 75<\/p>\n\n\n\n

Consider the fraction 48\/60<\/p>\n\n\n\n

By dividing the fraction by 12<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

48\/60 = 4\/5<\/p>\n\n\n\n

Consider the fraction 60\/75<\/p>\n\n\n\n

By dividing the fraction by 15<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

60\/75 = 4\/5<\/p>\n\n\n\n

Hence, 48: 60 :: 60: 75.<\/p>\n\n\n\n

(iii) 16, 84, 441<\/p>\n\n\n\n

Consider the fraction 16\/84<\/p>\n\n\n\n

By dividing the fraction by 4<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

16\/84 = 4\/21<\/p>\n\n\n\n

Consider the fraction 84\/441<\/p>\n\n\n\n

By dividing the fraction by 21<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

84\/441 = 4\/21<\/p>\n\n\n\n

Hence, 16: 84 :: 84: 441.<\/p>\n\n\n\n

9. The ratio of the length of a school ground to its width is 5: 2. Find its length if the width is 40 metres.<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Ratio of length of a school ground to its width = 5: 2<\/p>\n\n\n\n

Width of the school ground = 40 m<\/p>\n\n\n\n

So the length of the school ground = 5\/2 (40) = 100 m<\/p>\n\n\n\n

Hence, the length of the school ground is 100 m.<\/p>\n\n\n\n

10. The ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop was 2: 9. If the total sale of eggs in the same week was Rs 360, find the sale of eggs on Sunday.<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Ratio of the sale of eggs on a Sunday to that of the whole week of a grocery shop = 2: 9<\/p>\n\n\n\n

We know that the sale of eggs in a week is Rs 9 and on Sunday is Rs 2<\/p>\n\n\n\n

If eggs of Rs 1 is sold in a week, the cost of eggs on Sunday = Rs 2\/9<\/p>\n\n\n\n

If the total sale of eggs in the same week was Rs 360, the sale of eggs on Sunday = 2\/9 (360) = Rs 80<\/p>\n\n\n\n

Hence, the sale of eggs on Sunday is Rs 80.<\/p>\n\n\n\n

11. The ratio of copper and zinc in an alloy is 9: 7. If the weight of zinc in the alloy is 9.8 kg, find the weight of copper in the alloy.<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Ratio of copper and zinc in an alloy = 9: 7<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

If the weight of zinc is 7 kg then the weight of copper is 9 kg<\/p>\n\n\n\n

If the weight of zinc is 1 kg then the weight of copper = 9\/7 kg<\/p>\n\n\n\n

So if the weight of zinc is 9.8 kg then the weight of copper = 9\/7 (9.8) = 12.6 kg<\/p>\n\n\n\n

Hence, the weight of copper in the alloy is 12.6 kg.<\/p>\n\n\n\n

12. The ratio of the 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

It is given that<\/p>\n\n\n\n

Ratio of the income to the expenditure of a family = 7: 6<\/p>\n\n\n\n

We know that saving = total income \u2013 expenditure<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

Ratio of saving to the income = [7 \u2013 6]: 7 = 1: 7<\/p>\n\n\n\n

It is given that income = Rs 1400<\/p>\n\n\n\n

So the saving of the family = 1\/7 (1400) = Rs 200<\/p>\n\n\n\n

Hence, the saving of the family is Rs 200.<\/p>\n\n\n\n

13. The ratio of story books in a library to other books is 1: 7. The total number of story books is 800. Find the total number of books in the library.<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Ratio of story books in a library to other books = 1: 7<\/p>\n\n\n\n

Consider the ratio as x<\/p>\n\n\n\n

So the number of story books = x<\/p>\n\n\n\n

Number of other books = 7x<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Total number of story books = 800<\/p>\n\n\n\n

Number of other books = 7 \u00d7 800 = 5600<\/p>\n\n\n\n

Total number of books = 5600 + 800 = 6400<\/p>\n\n\n\n

Hence, the total number of books in the library is 6400.<\/p>\n\n\n\n

\n\n\n\n

Exercise 9.4 PAGE: 9.18<\/h4>\n\n\n\n

1. The price of 3 metres of cloth is Rs 79.50. Find the price of 15 metres of such cloth.<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Price of 3 m of cloth = Rs 79.50<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

Price of 1 m of cloth = 79.50\/3 = Rs 26.5<\/p>\n\n\n\n

So the price of 15 m of cloth = 26.5 (15) = Rs 397.50<\/p>\n\n\n\n

Hence, the price of 15 m of such cloth is Rs 397.50.<\/p>\n\n\n\n

2. The cost of 17 chairs is Rs 9605. Find the number of chairs that can be purchased in Rs 56500.<\/strong><\/p>\n\n\n\n

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

No. of chairs purchased for Rs 9605 = 17<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

No. of chairs purchased for Rs 1 = 17\/9605<\/p>\n\n\n\n

So the number of chairs purchased for Rs 56500 = 17\/9605 (56500) = 100<\/p>\n\n\n\n

Hence, 100 chairs can be purchased in Rs 56500.<\/p>\n\n\n\n

3. Three ferryloads are needed to carry 150 people across a river. How many people will be carried on 4 ferryloads?<\/strong><\/p>\n\n\n\n

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

We know that<\/p>\n\n\n\n

No. of people required to carry 3 ferryloads = 150<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

No. of people required to carry 1 ferryload = 150\/3 = 50<\/p>\n\n\n\n

So the number of people required to carry 4 ferryloads = 4 (50) = 200<\/p>\n\n\n\n

Hence, 200 people are required to carry 4 ferryloads.<\/p>\n\n\n\n

4. If 9 kg of rice costs Rs 120.60, what will 50 kg of such a quality of rice cost?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Cost of 9 kg rice = Rs 120.60<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Cost of 1 kg rice = 120.60\/9 = Rs 13.4<\/p>\n\n\n\n

So the cost of 50 kg rice = 13.4 (50) = Rs 670<\/p>\n\n\n\n

Hence, 50 kg of such a quality of rice costs Rs 670.<\/p>\n\n\n\n

5. A train runs 200 kilometres in 5 hours. How many kilometres does it run in 7 hours?<\/strong><\/p>\n\n\n\n

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

Distance travelled by train in 5 hours = 200 km<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Distance travelled by train in 1 hour = 200\/5 = 40 km<\/p>\n\n\n\n

So the distance travelled by train in 7 hours = 40 (7) = 280 km<\/p>\n\n\n\n

Hence, the train runs 280 km in 7 hours.<\/p>\n\n\n\n

6. 10 boys can dig a pitch in 12 hours. How long will 8 boys take to do it?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

10 boys can dig a pitch in 12 hours<\/p>\n\n\n\n

We know that the time taken by one boy = 10 (12) = 120 hours<\/p>\n\n\n\n

So the time taken by 8 boys to dig the pitch = 120\/8 = 15 hours<\/p>\n\n\n\n

Hence, 8 boys will take 15 hours to dig the pitch.<\/p>\n\n\n\n

7. A man can work 8 hours daily and finishes a work in 12 days. If he works 6 hours daily, in how many days will the same work be finished?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

A man can work 8 hours daily and finishes a work in 12 days<\/p>\n\n\n\n

If he works for one hour, then the time taken to finish the work = 8 \u00d7 12 = 96 days<\/p>\n\n\n\n

If he works 6 hours daily, the days required to finish the work = 96\/6 = 16 days<\/p>\n\n\n\n

Hence, the man requires 16 days to finish the same work.<\/p>\n\n\n\n

8. Fifteen post cards cost Rs 2.25. What will be the cost of 36 post cards? How many postcards can be bought in Rs 45?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Cost of fifteen post cards = Rs 2.25<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Cost of one post card = Rs 2.25\/15<\/p>\n\n\n\n

So the cost of 36 post cards = 2.25\/15 (36) = Rs 5.40<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

No. of postcards that can be purchased in Rs 1 = 15\/2.25<\/p>\n\n\n\n

Number of postcards that can be bought in Rs 45 = 15\/2.25 (45) = 300<\/p>\n\n\n\n

Hence, the cost of 36 post cards is Rs 5.40 and 300 post cards can be bought in Rs 45.<\/p>\n\n\n\n

9. A rail journey of 75 km costs Rs 215. How much will a journey of 120 km cost?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Cost of rail journey of 75 km = Rs 215<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Cost of rail journey of 1 km = Rs 215\/75<\/p>\n\n\n\n

So the cost of rail journey of 120 km = 215\/75 (120) = Rs 344<\/p>\n\n\n\n

Hence, the cost of rail journey of 120 km is Rs 344.<\/p>\n\n\n\n

10. If the sales tax on a purchase worth Rs 60 is Rs 4.20. What will be the sales tax on the purchase worth Rs 150?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Sales tax on a purchase worth Rs 60 = Rs 4.20<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Sales tax on a purchase worth Rs 1 = Rs 4.20\/60<\/p>\n\n\n\n

So the sales tax on the purchase worth Rs 150 = 4.20\/60 (150) = Rs 10.50<\/p>\n\n\n\n

Hence, the sales tax on the purchase worth Rs 150 is Rs 10.50.<\/p>\n\n\n\n

11. The cost of 17 chairs is Rs 19210. Find the number of such chairs that can be purchased in Rs 113000?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

No. of chairs purchased in Rs 19210 = 17<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

No. of chairs purchased in Rs 1 = 17\/19210<\/p>\n\n\n\n

So the number of chairs that can be purchased in Rs 113000 = 17\/19210 (113000) = 100<\/p>\n\n\n\n

Hence, 100 chairs can be purchased in Rs 113000.<\/p>\n\n\n\n

12. A car travels 165 km in 3 hours<\/strong><\/p>\n\n\n\n

(i) How long will it take to travel 440 km?<\/strong><\/p>\n\n\n\n

(ii) How far will it travel in 7 hours?<\/strong><\/p>\n\n\n\n

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

Distance travelled by car = 165 km in 3 hours<\/p>\n\n\n\n

So the speed of car = Distance\/ time = 165\/3 = 55 km per hour<\/p>\n\n\n\n

(i) Time taken to travel 440 km = 440\/55 = 8 hours<\/p>\n\n\n\n

(ii) Distance covered in 7 hours = 55 (7) = 385 km<\/p>\n\n\n\n

13. 2 dozens of oranges cost Rs 60. Find the cost of 120 similar oranges?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Cost of 2 dozens of oranges = Rs 60<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Cost of 1 orange = Rs 60\/24<\/p>\n\n\n\n

So the cost of 120 similar oranges = 60\/24 (120) = Rs 300<\/p>\n\n\n\n

Hence, the cost of 120 similar oranges is Rs 300.<\/p>\n\n\n\n

14. A family of 4 members consumes 6 kg of sugar in a month. What will be the monthly consumption of sugar, if the number of family members becomes 6?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

Amount of sugar used by a 4 members family = 6 kg<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Amount of sugar used by 1 member = 6\/4 kg<\/p>\n\n\n\n

So the sugar consumed by 6 members of a family = 6\/4 (6) = 9kg<\/p>\n\n\n\n

Hence, 9 kg is the monthly consumption of sugar, if the number of family members becomes 6.<\/p>\n\n\n\n

15. The weight of 45 folding chairs is 18 kg. How many such chairs can be loaded on a truck having a capacity of carrying 4000 kg load?<\/strong><\/p>\n\n\n\n

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

It is given that<\/p>\n\n\n\n

No. of folding chairs weighing 18 kg = 45<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

No. of folding chairs weighing 1 kg = 45\/18<\/p>\n\n\n\n

So the number of folding chairs weighing 4000 kg = 45\/18 (4000) = 10000<\/p>\n\n\n\n

Hence, 10000 chairs can be loaded on a truck having a capacity of carrying 4000 kg load.<\/p>\n\n\n\n

\n\n\n\n

Objective Type Questions page: 9.19<\/h4>\n\n\n\n

Mark the correct alternative in each of the following:<\/strong><\/p>\n\n\n\n

1. A ratio equivalent of 2 : 3 is
(a) 4 : 3
(b) 2 : 6
(c) 6 : 9
(d) 10 : 9<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

We know that 6: 9 when divided by 3 we get 2: 3.<\/p>\n\n\n\n

2. The angles of a triangle are in the ratio 1 : 2 : 3. The measure of the largest angle is
(a) 30\u00b0
(b) 60\u00b0
(c) 90\u00b0
(d) 120\u00b0<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

We know that the sum of all the angles = 180\u00b0<\/p>\n\n\n\n

So the largest angle = 3\/ (1 + 2 + 3) \u00d7 180<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

Largest angle = 3\/6 \u00d7 180 = 90\u00b0<\/p>\n\n\n\n

3. The sides of a triangle are in the ratio 2 : 3 : 5. If its perimeter is 100 cm, the length of its smallest side is
(a) 2 cm
(b) 20 cm
(c) 3 cm
(d) 5 cm<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that the length of smallest side = 100 \u00d7 2\/ (2 + 3 + 5) = 200\/10 = 20 cm<\/p>\n\n\n\n

4. Two numbers are in the ratio 7 : 9. If the sum of the numbers is 112, then the larger number is
(a) 63
(b) 42
(c) 49
(d) 72<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

Consider x as the largest number<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

7x + 9x = 112<\/p>\n\n\n\n

16x = 112<\/p>\n\n\n\n

x = 112\/16 = 7<\/p>\n\n\n\n

Here<\/p>\n\n\n\n

7x = 7 \u00d7 7 = 49<\/p>\n\n\n\n

9x = 9 \u00d7 7 = 63<\/p>\n\n\n\n

Hence, the largest number is 63.<\/p>\n\n\n\n

5. Two ratio 384 : 480 in its simplest form is
(a) 3 : 5
(b) 5 : 4
(c) 4 : 5
(d) 2 : 5<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

384: 480 can be written as<\/p>\n\n\n\n

384\/480 = 4\/5 when divided by 96<\/p>\n\n\n\n

6. If A, <\/em>B, <\/em>C, divide Rs 1200 in the ratio 2 : 3 : 5, then B\u2019s share is
(a) Rs 240
(b) Rs 600
(c) Rs 380
(d) Rs 360<\/strong><\/p>\n\n\n\n

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

The option (d) is correct answer.<\/p>\n\n\n\n

So B\u2019s share = 1200 \u00d7 3\/ (2 + 3 + 5)<\/p>\n\n\n\n

On further calculation<\/p>\n\n\n\n

B\u2019s share = 1200 \u00d7 3\/10 = Rs 360<\/p>\n\n\n\n

7. If a bus travels 126 km in 3 hours and a train travels 315 km in 5 hours, then the ratio of their speeds is
(a) 2 : 5
(b) 2 : 3
(c) 5 : 2
(d) 25 : 6<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

We know that speed = distance\/time<\/p>\n\n\n\n

So the speed of bus = 126\/3 = 42 km\/h<\/p>\n\n\n\n

Speed of train = 315\/5 = 63 km\/h<\/p>\n\n\n\n

So the ratio of their speeds = 42: 63 = 2: 3<\/p>\n\n\n\n

8. The ratio of male and female employees in a multinational company is 5 : 3. If there are 115 male employees in the company, then the number of female employees is
(a) 96
(b) 52
(c) 69
(d) 66<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

Consider x as the number of female employees<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

5\/3 = 115\/x<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

5x = 115 \u00d7 3 = 345<\/p>\n\n\n\n

By division<\/p>\n\n\n\n

x = 345\/5 = 69<\/p>\n\n\n\n

9. Length and width of a field are in the ratio 5 : 3. If the width of the field is 42 m, then its length is
(a) 50 m
(b) 70 m
(c) 80 m
(d) 100 m<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It is given that length and width of a field = 5: 3<\/p>\n\n\n\n

Consider x m as the length<\/p>\n\n\n\n

Width of the filed = 42 m<\/p>\n\n\n\n

So the length can be written as<\/p>\n\n\n\n

5\/3 = x\/42<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

3x = 42 \u00d7 5 = 210<\/p>\n\n\n\n

By division<\/p>\n\n\n\n

x = 210\/3 = 70<\/p>\n\n\n\n

10. If 57 : x = 51 : 85, then the value of x is
(a) 95
(b) 76
(c) 114
(d) None of these<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

57\/x = 51\/85<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

57 \u00d7 85\/51 = x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 95<\/p>\n\n\n\n

11. The ratio of boys and girls in a school is 12 : 5. If there are 840 girls in the school, then the number of boys is
(a) 1190
(b) 2380
(c) 2856
(d) 2142<\/strong><\/p>\n\n\n\n

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

The options are not correct.<\/p>\n\n\n\n

Consider x as the number of boys<\/p>\n\n\n\n

Ratio of boys and girls = 12: 5<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

12\/5 = x\/840<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

x = 12\/5 \u00d7 840 = 2016<\/p>\n\n\n\n

12. If 4, a, <\/em>a,<\/em> 36 are in proportion, then a =
(a) 24
(b) 12
(c) 3
(d) 24<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It is given that 4, a, a, 36 are in proportion<\/p>\n\n\n\n

We can write it as 4 : a :: a : 36<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

4\/a = a\/36<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

4 \u00d7 36 = a \u00d7 a<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

a2<\/sup> = 144<\/p>\n\n\n\n

So a = 12<\/p>\n\n\n\n

13. If 5 : 4 : : 30 : x, then the value of x is
(a) 24
(b) 12
(c) 3\/2
(d) 6<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

5\/4 = 30\/x<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

x = 30 \u00d7 4\/5 = 24<\/p>\n\n\n\n

14. If a, <\/em>b, <\/em>c, <\/em>d are in proportion, then
(a) ab = cd
(b) ac = bd
(c) ad = bc
(d) None of these<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

It is given that a, b, c, d are in proportion<\/p>\n\n\n\n

We can write it as a : b :: c : d<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a\/b = c\/d<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

ad = bc<\/p>\n\n\n\n

15. If a, b, c, are in proportion, then
(a) a2<\/sup> = bc
(b) b2 <\/sup>= ac
(c) c2<\/sup> = ab
<\/em>(d) None of these<\/strong><\/p>\n\n\n\n

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

The option (b) is correct answer.<\/p>\n\n\n\n

It is given that a, b, c are in proportion<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

a : b :: b : c<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

a\/b = b\/c<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

b2<\/sup> = ac<\/p>\n\n\n\n

16. If the cost of 5 bars of a soap is Rs. 30, then the cost of one dozen bars is
(a) Rs 60
(b) Rs 120
(c) Rs 72
(d) Rs 140<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

Consider Rs x as the cost of one dozen bars<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

30\/5 = x\/12<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 30\/5 \u00d7 12 = Rs 72<\/p>\n\n\n\n

17. 12 men can finish a piece of work in 25 days. The number of days in which the same piece of work can be done by 20 men, is
(a) 10 days
(b) 12 days
(c) 15 days
(d) 14 days<\/strong><\/p>\n\n\n\n

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

The option (c) is correct answer.<\/p>\n\n\n\n

Consider x days required by 20 men to do the same work<\/p>\n\n\n\n

20\/12 = 25\/x<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

x = 12 \u00d7 25\/20 = 15 days<\/p>\n\n\n\n

18. If the cost of 25 packets of 12 pencils each is Rs 750, then the cost of 30 packets of 8 pencils each is<\/strong><\/p>\n\n\n\n

(a) Rs 600<\/strong><\/p>\n\n\n\n

(b) Rs 720<\/strong><\/p>\n\n\n\n

(c) Rs 640<\/strong><\/p>\n\n\n\n

(d) None of these<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

We know that<\/p>\n\n\n\n

Cost of 300 pencils = Rs 750<\/p>\n\n\n\n

So consider Rs x as the cost of 240 pencils<\/p>\n\n\n\n

It can be written as<\/p>\n\n\n\n

750: 300 :: x: 240<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

Cost of 240 pencils = 750\/300 \u00d7 240 = Rs 600<\/p>\n\n\n\n

19. If a, b, c are in proportion, then
(a) a : b : : b : c
(b) a : b : : c : a
(c) a : b : : c : b
(d) a : c : : b : c<\/strong><\/p>\n\n\n\n

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

The option (a) is correct answer.<\/p>\n\n\n\n

We know that a, b, c are in proportion<\/p>\n\n\n\n

So we get a: b :: b: c<\/p>\n\n\n\n

It can be written as ac = b2<\/sup><\/p>\n\n\n\n

20. The first, second and fourth terms of a proportion are 16, 24 and 54 respectively. The third term is
(a) 32
(b) 48
(c) 28
(d) 36<\/strong><\/p>\n\n\n\n

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

The option (d) is correct answer.<\/p>\n\n\n\n

Consider x as the third term<\/p>\n\n\n\n

We can write it as<\/p>\n\n\n\n

16: 24 = x: 54<\/p>\n\n\n\n

So we get<\/p>\n\n\n\n

16\/24 = x\/54<\/p>\n\n\n\n

By cross multiplication<\/p>\n\n\n\n

x = 16\/24 \u00d7 54<\/p>\n\n\n\n

We get<\/p>\n\n\n\n

x = 36<\/p>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 6 Maths Chapter 9\u2013Ratio, Proportion and Unitary Method<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 6 Maths Chapter 9\u2013Ratio, Proportion and Unitary Method PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 6 Maths :<\/strong><\/h2>\n\n\n\n