{"id":598185,"date":"2022-05-02T11:15:06","date_gmt":"2022-05-02T11:15:06","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=598185"},"modified":"2022-05-10T06:06:49","modified_gmt":"2022-05-10T06:06:49","slug":"selina-class-7-icse-solutions-mathematics-chapter-3-fraction-including-problems","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/selina-class-7-icse-solutions-mathematics-chapter-3-fraction-including-problems\/","title":{"rendered":"Selina Class 7 ICSE Solutions Mathematics : Chapter 3-\u00a0Fraction (Including Problems)"},"content":{"rendered":"\n

Class 7: Maths Chapter 3 solutions. Complete Class 7 Maths Chapter 3 Notes.<\/p>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 3- Fraction (Including Problems)<\/h2>\n\n\n\n

Selina 7th Maths Chapter 3, Class 7 Maths Chapter 3 solutions<\/p>\n\n\n\n

Exercise 3A page: 39<\/h4>\n\n\n\n

1. Classify each fraction given below as decimal or vulgar fraction, proper or improper fraction and mixed fraction:<\/strong><\/p>\n\n\n\n

(i) 3\/5<\/strong><\/p>\n\n\n\n

(ii) 11\/10<\/strong><\/p>\n\n\n\n

(iii) 13\/20<\/strong><\/p>\n\n\n\n

(iv) 18\/7<\/strong><\/p>\n\n\n\n

(v) 3 2\/9<\/strong><\/p>\n\n\n\n

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

(i) 3\/5 is a vulgar and proper fraction.<\/p>\n\n\n\n

(ii) 11\/10 is a decimal and improper fraction.<\/p>\n\n\n\n

(iii) 13\/20 is a decimal and proper fraction.<\/p>\n\n\n\n

(iv) 18\/7 is a vulgar and improper fraction.<\/p>\n\n\n\n

(v) 3 2\/9 is a mixed fraction.<\/p>\n\n\n\n

2. Express the following improper fractions as mixed fractions:<\/strong><\/p>\n\n\n\n

(i) 18\/5<\/strong><\/p>\n\n\n\n

(ii) 7\/4<\/strong><\/p>\n\n\n\n

(iii) 25\/6<\/strong><\/p>\n\n\n\n

(iv) 38\/5<\/strong><\/p>\n\n\n\n

(v) 22\/5<\/strong><\/p>\n\n\n\n

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

(i) 18\/5 can be expressed as mixed fractions as 3 3\/5.<\/p>\n\n\n\n

(ii) 7\/4 can be expressed as mixed fractions as 1 3\/4.<\/p>\n\n\n\n

(iii) 25\/6 can be expressed as mixed fractions as 4 1\/6.<\/p>\n\n\n\n

(iv) 38\/5 can be expressed as mixed fractions as 7 3\/5.<\/p>\n\n\n\n

(v) 22\/5 can be expressed as mixed fractions as 4 2\/5.<\/p>\n\n\n\n

3. Express the following mixed fractions as improper fractions:<\/strong><\/p>\n\n\n\n

(i) 2 4\/9<\/strong><\/p>\n\n\n\n

(ii) 7 5\/13<\/strong><\/p>\n\n\n\n

(iii) 3 1\/4<\/strong><\/p>\n\n\n\n

(iv) 2 5\/48<\/strong><\/p>\n\n\n\n

(v) 12 7\/11<\/strong><\/p>\n\n\n\n

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

(i) 2 4\/9<\/p>\n\n\n\n

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

= (2 \u00d7 9 + 4)\/ 9<\/p>\n\n\n\n

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

= (18 + 4)\/ 9<\/p>\n\n\n\n

= 22\/9<\/p>\n\n\n\n

(ii) 7 5\/13<\/p>\n\n\n\n

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

= (7 \u00d7 13 + 5)\/ 13<\/p>\n\n\n\n

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

= (91 + 5)\/ 13<\/p>\n\n\n\n

= 96\/13<\/p>\n\n\n\n

(iii) 3 1\/4<\/p>\n\n\n\n

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

= (3 \u00d7 4 + 1)\/ 4<\/p>\n\n\n\n

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

= (12 + 1)\/ 4<\/p>\n\n\n\n

= 13\/4<\/p>\n\n\n\n

(iv) 2 5\/48<\/p>\n\n\n\n

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

= (2 \u00d7 48 + 5)\/ 48<\/p>\n\n\n\n

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

= (96 + 5)\/ 48<\/p>\n\n\n\n

= 101\/48<\/p>\n\n\n\n

(v) 12 7\/11<\/p>\n\n\n\n

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

= (12 \u00d7 11 + 7)\/ 11<\/p>\n\n\n\n

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

= (132 + 7)\/ 11<\/p>\n\n\n\n

= 139\/11<\/p>\n\n\n\n

4. Reduce the given fractions to lowest terms:<\/strong><\/p>\n\n\n\n

(i) 8\/18<\/strong><\/p>\n\n\n\n

(ii) 27\/36<\/strong><\/p>\n\n\n\n

(iii) 18\/42<\/strong><\/p>\n\n\n\n

(iv) 35\/75<\/strong><\/p>\n\n\n\n

(v) 18\/45<\/strong><\/p>\n\n\n\n

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

(i) 8\/18<\/p>\n\n\n\n

Here the HCF of 8 and 18 is 2<\/p>\n\n\n\n

So by dividing numerator and denominator by 2<\/p>\n\n\n\n

= (8 \u00f7 2)\/ (18 \u00f7 2)<\/p>\n\n\n\n

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

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

(ii) 27\/36<\/p>\n\n\n\n

Here the HCF of 27 and 36 is 9<\/p>\n\n\n\n

So by dividing numerator and denominator by 9<\/p>\n\n\n\n

= (27 \u00f7 9)\/ (36 \u00f7 9)<\/p>\n\n\n\n

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

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

(iii) 18\/42<\/p>\n\n\n\n

Here the HCF of 18 and 42 is 6<\/p>\n\n\n\n

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

= (18 \u00f7 6)\/ (42 \u00f7 6)<\/p>\n\n\n\n

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

= 3\/7<\/p>\n\n\n\n

(iv) 35\/75<\/p>\n\n\n\n

Here the HCF of 35 and 75 is 5<\/p>\n\n\n\n

So by dividing both numerator and denominator by 5<\/p>\n\n\n\n

= (35 \u00f7 5)\/ (75 \u00f7 5)<\/p>\n\n\n\n

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

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

(v) 18\/45<\/p>\n\n\n\n

Here the HCF of 18 and 45 is 9<\/p>\n\n\n\n

So by dividing both numerator and denominator by 9<\/p>\n\n\n\n

= (18 \u00f7 9)\/ (45 \u00f7 9)<\/p>\n\n\n\n

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

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

5. State true or false:<\/strong><\/p>\n\n\n\n

(i) 30\/40 and 12\/16 are equivalent fractions.<\/strong><\/p>\n\n\n\n

(ii) 10\/25 and 25\/10 are equivalent fractions.<\/strong><\/p>\n\n\n\n

(iii) 35\/49, 20\/28, 45\/63 and 100\/140 are equivalent fractions.<\/strong><\/p>\n\n\n\n

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

(i) True.<\/p>\n\n\n\n

Here 30\/40 = 3\/4 and 12\/16 = 3\/4<\/p>\n\n\n\n

(ii) False.<\/p>\n\n\n\n

Here 10\/25 = 2\/5 and 25\/10 = 5\/2<\/p>\n\n\n\n

(iii) True.<\/p>\n\n\n\n

35\/49 = 5\/7, 20\/28 = 5\/7, 45\/63 = 5\/7 and 100\/140 = 5\/7 where all are equal.<\/p>\n\n\n\n

6. Distinguish each of the fractions, given below, as a simple fraction or a complex fraction:<\/strong><\/p>\n\n\n\n

(i) 0\/8<\/strong><\/p>\n\n\n\n

(ii) -3\/-8<\/strong><\/p>\n\n\n\n

(iii) 5\/-7<\/strong><\/p>\n\n\n\n

(iv) 3 3\/5\/ 18<\/strong><\/p>\n\n\n\n

(v) -6\/ 2 2\/5<\/strong><\/p>\n\n\n\n

(vi) 3 1\/3\/ 7 2\/7<\/strong><\/p>\n\n\n\n

(vii) \u2013 5 2\/9\/ 5<\/strong><\/p>\n\n\n\n

(viii) -8\/0<\/strong><\/p>\n\n\n\n

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

(i) 0\/8 is a simple fraction.<\/p>\n\n\n\n

(ii) -3\/-8 is a simple fraction.<\/p>\n\n\n\n

(iii) 5\/-7 is a simple fraction.<\/p>\n\n\n\n

(iv) 3 3\/5\/ 18 is a complex fraction.<\/p>\n\n\n\n

(v) -6\/ 2 2\/5 is a complex fraction.<\/p>\n\n\n\n

(vi) 3 1\/3\/ 7 2\/7 is a complex fraction.<\/p>\n\n\n\n

(vii) \u2013 5 2\/9\/ 5 is a complex fraction.<\/p>\n\n\n\n

(viii) -8\/0 is neither a complex nor a simple fraction.<\/p>\n\n\n\n

Exercise 3B page: 43<\/h4>\n\n\n\n

1. For each pair, given below, state whether it forms like fractions or unlike fractions:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 7\/8<\/strong><\/p>\n\n\n\n

(ii) 8\/15 and 8\/21<\/strong><\/p>\n\n\n\n

(iii) 4\/9 and 9\/4<\/strong><\/p>\n\n\n\n

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

(i) 5\/8 and 7\/8 are like fractions.<\/p>\n\n\n\n

(ii) 8\/15 and 8\/21 are unlike fractions.<\/p>\n\n\n\n

(iii) 4\/9 and 9\/4 are unlike fractions.<\/p>\n\n\n\n

2. Convert given fractions into fractions with equal denominators:<\/strong><\/p>\n\n\n\n

(i) 5\/6 and 7\/9<\/strong><\/p>\n\n\n\n

(ii) 2\/3, 5\/6 and 7\/12<\/strong><\/p>\n\n\n\n

(iii) 4\/5, 17\/20, 23\/20 and 11\/16<\/strong><\/p>\n\n\n\n

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

(i) 5\/6 and 7\/9<\/p>\n\n\n\n

Here the LCM of 6 and 9 is 18<\/p>\n\n\n\n

5\/6 = (5 \u00d7 3)\/ (6 \u00d7 3) = 15\/18<\/p>\n\n\n\n

7\/9 = (7 \u00d7 2)\/ (9 \u00d7 2) = 14\/18<\/p>\n\n\n\n

Therefore, 15\/18 and 14\/18 are the required fractions.<\/p>\n\n\n\n

(ii) 2\/3, 5\/6 and 7\/12<\/p>\n\n\n\n

Here the LCM of 3, 6 and 12 is 12<\/p>\n\n\n\n

2\/3 = (2 \u00d7 4)\/ (3 \u00d7 4) = 8\/12<\/p>\n\n\n\n

5\/6 = (5 \u00d7 2)\/ (6 \u00d7 2) = 10\/12<\/p>\n\n\n\n

7\/12 = 7\/12<\/p>\n\n\n\n

Therefore, 8\/12, 10\/12 and 7\/12 are the required fractions.<\/p>\n\n\n\n

(iii) 4\/5, 17\/20, 23\/40 and 11\/16<\/p>\n\n\n\n

Here the LCM of 5, 20, 40 and 16 is 80<\/p>\n\n\n\n

4\/5 = (4 \u00d7 16)\/ (5 \u00d7 16) = 64\/80<\/p>\n\n\n\n

17\/20 = (17 \u00d7 4)\/ (20 \u00d7 4) = 68\/80<\/p>\n\n\n\n

23\/40 = (23 \u00d7 2)\/ (40 \u00d7 2) = 46\/80<\/p>\n\n\n\n

11\/16 = (11 \u00d7 5)\/ (16 \u00d7 5) = 55\/80<\/p>\n\n\n\n

Therefore, 64\/80, 68\/80, 46\/80 and 55\/80 are the required fractions.<\/p>\n\n\n\n

3. Convert given fractions into fractions with equal numerators:<\/strong><\/p>\n\n\n\n

(i) 8\/9 and 12\/17<\/strong><\/p>\n\n\n\n

(ii) 6\/13, 15\/23 and 12\/17<\/strong><\/p>\n\n\n\n

(iii) 15\/19, 25\/28, 9\/11 and 45\/47<\/strong><\/p>\n\n\n\n

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

(i) 8\/9 and 12\/17<\/p>\n\n\n\n

Here the LCM of 8 and 12 is 24<\/p>\n\n\n\n

8\/9 = (8 \u00d7 3)\/ (9 \u00d7 3) = 24\/27<\/p>\n\n\n\n

12\/17 = (12 \u00d7 2)\/ (17 \u00d7 2) = 24\/34<\/p>\n\n\n\n

Therfore, 24\/27 and 24\/34 are the required fractions.<\/p>\n\n\n\n

(ii) 6\/13, 15\/23 and 12\/17<\/p>\n\n\n\n

Here the LCM of 6, 15 and 12 is 60<\/p>\n\n\n\n

6\/13 = (6 \u00d7 10)\/ (13 \u00d7 10) = 60\/130<\/p>\n\n\n\n

15\/23 = (15 \u00d7 4)\/ (23 \u00d7 4) = 60\/92<\/p>\n\n\n\n

12\/17 = (12 \u00d7 5)\/ (17 \u00d7 5) = 60\/85<\/p>\n\n\n\n

Therefore, 60\/130, 60\/92 and 60\/85 are the required fractions.<\/p>\n\n\n\n

(iii) 15\/19, 25\/28, 9\/11 and 45\/47<\/p>\n\n\n\n

Here the LCM of 15, 25, 9 and 45 is 225<\/p>\n\n\n\n

15\/19 = (15 \u00d7 15)\/ (19 \u00d7 15) = 225\/285<\/p>\n\n\n\n

25\/28 = (25 \u00d7 9)\/ (28 \u00d7 9) = 225\/252<\/p>\n\n\n\n

9\/11 = (9 \u00d7 25)\/ (11 \u00d7 25) = 225\/275<\/p>\n\n\n\n

45\/47 = (45 \u00d7 5)\/ (47 \u00d7 5) = 225\/ 235<\/p>\n\n\n\n

Therefore, 225\/285, 225\/252, 225\/275 and 225\/235 are the required fractions.<\/p>\n\n\n\n

4. Put the given fractions in ascending order by making denominators equal:<\/strong><\/p>\n\n\n\n

(i) 1\/3, 2\/5, 3\/4 and 1\/6<\/strong><\/p>\n\n\n\n

(ii) 5\/6, 7\/8, 11\/12 and 3\/10<\/strong><\/p>\n\n\n\n

(iii) 5\/7, 3\/8, 9\/14 and 20\/21<\/strong><\/p>\n\n\n\n

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

(i) 1\/3, 2\/5, 3\/4 and 1\/6<\/p>\n\n\n\n

Here the LCM of 3, 5, 4 and 6 is 60<\/p>\n\n\n\n

1\/3 = (1 \u00d7 20)\/ (3 \u00d7 20) = 20\/60<\/p>\n\n\n\n

2\/5 = (2 \u00d7 12)\/ (5 \u00d7 12) = 24\/60<\/p>\n\n\n\n

3\/4 = (3 \u00d7 15)\/ (4 \u00d7 15) = 45\/60<\/p>\n\n\n\n

1\/6 = (1 \u00d7 10)\/ (6 \u00d7 10) = 10\/60<\/p>\n\n\n\n

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

10\/60 < 20\/60 < 24\/60 < 45\/60<\/p>\n\n\n\n

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

1\/6 < 1\/3 < 2\/5 < 3\/4<\/p>\n\n\n\n

Therefore, 1\/6, 1\/3, 2\/5 and 3\/4 are in ascending order.<\/p>\n\n\n\n

(ii) 5\/6, 7\/8, 11\/12 and 3\/10<\/p>\n\n\n\n

Here the LCM of 6, 8, 12 and 10 is 240<\/p>\n\n\n\n

5\/6 = (5 \u00d7 40)\/ (6 \u00d7 40) = 200\/240<\/p>\n\n\n\n

7\/8 = (7 \u00d7 30)\/ (8 \u00d7 30) = 210\/240<\/p>\n\n\n\n

11\/12 = (11 \u00d7 20)\/ (12 \u00d7 20) = 220\/240<\/p>\n\n\n\n

3\/10 = (3 \u00d7 24)\/ (10 \u00d7 24) = 72\/240<\/p>\n\n\n\n

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

72\/240 < 200\/240 < 210\/240 < 220\/240<\/p>\n\n\n\n

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

3\/10 < 5\/6 < 7\/8 < 11\/12<\/p>\n\n\n\n

Therefore, 3\/10, 5\/6, 7\/8 and 11\/12 are in ascending order.<\/p>\n\n\n\n

(iii) 5\/7, 3\/8, 9\/14 and 20\/21<\/p>\n\n\n\n

Here the LCM of 7, 8, 14 and 21 is 168<\/p>\n\n\n\n

5\/7 = (5 \u00d7 24)\/ (7 \u00d7 24) = 120\/168<\/p>\n\n\n\n

3\/8 = (3 \u00d7 21)\/ (8 \u00d7 21) = 63\/168<\/p>\n\n\n\n

9\/14 = (9 \u00d7 12)\/ (14 \u00d7 12) = 108\/168<\/p>\n\n\n\n

20\/21 = (20 \u00d7 8)\/ (21 \u00d7 8) = 160\/168<\/p>\n\n\n\n

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

63\/168 < 108\/ 168 < 120\/168 < 160\/168<\/p>\n\n\n\n

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

3\/8 < 9\/14 < 5\/7 < 20\/21<\/p>\n\n\n\n

Therefore, 3\/8, 9\/14, 5\/7 and 20\/21 are in ascending order.<\/p>\n\n\n\n

5. Arrange the given fractions in descending order by making numerators equal:<\/strong><\/p>\n\n\n\n

(i) 5\/6, 4\/15, 8\/9 and 1\/3<\/strong><\/p>\n\n\n\n

(ii) 3\/7, 4\/9, 5\/7 and 8\/11<\/strong><\/p>\n\n\n\n

(iii) 1\/10, 6\/11, 8\/11 and 3\/5<\/strong><\/p>\n\n\n\n

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

(i) 5\/6, 4\/15, 8\/9 and 1\/3<\/p>\n\n\n\n

Here the LCM of 5, 4, 8 and 1 is 40<\/p>\n\n\n\n

5\/6 = (5 \u00d7 8)\/ (6 \u00d7 8) = 40\/48<\/p>\n\n\n\n

4\/15 = (4 \u00d7 10)\/ (15 \u00d7 10) = 40\/150<\/p>\n\n\n\n

8\/9 = (8 \u00d7 5)\/ (9 \u00d7 5) = 40\/45<\/p>\n\n\n\n

1\/3 = (1 \u00d7 40)\/ (3 \u00d7 40) = 40\/120<\/p>\n\n\n\n

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

40\/45 > 40\/48 > 40\/120 > 40\/150<\/p>\n\n\n\n

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

8\/9 > 5\/6 > 1\/3 > 4\/15<\/p>\n\n\n\n

Therefore, 8\/9, 5\/6, 1\/3 and 4\/15 are in descending order.<\/p>\n\n\n\n

(ii) 3\/7, 4\/9, 5\/7 and 8\/11<\/p>\n\n\n\n

Here the LCM of 3, 4, 5 and 8 is 120<\/p>\n\n\n\n

3\/7 = (3 \u00d7 40)\/ (7 \u00d7 40) = 120\/280<\/p>\n\n\n\n

4\/9 = (4 \u00d7 30)\/ (9 \u00d7 30) = 120\/270<\/p>\n\n\n\n

5\/7 = (5 \u00d7 24)\/ (7 \u00d7 24) = 120\/168<\/p>\n\n\n\n

8\/11 = (8 \u00d7 15)\/ (11 \u00d7 15) = 120\/165<\/p>\n\n\n\n

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

120\/165 > 120\/168 > 120\/270 > 120\/280<\/p>\n\n\n\n

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

8\/11 > 5\/7 > 4\/9 > 3\/7<\/p>\n\n\n\n

Therefore, 8\/11, 5\/7, 4\/9 and 3\/7 are in descending order.<\/p>\n\n\n\n

(iii) 1\/10, 6\/11, 8\/11 and 3\/5<\/p>\n\n\n\n

Here the LCM of 1, 6, 8 and 3 is 24<\/p>\n\n\n\n

1\/10 = (1 \u00d7 24)\/ (10 \u00d7 24) = 24\/240<\/p>\n\n\n\n

6\/11 = (6 \u00d7 4)\/ (11 \u00d7 4) = 24\/44<\/p>\n\n\n\n

8\/11 = (8 \u00d7 3)\/ (11 \u00d7 3) = 24\/33<\/p>\n\n\n\n

3\/5 = (3 \u00d7 8)\/ (5 \u00d7 8) = 24\/40<\/p>\n\n\n\n

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

24\/33 > 24\/40 > 24\/44 > 24\/240<\/p>\n\n\n\n

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

8\/11 > 3\/5 > 6\/11 > 1\/10<\/p>\n\n\n\n

Therefore, 8\/11, 3\/5, 6\/11 and 1\/10 are in descending order.<\/p>\n\n\n\n

6. Find the greater fraction:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 11\/15<\/strong><\/p>\n\n\n\n

(ii) 4\/5 and 3\/10<\/strong><\/p>\n\n\n\n

(iii) 6\/7 and 5\/9<\/strong><\/p>\n\n\n\n

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

(i) 3\/5 and 11\/15<\/p>\n\n\n\n

Here the LCM of 5 and 15 is 15<\/p>\n\n\n\n

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

11\/15 = 11\/15<\/p>\n\n\n\n

So we get, 11\/15 > 9\/15<\/p>\n\n\n\n

Therefore, 11\/15 is greater.<\/p>\n\n\n\n

(ii) 4\/5 and 3\/10<\/p>\n\n\n\n

Here the LCM of 5 and 10 is 10<\/p>\n\n\n\n

4\/5 = (4 \u00d7 2)\/ (5 \u00d7 2) = 8\/10<\/p>\n\n\n\n

3\/10 = 3\/10<\/p>\n\n\n\n

So we get, 8\/10 > 3\/10<\/p>\n\n\n\n

4\/5 > 3\/10<\/p>\n\n\n\n

Therefore, 4\/5 is greater.<\/p>\n\n\n\n

(iii) 6\/7 and 5\/9<\/p>\n\n\n\n

Here LCM of 7 and 9 is 63<\/p>\n\n\n\n

6\/7 = (6 \u00d7 9)\/ (7 \u00d7 9) = 54\/63<\/p>\n\n\n\n

5\/9 = (5 \u00d7 7)\/ (9 \u00d7 7) = 35\/63<\/p>\n\n\n\n

So we get, 54\/63 > 35\/63<\/p>\n\n\n\n

6\/7 > 35\/63<\/p>\n\n\n\n

Therefore, 6\/7 is greater.<\/p>\n\n\n\n

7. Insert one fraction between:
(i) 3\/7 and 4\/9<\/strong><\/p>\n\n\n\n

(ii) 2 and 8\/3<\/strong><\/p>\n\n\n\n

(iii) 9\/17 and 6\/13<\/strong><\/p>\n\n\n\n

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

(i) 3\/7 and 4\/9<\/p>\n\n\n\n

So the fraction between 3\/7 and 4\/9<\/p>\n\n\n\n

= (3 + 4)\/ (7 + 9)<\/p>\n\n\n\n

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

(ii) 2 and 8\/3<\/p>\n\n\n\n

So the fraction between 2 and 8\/3<\/p>\n\n\n\n

= (2 + 8)\/ (1 + 3)<\/p>\n\n\n\n

= 10\/4<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

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

= 2 \u00bd<\/p>\n\n\n\n

(iii) 9\/17 and 6\/13<\/p>\n\n\n\n

So the fraction between 9\/17 and 6\/13<\/p>\n\n\n\n

= (9 + 6)\/ (17 + 13)<\/p>\n\n\n\n

= 15\/30<\/p>\n\n\n\n

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

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

8. Insert three fractions between:<\/strong><\/p>\n\n\n\n

(i) 2\/5 and 4\/9<\/strong><\/p>\n\n\n\n

(ii) 1\/2 and 5\/7<\/strong><\/p>\n\n\n\n

(iii) 3\/8 and 6\/11<\/strong><\/p>\n\n\n\n

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

(i) 2\/5 and 4\/9<\/p>\n\n\n\n

So the fraction between 2\/5 and 4\/9<\/p>\n\n\n\n

= (2 + 4)\/ (5 + 9)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/14<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

= 3\/7<\/p>\n\n\n\n

Fraction between 2\/5 and 3\/7<\/p>\n\n\n\n

= (2 + 3)\/ (5 + 7)<\/p>\n\n\n\n

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

Fraction between 3\/7 and 4\/9<\/p>\n\n\n\n

= (3 + 4)\/ (7 + 9)<\/p>\n\n\n\n

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

Therefore, three fractions between 2\/5 and 4\/9 will be 5\/12, 3\/7 and 7\/16.<\/p>\n\n\n\n

(ii) 1\/2 and 5\/7<\/p>\n\n\n\n

So the fraction between 1\/2 and 5\/7<\/p>\n\n\n\n

= (1 + 5)\/ (2 + 7)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/ 9<\/p>\n\n\n\n

Dividing by 3<\/p>\n\n\n\n

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

Fraction between 1\/2 and 2\/3<\/p>\n\n\n\n

= (1 + 2)\/ (2 + 3)<\/p>\n\n\n\n

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

Fraction between 2\/3 and 5\/7<\/p>\n\n\n\n

= (2 + 5)\/ (3 + 7)<\/p>\n\n\n\n

= 7\/ 10<\/p>\n\n\n\n

Therefore, three fractions between 1\/2 and 5\/7 will be 3\/5, 2\/3 and 7\/10.<\/p>\n\n\n\n

(iii) 3\/8 and 6\/11<\/p>\n\n\n\n

So the fraction between 3\/8 and 6\/11<\/p>\n\n\n\n

= (3 + 6)\/ (8 + 11)<\/p>\n\n\n\n

= 9\/ 19<\/p>\n\n\n\n

Fraction between 3\/8 and 9\/19<\/p>\n\n\n\n

= (3 + 9)\/ (8 + 19)<\/p>\n\n\n\n

= 12\/27<\/p>\n\n\n\n

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

Fraction between 9\/19 and 6\/11<\/p>\n\n\n\n

= (9 + 6)\/ (19 + 11)<\/p>\n\n\n\n

= 15\/30<\/p>\n\n\n\n

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

Therefore, three fractions between 3\/8 and 6\/11 will be 4\/9, 9\/19 and 1\/2.<\/p>\n\n\n\n

9. Insert two fractions between:<\/strong><\/p>\n\n\n\n

(i) 1 and 3\/11<\/strong><\/p>\n\n\n\n

(ii) 5\/9 and 1\/4<\/strong><\/p>\n\n\n\n

(iii) 5\/6 and 1 1\/5<\/strong><\/p>\n\n\n\n

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

(i) 1 and 3\/11<\/p>\n\n\n\n

Fraction between 1 and 3\/11<\/p>\n\n\n\n

= (1 + 3)\/ (1 + 11)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 4\/12<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

Fraction between 1\/3 and 3\/11<\/p>\n\n\n\n

= (1 + 3)\/ (3 + 11)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 4\/14<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

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

Therefore, two fractions between 1 and 3\/11 will be 1\/3 and 2\/7.<\/p>\n\n\n\n

(ii) 5\/9 and 1\/4<\/p>\n\n\n\n

Fraction between 5\/9 and 1\/4<\/p>\n\n\n\n

= (5 + 1)\/ (9 + 4)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/13<\/p>\n\n\n\n

Fraction between 6\/13 and 1\/4<\/p>\n\n\n\n

= (6 + 1)\/ (13 + 4)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 7\/17<\/p>\n\n\n\n

Therefore, two fractions between 5\/9 and 1\/4 will be 6\/13 and 7\/17.<\/p>\n\n\n\n

(iii) 5\/6 and 1 1\/5<\/p>\n\n\n\n

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

= 5\/6 + 6\/5<\/p>\n\n\n\n

Fraction between 5\/6 and 6\/5<\/p>\n\n\n\n

= (5 + 6)\/ (6 + 5)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

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

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

Fraction between 1 and 6\/5<\/p>\n\n\n\n

= (1 + 6)\/ (1 + 5)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 7\/6<\/p>\n\n\n\n

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

Therefore, two fractions between 5\/6 and 1 1\/5 will be 1 and 1 1\/6.<\/p>\n\n\n\n

Exercise 3C page: 46<\/h4>\n\n\n\n

1. Reduce to a single fraction:<\/strong><\/p>\n\n\n\n

(i) 1\/2 + 2\/3<\/strong><\/p>\n\n\n\n

(ii) 3\/5 \u2013 1\/10<\/strong><\/p>\n\n\n\n

(iii) 2\/3 \u2013 1\/6<\/strong><\/p>\n\n\n\n

(iv) 1 1\/3 + 2 1\/4<\/strong><\/p>\n\n\n\n

(v) 1\/4 + 5\/6 \u2013 1\/12<\/strong><\/p>\n\n\n\n

(vi) 2\/3 \u2013 3\/5 + 3 \u2013 1\/5<\/strong><\/p>\n\n\n\n

(vii) 2\/3 \u2013 1\/5 + 1\/10<\/strong><\/p>\n\n\n\n

(viii) 2 1\/2 + 2 1\/3 \u2013 1 1\/4<\/strong><\/p>\n\n\n\n

(ix) 2 5\/8 \u2013 2 1\/6 + 4 3\/4<\/strong><\/p>\n\n\n\n

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

(i) 1\/2 + 2\/3<\/p>\n\n\n\n

Here the LCM of 2 and 3 is 6<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (2 \u00d7 3) + (2 \u00d7 2)\/ (3 \u00d7 2)<\/p>\n\n\n\n

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

= 3\/6 + 4\/6<\/p>\n\n\n\n

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

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

= 7\/6<\/p>\n\n\n\n

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

(ii) 3\/5 \u2013 1\/10<\/p>\n\n\n\n

Here the LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (3 \u00d7 2)\/ (5 \u00d7 2) \u2013 1\/10<\/p>\n\n\n\n

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

= 6\/10 \u2013 1\/10<\/p>\n\n\n\n

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

= (6 \u2013 1)\/ 10<\/p>\n\n\n\n

= 5\/10<\/p>\n\n\n\n

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

(iii) 2\/3 \u2013 1\/6<\/p>\n\n\n\n

Here the LCM of 3 and 6 is 6<\/p>\n\n\n\n

= (2 \u00d7 2)\/ (3 \u00d7 2) \u2013 1\/6<\/p>\n\n\n\n

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

= 4\/6 \u2013 1\/6<\/p>\n\n\n\n

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

= (4 \u2013 1)\/ 6<\/p>\n\n\n\n

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

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

(iv) 1 1\/3 + 2 1\/4<\/p>\n\n\n\n

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

= 4\/3 + 9\/4<\/p>\n\n\n\n

Here LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (4 \u00d7 4)\/ (3 \u00d7 4) + (9 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

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

= 16\/12 + 27\/12<\/p>\n\n\n\n

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

= (16 + 27)\/ 12<\/p>\n\n\n\n

= 43\/12<\/p>\n\n\n\n

= 3 7\/12<\/p>\n\n\n\n

(v) 1\/4 + 5\/6 \u2013 1\/12<\/p>\n\n\n\n

Here LCM of 4, 6 and 12 is 12<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (4 \u00d7 3) + (5 \u00d7 2)\/ (6 \u00d7 2) \u2013 1\/12<\/p>\n\n\n\n

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

= 3\/12 + 10\/12 \u2013 1\/12<\/p>\n\n\n\n

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

= (3 + 10 \u2013 1)\/ 12<\/p>\n\n\n\n

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

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

(vi) 2\/3 \u2013 3\/5 + 3 \u2013 1\/5<\/p>\n\n\n\n

Here LCM of 3 and 5 is 15<\/p>\n\n\n\n

= (2 \u00d7 5)\/ (3 \u00d7 5) \u2013 (3 \u00d7 3)\/ (5 \u00d7 3) + (3 \u00d7 15)\/ 15 \u2013 (1 \u00d7 3)\/ (5 \u00d7 3)<\/p>\n\n\n\n

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

= 10\/15 \u2013 9\/15 + 45\/15 \u2013 3\/15<\/p>\n\n\n\n

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

= (10 \u2013 9 + 45 \u2013 3)\/ 15<\/p>\n\n\n\n

= (55 \u2013 12)\/ 15<\/p>\n\n\n\n

= 43\/15<\/p>\n\n\n\n

= 2 13\/15<\/p>\n\n\n\n

(vii) 2\/3 \u2013 1\/5 + 1\/10<\/p>\n\n\n\n

Here the LCM of 3, 5 and 10 is 30<\/p>\n\n\n\n

= (2 \u00d7 10)\/ (3 \u00d7 10) \u2013 (1 \u00d7 6)\/ (5 \u00d7 6) + (1 \u00d7 3)\/ (10 \u00d7 3)<\/p>\n\n\n\n

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

= 20\/30 \u2013 6\/30 + 3\/30<\/p>\n\n\n\n

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

= (20 \u2013 6 + 3)\/ 30<\/p>\n\n\n\n

= (23 \u2013 6)\/ 30<\/p>\n\n\n\n

= 17\/30<\/p>\n\n\n\n

(viii) 2 1\/2 + 2 1\/3 \u2013 1 \u00bc<\/p>\n\n\n\n

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

= 5\/2 + 7\/3 \u2013 5\/4<\/p>\n\n\n\n

Here the LCM of 2, 3 and 4 is 12<\/p>\n\n\n\n

= (5 \u00d7 6)\/ (2 \u00d7 6) + (7 \u00d7 4)\/ (3 \u00d7 4) \u2013 (5 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

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

= 30\/12 + 28\/12 \u2013 15\/12<\/p>\n\n\n\n

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

= (30 + 28 \u2013 15)\/ 12<\/p>\n\n\n\n

= (58 \u2013 15)\/ 12<\/p>\n\n\n\n

= 43\/12<\/p>\n\n\n\n

= 3 7\/12<\/p>\n\n\n\n

(ix) 2 5\/8 \u2013 2 1\/6 + 4 3\/4<\/p>\n\n\n\n

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

= 21\/8 \u2013 13\/6 + 19\/4<\/p>\n\n\n\n

Here the LCM of 8, 6 and 4 is 24<\/p>\n\n\n\n

= (21 \u00d7 3)\/ (8 \u00d7 3) \u2013 (13 \u00d7 4)\/ (6 \u00d7 4) + (19 \u00d7 6)\/ (4 \u00d7 6)<\/p>\n\n\n\n

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

= 63\/24 \u2013 52\/24 + 114\/24<\/p>\n\n\n\n

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

= (63 \u2013 52 + 114)\/ 24<\/p>\n\n\n\n

= (177 \u2013 52)\/ 24<\/p>\n\n\n\n

= 125\/24<\/p>\n\n\n\n

= 5 5\/24<\/p>\n\n\n\n

2. Simplify:<\/strong><\/p>\n\n\n\n

(i) 3\/4 \u00d7 6<\/strong><\/p>\n\n\n\n

(ii) 2\/3 \u00d7 15<\/strong><\/p>\n\n\n\n

(iii) 3\/4 \u00d7 1\/2<\/strong><\/p>\n\n\n\n

(iv) 9\/12 \u00d7 4\/7<\/strong><\/p>\n\n\n\n

(v) 45 \u00d7 2 1\/3<\/strong><\/p>\n\n\n\n

(vi) 36 \u00d7 3 1\/4<\/strong><\/p>\n\n\n\n

(vii) 2 \u00f7 1\/3<\/strong><\/p>\n\n\n\n

(viii) 3 \u00f7 2\/5<\/strong><\/p>\n\n\n\n

(ix) 1 \u00f7 3\/5<\/strong><\/p>\n\n\n\n

(x) 1\/3 \u00f7 1\/4<\/strong><\/p>\n\n\n\n

(xi) -5\/8 \u00f7 3\/4<\/strong><\/p>\n\n\n\n

(xii) 3 3\/7 \u00f7 1 1\/14<\/strong><\/p>\n\n\n\n

(xiii) 3 \u00be \u00d7 1 1\/5 \u00d7 20\/21<\/strong><\/p>\n\n\n\n

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

(i) 3\/4 \u00d7 6<\/p>\n\n\n\n

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

= 3\/4 \u00d7 6\/1<\/p>\n\n\n\n

= (3 \u00d7 6)\/ (4 \u00d7 1)<\/p>\n\n\n\n

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

= 18\/4<\/p>\n\n\n\n

HCF of 18 and 4 is 2<\/p>\n\n\n\n

Dividing both numerator and denominator by 2<\/p>\n\n\n\n

= (18 \u00f7 2)\/ (4 \u00f7 2)<\/p>\n\n\n\n

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

= 4 1\/2<\/p>\n\n\n\n

(ii) 2\/3 \u00d7 15<\/p>\n\n\n\n

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

= 2\/3 \u00d7 15\/1<\/p>\n\n\n\n

= (2 \u00d7 15)\/ (3 \u00d7 1)<\/p>\n\n\n\n

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

= 30\/3<\/p>\n\n\n\n

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

(iii) 3\/4 \u00d7 1\/2<\/p>\n\n\n\n

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

= (3 \u00d7 1)\/ (4 \u00d7 2)<\/p>\n\n\n\n

= 3\/8<\/p>\n\n\n\n

(iv) 9\/12 \u00d7 4\/7<\/p>\n\n\n\n

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

= (9 \u00d7 4)\/ (12 \u00d7 7)<\/p>\n\n\n\n

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

= 36\/84<\/p>\n\n\n\n

Here the HCF of 36 and 84 is 12<\/p>\n\n\n\n

Dividing both numerator and denominator by 12<\/p>\n\n\n\n

= (36 \u00f7 12)\/ (84 \u00f7 12)<\/p>\n\n\n\n

= 3\/7<\/p>\n\n\n\n

(v) 45 \u00d7 2 1\/3<\/p>\n\n\n\n

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

= 45\/1 \u00d7 7\/3<\/p>\n\n\n\n

= (45 \u00d7 7)\/ (1 \u00d7 3)<\/p>\n\n\n\n

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

= 315\/3<\/p>\n\n\n\n

= 105<\/p>\n\n\n\n

(vi) 36 \u00d7 3 1\/4<\/p>\n\n\n\n

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

= 36\/1 \u00d7 13\/4<\/p>\n\n\n\n

= (36 \u00d7 13)\/ (4 \u00d7 1)<\/p>\n\n\n\n

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

= 468\/4<\/p>\n\n\n\n

= 117<\/p>\n\n\n\n

(vii) 2 \u00f7 1\/3<\/p>\n\n\n\n

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

= 2\/1 \u00d7 3\/1<\/p>\n\n\n\n

= (2 \u00d7 3)\/ (1 \u00d7 1)<\/p>\n\n\n\n

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

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

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

(viii) 3 \u00f7 2\/5<\/p>\n\n\n\n

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

= 3\/1 \u00d7 5\/2<\/p>\n\n\n\n

= (3 \u00d7 5)\/ (1 \u00d7 2)<\/p>\n\n\n\n

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

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

= 7 \u00bd<\/p>\n\n\n\n

(ix) 1 \u00f7 3\/5<\/p>\n\n\n\n

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

= 1 \u00d7 5\/3<\/p>\n\n\n\n

= (1 \u00d7 5)\/ 3<\/p>\n\n\n\n

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

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

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

(x) 1\/3 \u00f7 1\/4<\/p>\n\n\n\n

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

= 1\/3 \u00d7 4\/1<\/p>\n\n\n\n

= (1 \u00d7 4)\/ (3 \u00d7 1)<\/p>\n\n\n\n

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

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

= 1 1\/3<\/p>\n\n\n\n

(xi) -5\/8 \u00f7 3\/4<\/p>\n\n\n\n

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

= -5\/8 \u00d7 4\/3<\/p>\n\n\n\n

= (-5 \u00d7 4)\/ (8 \u00d7 3)<\/p>\n\n\n\n

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

= \u2013 20\/24<\/p>\n\n\n\n

Here the HCF of 20 and 24 is 4<\/p>\n\n\n\n

So by dividing both numerator and denominator by 4<\/p>\n\n\n\n

= (-20 \u00f7 4)\/ (24 \u00f7 4)<\/p>\n\n\n\n

= -5\/6<\/p>\n\n\n\n

(xii) 3 3\/7 \u00f7 1 1\/14<\/p>\n\n\n\n

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

= 24\/7 \u00d7 14\/15<\/p>\n\n\n\n

= (24 \u00d7 14)\/ (7 \u00d7 15)<\/p>\n\n\n\n

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

= 336\/105<\/p>\n\n\n\n

Here the HCF of 336 and 105 is 21<\/p>\n\n\n\n

So by dividing both numerator and denominator by 21<\/p>\n\n\n\n

= (336 \u00f7 21)\/ (105 \u00f7 21)<\/p>\n\n\n\n

= 16\/5<\/p>\n\n\n\n

= 3 1\/5<\/p>\n\n\n\n

(xiii) 3 \u00be \u00d7 1 1\/5 \u00d7 20\/21<\/p>\n\n\n\n

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

= 15\/4 \u00d7 6\/5 \u00d7 20\/21<\/p>\n\n\n\n

= (15 \u00d7 6 \u00d7 20)\/ (4 \u00d7 5 \u00d7 21)<\/p>\n\n\n\n

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

= 1800\/420<\/p>\n\n\n\n

Here the HCF of 1800 and 420 is 60<\/p>\n\n\n\n

So by dividing both numerator and denominator by 60<\/p>\n\n\n\n

= (1800 \u00f7 60)\/ (420 \u00f7 60)<\/p>\n\n\n\n

= 30\/7<\/p>\n\n\n\n

= 4 2\/7<\/p>\n\n\n\n

3. Subtract:<\/strong><\/p>\n\n\n\n

(i) 2 from 2\/3<\/strong><\/p>\n\n\n\n

(ii) 1\/8 from 5\/8<\/strong><\/p>\n\n\n\n

(iii) -2\/5 and 2\/5<\/strong><\/p>\n\n\n\n

(iv) \u2013 3\/7 from 3\/7<\/strong><\/p>\n\n\n\n

(v) 0 from \u2013 4\/5<\/strong><\/p>\n\n\n\n

(vi) 2\/9 from 4\/5<\/strong><\/p>\n\n\n\n

(vii) \u2013 4\/7 from -6\/11<\/strong><\/p>\n\n\n\n

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

(i) 2 from 2\/3<\/p>\n\n\n\n

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

= 2\/3 \u2013 2\/1<\/p>\n\n\n\n

LCM of 3 and 1 is 3<\/p>\n\n\n\n

= 2\/3 \u2013 (2 \u00d7 3)\/ 3<\/p>\n\n\n\n

= 2\/3 \u2013 6\/3<\/p>\n\n\n\n

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

= (2 \u2013 6)\/ 3<\/p>\n\n\n\n

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

= \u2013 1 1\/3<\/p>\n\n\n\n

(ii) 1\/8 from 5\/8<\/p>\n\n\n\n

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

= 5\/8 \u2013 1\/8<\/p>\n\n\n\n

= (5 \u2013 1)\/ 8<\/p>\n\n\n\n

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

= 4\/8<\/p>\n\n\n\n

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

(iii) -2\/5 and 2\/5<\/p>\n\n\n\n

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

= 2\/5 \u2013 (-2\/5)<\/p>\n\n\n\n

= 2\/5 + 2\/5<\/p>\n\n\n\n

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

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

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

(iv) \u2013 3\/7 from 3\/7<\/p>\n\n\n\n

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

= 3\/7 \u2013 (-3\/7)<\/p>\n\n\n\n

= 3\/7 + 3\/7<\/p>\n\n\n\n

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

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

= 6\/7<\/p>\n\n\n\n

(v) 0 from \u2013 4\/5<\/p>\n\n\n\n

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

= -4\/5 \u2013 0<\/p>\n\n\n\n

= \u2013 4\/5<\/p>\n\n\n\n

(vi) 2\/9 from 4\/5<\/p>\n\n\n\n

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

= 4\/5 \u2013 2\/9<\/p>\n\n\n\n

LCM of 5 and 9 is 45<\/p>\n\n\n\n

= (4 \u00d7 9)\/ (5 \u00d7 9) \u2013 (2 \u00d7 5)\/ (9 \u00d7 5)<\/p>\n\n\n\n

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

= 36\/45 \u2013 10\/45<\/p>\n\n\n\n

= (36 \u2013 10)\/ 45<\/p>\n\n\n\n

= 26\/45<\/p>\n\n\n\n

(vii) \u2013 4\/7 from -6\/11<\/p>\n\n\n\n

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

= -6\/11 \u2013 (-4\/7)<\/p>\n\n\n\n

= \u2013 6\/11 + 4\/7<\/p>\n\n\n\n

Here LCM of 7 and 11 is 77<\/p>\n\n\n\n

= (- 6 \u00d7 7)\/ (11 \u00d7 7) + (4 \u00d7 11)\/ (7 \u00d7 11)<\/p>\n\n\n\n

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

= \u2013 42\/77 + 44\/77<\/p>\n\n\n\n

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

= (-42 + 44)\/ 77<\/p>\n\n\n\n

= 2\/77<\/p>\n\n\n\n

4. Find the value of:<\/strong><\/p>\n\n\n\n

(i) \u00bd and 10 kg<\/strong><\/p>\n\n\n\n

(ii) 3\/5 of 1 hour<\/strong><\/p>\n\n\n\n

(iii) 4\/7 of 2 1\/3 kg<\/strong><\/p>\n\n\n\n

(iv) 3 \u00bd times of 2 metre<\/strong><\/p>\n\n\n\n

(v) 1\/2 of 2 2\/3<\/strong><\/p>\n\n\n\n

(vi) 5\/11 of 4\/5 of 22 kg<\/strong><\/p>\n\n\n\n

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

(i) \u00bd and 10 kg<\/p>\n\n\n\n

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

= (1\/2 \u00d7 10) kg<\/p>\n\n\n\n

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

(ii) 3\/5 of 1 hour<\/p>\n\n\n\n

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

= (3\/5 \u00d7 60) minutes<\/p>\n\n\n\n

= 3 \u00d7 12<\/p>\n\n\n\n

= 36 minutes<\/p>\n\n\n\n

(iii) 4\/7 of 2 1\/3 kg<\/p>\n\n\n\n

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

= (4\/7 \u00d7 7\/3) kg<\/p>\n\n\n\n

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

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

= 1 1\/3 kg<\/p>\n\n\n\n

(iv) 3 \u00bd times of 2 metre<\/p>\n\n\n\n

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

= (7\/2 \u00d7 2) metres<\/p>\n\n\n\n

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

(v) 1\/2 of 2 2\/3<\/p>\n\n\n\n

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

= 1\/2 \u00d7 8\/3<\/p>\n\n\n\n

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

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

= 1 1\/3<\/p>\n\n\n\n

(vi) 5\/11 of 4\/5 of 22 kg<\/p>\n\n\n\n

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

= (5\/11 \u00d7 4\/5 \u00d7 22\/1) kg<\/p>\n\n\n\n

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

= 4 \u00d7 2<\/p>\n\n\n\n

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

5. Simplify and reduce to a simple fraction:<\/strong><\/p>\n\n\n\n

(i) 3\/ 3 \u00be<\/strong><\/p>\n\n\n\n

(ii) 3\/5\/ 7<\/strong><\/p>\n\n\n\n

(iii) 3\/ 5\/7<\/strong><\/p>\n\n\n\n

(iv) 2 1\/5\/ 1 1\/10<\/strong><\/p>\n\n\n\n

(v) 2\/5 of 6\/11 \u00d7 1 \u00bc<\/strong><\/p>\n\n\n\n

(vi) 2 \u00bc \u00f7 1\/7 \u00d7 1\/3<\/strong><\/p>\n\n\n\n

(vii) 1\/3 \u00d7 4 2\/3 \u00f7 3 \u00bd \u00d7 1\/2<\/strong><\/p>\n\n\n\n

(viii) 2\/3 \u00d7 1 \u00bc \u00f7 3\/7 of 2 5\/8<\/strong><\/p>\n\n\n\n

(ix) 0 \u00f7 8\/11<\/strong><\/p>\n\n\n\n

(x) 4\/5 \u00f7 7\/15 of 8\/9<\/strong><\/p>\n\n\n\n

(xi) 4\/5 \u00f7 7\/15 \u00d7 8\/9<\/strong><\/p>\n\n\n\n

(xii) 4\/5 of 7\/15 \u00f7 8\/9<\/strong><\/p>\n\n\n\n

(xiii) 1\/2 of 3\/4 \u00d7 1\/2 \u00f7 2\/3<\/strong><\/p>\n\n\n\n

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

(i) 3\/ 3 \u00be<\/p>\n\n\n\n

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

= 3 \/ 15\/4<\/p>\n\n\n\n

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

= (3 \u00d7 4)\/ 15<\/p>\n\n\n\n

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

(ii) 3\/5\/ 7<\/p>\n\n\n\n

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

= 3\/5 \u00d7 1\/7<\/p>\n\n\n\n

= 3\/35<\/p>\n\n\n\n

(iii) 3\/ 5\/7<\/p>\n\n\n\n

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

= 3 \u00d7 7\/5<\/p>\n\n\n\n

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

= 21\/5<\/p>\n\n\n\n

= 4 1\/5<\/p>\n\n\n\n

(iv) 2 1\/5\/ 1 1\/10<\/p>\n\n\n\n

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

= 11\/5 \/ 11\/10<\/p>\n\n\n\n

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

= 11\/5 \u00d7 10\/11<\/p>\n\n\n\n

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

(v) 2\/5 of 6\/11 \u00d7 1 \u00bc<\/p>\n\n\n\n

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

= 2\/5 of 6\/11 \u00d7 5\/4<\/p>\n\n\n\n

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

= 12\/55 \u00d7 5\/4<\/p>\n\n\n\n

= 3\/11<\/p>\n\n\n\n

(vi) 2 \u00bc \u00f7 1\/7 \u00d7 1\/3<\/p>\n\n\n\n

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

= 9\/4 \u00f7 1\/7 \u00d7 1\/3<\/p>\n\n\n\n

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

= 9\/4 \u00d7 7\/1 \u00d7 1\/3<\/p>\n\n\n\n

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

= 5 \u00bc<\/p>\n\n\n\n

(vii) 1\/3 \u00d7 4 2\/3 \u00f7 3 \u00bd \u00d7 1\/2<\/p>\n\n\n\n

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

= 1\/3 \u00d7 14\/3 \u00f7 7\/2 \u00d7 1\/2<\/p>\n\n\n\n

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

= 1\/3 \u00d7 14\/3 \u00d7 2\/7 \u00d7 1\/2<\/p>\n\n\n\n

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

(viii) 2\/3 \u00d7 1 \u00bc \u00f7 3\/7 of 2 5\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00f7 3\/7 of 21\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00f7 9\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00d7 8\/9<\/p>\n\n\n\n

= 20\/27<\/p>\n\n\n\n

(ix) 0 \u00f7 8\/11<\/p>\n\n\n\n

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

= 0 \u00d7 11\/8<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

(x) 4\/5 \u00f7 7\/15 of 8\/9<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 4\/5 \u00f7 156\/135<\/p>\n\n\n\n

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

= 4\/5 \u00d7 135\/56<\/p>\n\n\n\n

= 27\/14<\/p>\n\n\n\n

= 1 13\/14<\/p>\n\n\n\n

(xi) 4\/5 \u00f7 7\/15 \u00d7 8\/9<\/p>\n\n\n\n

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

= 4\/5 \u00d7 15\/7 \u00d7 8\/9<\/p>\n\n\n\n

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

= 32\/21<\/p>\n\n\n\n

= 1 11\/21<\/p>\n\n\n\n

(xii) 4\/5 of 7\/15 \u00f7 8\/9<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 28\/75 \u00f7 8\/9<\/p>\n\n\n\n

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

= 28\/75 \u00d7 9\/8<\/p>\n\n\n\n

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

= (7 \u00d7 3)\/ (25 \u00d7 2)<\/p>\n\n\n\n

= 21\/50<\/p>\n\n\n\n

(xiii) 1\/2 of 3\/4 \u00d7 1\/2 \u00f7 2\/3<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 3\/8 \u00d7 1\/2 \u00f7 2\/3<\/p>\n\n\n\n

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

= 3\/8 \u00d7 1\/2 \u00d7 3\/2<\/p>\n\n\n\n

= 9\/32<\/p>\n\n\n\n

6. A bought 3 \u00be kg of wheat and 2 \u00bd kg of rice. Find the total weight wheat and rice bought.<\/strong><\/p>\n\n\n\n

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

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

Weight of wheat = 3 \u00be kg = 15\/4 kg<\/p>\n\n\n\n

Weight of rice = 2 \u00bd kg = 5\/2 kg<\/p>\n\n\n\n

So the total weight of wheat and rice = 15\/4 + 5\/2<\/p>\n\n\n\n

Here the LCM of 4 and 2 is 4<\/p>\n\n\n\n

= (15 \u00d7 1)\/ (4 \u00d7 1) + (5 \u00d7 2)\/ (2 \u00d7 2)<\/p>\n\n\n\n

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

= (15 + 10)\/ 4<\/p>\n\n\n\n

= 25\/4 kg<\/p>\n\n\n\n

= 6 \u00bc kg<\/p>\n\n\n\n

7. Which is greater, 3\/5 or 7\/10 and by how much?<\/strong><\/p>\n\n\n\n

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

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

3 \u00d7 10 = 30 and 7 \u00d7 5 = 35<\/p>\n\n\n\n

Here 30 is smaller than 35<\/p>\n\n\n\n

So 3\/5 < 7\/10<\/p>\n\n\n\n

We know that difference between 7\/10 and 3\/5<\/p>\n\n\n\n

= 7\/10 \u2013 3\/5<\/p>\n\n\n\n

LCM of 10 and 5 is 10<\/p>\n\n\n\n

= (7 \u00d7 1)\/ (10 \u00d7 1) \u2013 (3 \u00d7 2)\/ (5 \u00d7 2)<\/p>\n\n\n\n

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

= (7 \u2013 6)\/ 10<\/p>\n\n\n\n

= 1\/10<\/p>\n\n\n\n

Therefore, 7\/10 is greater than 3\/5 by 1\/10.<\/p>\n\n\n\n

8. What number should be added to 8 2\/3 to get 12 5\/6?<\/strong><\/p>\n\n\n\n

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

To find the fraction we must subtract 8 2\/3 from 12 5\/6<\/p>\n\n\n\n

So the required number = 12 5\/6 \u2013 8 2\/3<\/p>\n\n\n\n

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

= 77\/6 \u2013 26\/3<\/p>\n\n\n\n

Here the LCM of 3 and 6 is 6<\/p>\n\n\n\n

= (77 \u00d7 1)\/ (6 \u00d7 1) \u2013 (26 \u00d7 2)\/ (3 \u00d7 2)<\/p>\n\n\n\n

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

= (77 \u2013 52)\/ 6<\/p>\n\n\n\n

= 25\/6<\/p>\n\n\n\n

= 4 1\/6<\/p>\n\n\n\n

9. What should be subtracted from 8 \u00be to get 2 2\/3?<\/strong><\/p>\n\n\n\n

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

Required number = 8 \u00be \u2013 2 2\/3<\/p>\n\n\n\n

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

= 35\/4 \u2013 8\/3<\/p>\n\n\n\n

LCM of 4 and 3 is 12<\/p>\n\n\n\n

= (35 \u00d7 3)\/ (4 \u00d7 3) \u2013 (8 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

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

= (105 \u2013 32)\/ 12<\/p>\n\n\n\n

= 73\/12<\/p>\n\n\n\n

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

10. A rectangular field is 16 \u00bd m long and 12 2\/5 m wide. Find the perimeter of the field.<\/strong><\/p>\n\n\n\n

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

Dimensions of rectangular field are<\/p>\n\n\n\n

Length = 16 \u00bd m<\/p>\n\n\n\n

Breadth = 12 2\/5 m<\/p>\n\n\n\n

So the perimeter = 2 (l + b)<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 2 \u00d7 (16 \u00bd + 12 2\/5)<\/p>\n\n\n\n

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

= 2 \u00d7 (33\/2 + 62\/5)<\/p>\n\n\n\n

LCM of 2 and 5 is 10<\/p>\n\n\n\n

= 2 \u00d7 [(33 \u00d7 5)\/ (2 \u00d7 5) + (62 \u00d7 2)\/ (5 \u00d7 2)]<\/p>\n\n\n\n

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

= 2 \u00d7 [(165 + 124)\/ 10)]
So we get<\/p>\n\n\n\n

= 2 \u00d7 289\/10<\/p>\n\n\n\n

= 289\/5 m<\/p>\n\n\n\n

= 57 4\/5 m<\/p>\n\n\n\n

11. Sugar costs \u20b9 37 \u00bd per kg. Find the cost of 8 \u00be kg sugar.<\/strong><\/p>\n\n\n\n

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

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

Cost of 1 kg sugar = \u20b9 37 1\/2<\/p>\n\n\n\n

So the cost of 8 \u00be kg sugar = 37 \u00bd \u00d7 8 \u00be<\/p>\n\n\n\n

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

= 75\/2 \u00d7 35\/4<\/p>\n\n\n\n

= 2625\/8<\/p>\n\n\n\n

= \u20b9 328 1\/8<\/p>\n\n\n\n

12. A motor cycle runs 31 \u00bc km consuming 1 litre of petrol. How much distance will it run consuming 1 3\/5 litre of petrol?<\/strong><\/p>\n\n\n\n

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

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

Distance covered consuming 1 litre petrol = 31 \u00bc km = 125\/4 km<\/p>\n\n\n\n

So the distance covered consuming 1 3\/5 litre petrol = 125\/4 \u00d7 8\/5<\/p>\n\n\n\n

= 1000\/20<\/p>\n\n\n\n

= 50 km<\/p>\n\n\n\n

13. A rectangular park has length = 23 2\/5 m and breadth = 16 2\/3 m. Find the area of the park.<\/strong><\/p>\n\n\n\n

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

Dimensions of rectangular park are<\/p>\n\n\n\n

Length = 23 2\/5 m = 117\/5 m<\/p>\n\n\n\n

Breadth = 16 2\/3 m = 50\/3 m<\/p>\n\n\n\n

So the area = l \u00d7 b<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 117\/5 \u00d7 50\/3<\/p>\n\n\n\n

= 39 \u00d7 10<\/p>\n\n\n\n

= 390 m2<\/sup><\/p>\n\n\n\n

14. Each of 40 identical boxes weighs 4 4\/5 kg. Find the total weight of all the boxes.<\/strong><\/p>\n\n\n\n

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

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

Weight of one box = 4 4\/5 kg = 24\/5 kg<\/p>\n\n\n\n

So the weight of 40 boxes = 40 \u00d7 24\/5<\/p>\n\n\n\n

= 8 \u00d7 24<\/p>\n\n\n\n

= 192 kg<\/p>\n\n\n\n

15. Out of 24 kg of wheat, 5\/6th<\/sup> of wheat is consumed. Find, how much wheat is still left?<\/strong><\/p>\n\n\n\n

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

Wheat available = 24 kg<\/p>\n\n\n\n

Wheat consumed = 5\/6th<\/sup> of 24 kg<\/p>\n\n\n\n

= 5\/6 \u00d7 24<\/p>\n\n\n\n

= 20kg<\/p>\n\n\n\n

So the remaining wheat = 24 \u2013 20 kg = 4 kg<\/p>\n\n\n\n

16. A rod of length 2 2\/5 metre is divided into five equal parts. Find the length of each part so obtained.<\/strong><\/p>\n\n\n\n

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

Length of rod = 2\/5 m<\/p>\n\n\n\n

It is given that the length of rod should be divided into 5 equal parts<\/p>\n\n\n\n

So the length of each part of rod = 2 2\/5 \u00f7 5<\/p>\n\n\n\n

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

= 12\/5 \u00d7 1\/5<\/p>\n\n\n\n

= 12\/25 m<\/p>\n\n\n\n

17. If A = 3 3\/8 and B = 6 5\/8, find: (i) A \u00f7 B (ii) B \u00f7 A.<\/strong><\/p>\n\n\n\n

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

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

A = 3 3\/8 = 27\/8<\/p>\n\n\n\n

B = 6 5\/8 = 53\/8<\/p>\n\n\n\n

(i) A \u00f7 B<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 27\/8 \u00f7 53\/8<\/p>\n\n\n\n

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

= 27\/8 \u00d7 8\/53<\/p>\n\n\n\n

= 27\/53<\/p>\n\n\n\n

(ii) B \u00f7 A<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 53\/8 \u00f7 27\/8<\/p>\n\n\n\n

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

= 53\/8 \u00d7 8\/27<\/p>\n\n\n\n

= 53\/27<\/p>\n\n\n\n

= 1 26\/27<\/p>\n\n\n\n

18. Cost of 3 5\/7 litres of oil is \u20b9 83 \u00bd. Find the cost of one litre oil.<\/strong><\/p>\n\n\n\n

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

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

Cost of 3 5\/7 litres of oil = \u20b9 83 \u00bd<\/p>\n\n\n\n

So the cost of one litre oil = \u20b9 83 \u00bd \u00f7 3 5\/7<\/p>\n\n\n\n

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

= \u20b9 167\/2 \u00f7 26\/7<\/p>\n\n\n\n

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

= \u20b9 167\/2 \u00d7 7\/26<\/p>\n\n\n\n

= \u20b9 1169\/52<\/p>\n\n\n\n

= \u20b9 22 25\/52<\/p>\n\n\n\n

19. The product of two numbers is 20 5\/7. If one of these numbers is 6 2\/3, find the other.<\/strong><\/p>\n\n\n\n

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

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

Product of two numbers = 20 5\/7 = 145\/7<\/p>\n\n\n\n

One number = 6 2\/3 = 20\/3<\/p>\n\n\n\n

So the other number = 145\/7 \u00f7 20\/3<\/p>\n\n\n\n

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

= 145\/7 \u00d7 3\/20<\/p>\n\n\n\n

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

= 87\/28<\/p>\n\n\n\n

= 3 3\/28<\/p>\n\n\n\n

20. By what number should 5 5\/6 be multiplied to get 3 1\/3?<\/strong><\/p>\n\n\n\n

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

Here the required number = 3 1\/3 \u00f7 5 5\/6<\/p>\n\n\n\n

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

= 10\/3 \u00f7 35\/6<\/p>\n\n\n\n

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

= 10\/3 \u00d7 6\/35<\/p>\n\n\n\n

= 4\/7<\/p>\n\n\n\n

Exercise 3D page: 49<\/h4>\n\n\n\n

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

1. 6 + {4\/3 + (3\/4 \u2013 1\/3)}<\/strong><\/p>\n\n\n\n

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

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

6 + {4\/3 + (3\/4 \u2013 1\/3)}<\/p>\n\n\n\n

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

= 6 + {4\/3 + 3\/4 \u2013 1\/3}<\/p>\n\n\n\n

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

= 6\/1 + 4\/3 + 3\/4 \u2013 1\/3<\/p>\n\n\n\n

Here the LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (72 + 16 + 9 \u2013 4)\/ 12<\/p>\n\n\n\n

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

= (97 \u2013 4)\/ 12<\/p>\n\n\n\n

= 93\/12<\/p>\n\n\n\n

= 31\/4<\/p>\n\n\n\n

= 7 \u00be<\/p>\n\n\n\n

2. 8 \u2013 {3\/2 + (3\/5 \u2013 1\/2)}<\/strong><\/p>\n\n\n\n

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

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

8 \u2013 {3\/2 + (3\/5 \u2013 1\/2)}<\/p>\n\n\n\n

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

= 8 \u2013 {3\/2 + 3\/5 \u2013 1\/2}<\/p>\n\n\n\n

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

= 8\/1 \u2013 3\/2 \u2013 3\/5 + 1\/2<\/p>\n\n\n\n

Here the LCM of 2 and 5 is 10<\/p>\n\n\n\n

= (80 \u2013 15 \u2013 6 + 5)\/ 10<\/p>\n\n\n\n

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

= (85 \u2013 21)\/ 10<\/p>\n\n\n\n

= 64\/10<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

= 32\/5<\/p>\n\n\n\n

= 6 2\/5<\/p>\n\n\n\n

3. 1\/4 (1\/4 + 1\/3) \u2013 2\/5<\/strong><\/p>\n\n\n\n

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

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

1\/4 (1\/4 + 1\/3) \u2013 2\/5<\/p>\n\n\n\n

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

= 1\/4 [(3 + 4)\/ 12] \u2013 2\/5<\/p>\n\n\n\n

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

= 1\/4 \u00d7 7\/12 \u2013 2\/5<\/p>\n\n\n\n

= 7\/48 \u2013 2\/5<\/p>\n\n\n\n

Here the LCM of 48 and 5 is 240<\/p>\n\n\n\n

= (35 \u2013 96)\/ 240<\/p>\n\n\n\n

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

= -61\/240<\/p>\n\n\n\n

4. 2 3\/4 \u2013 [3 1\/8 \u00f7 {5 \u2013 (4 2\/3 \u2013 11\/12)}]<\/strong><\/p>\n\n\n\n

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

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

2 3\/4 \u2013 [3 1\/8 \u00f7 {5 \u2013 (4 2\/3 \u2013 11\/12)}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 (14\/3 \u2013 11\/12)}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 (56 \u2013 11)\/12}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 45\/12}]<\/p>\n\n\n\n

LCM of 12 and 1 is 12<\/p>\n\n\n\n

= 11\/4 \u2013 [25\/8 \u00f7 {(60 \u2013 45)\/12}]<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

= 11\/4 \u2013 [25\/8 \u00f7 15\/12]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00d7 12\/15]<\/p>\n\n\n\n

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

= 11\/4 \u2013 5\/2<\/p>\n\n\n\n

LCM of 4 and 2 is 4<\/p>\n\n\n\n

= (11 \u2013 10)\/ 4<\/p>\n\n\n\n

= 1\/4<\/p>\n\n\n\n

5. 12 1\/2 \u2013 [8 1\/2 + {9 \u2013 (5 \u2013 )<\/strong><\/p>\n\n\n\n

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

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

12 1\/2 \u2013 [8 1\/2 + {9 \u2013 (5 \u2013
\"Selina)<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + {9 \u2013 (5 \u2013 1)}]<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + {9 \u2013 4}]<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + 5]<\/p>\n\n\n\n

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

= 25\/2 \u2013 17\/2 \u2013 5\/1<\/p>\n\n\n\n

LCM of 2 and 1 is 2<\/p>\n\n\n\n

= (25 \u2013 17 \u2013 10)\/ 2<\/p>\n\n\n\n

= (25 \u2013 27)\/ 2<\/p>\n\n\n\n

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

= \u2013 2\/2<\/p>\n\n\n\n

= -1<\/p>\n\n\n\n

6. 1 1\/5 \u00f7 {2 1\/3 \u2013 (5 + )} \u2013 3 \u00bd<\/strong><\/p>\n\n\n\n

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

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

1 1\/5 \u00f7 {2 1\/3 \u2013 (5 +
\"Selina)} \u2013 3 \u00bd<\/p>\n\n\n\n

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

= 6\/5 \u00f7 {7\/3 \u2013 (5 \u2013 1)} \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00f7 {7\/3 \u2013 4} \u2013 7\/2<\/p>\n\n\n\n

LCM of 3 and 1 is 3<\/p>\n\n\n\n

= 6\/5 \u00f7 {(7 \u2013 12)\/ 3} \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00f7 -5\/3 \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00d7 3\/-5 \u2013 7\/2<\/p>\n\n\n\n

= \u2013 18\/25 \u2013 7\/2<\/p>\n\n\n\n

LCM of 25 and 2 is 50<\/p>\n\n\n\n

= (- 36 \u2013 175)\/ 50<\/p>\n\n\n\n

= \u2013 211\/50<\/p>\n\n\n\n

= \u2013 4 11\/50<\/p>\n\n\n\n

7. (1\/2 + 2\/3) \u00f7 (3\/4 \u2013 2\/9)<\/strong><\/p>\n\n\n\n

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

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

(1\/2 + 2\/3) \u00f7 (3\/4 \u2013 2\/9)<\/p>\n\n\n\n

LCM of 2 and 3 is 6 and 4 and 9 is 36<\/p>\n\n\n\n

= (3 + 4)\/ 6 \u00f7 (27 \u2013 8)\/ 36<\/p>\n\n\n\n

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

= 7\/6 \u00f7 19\/36<\/p>\n\n\n\n

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

= 7\/6 \u00d7 36\/19<\/p>\n\n\n\n

= 42\/19<\/p>\n\n\n\n

= 2 4\/19<\/p>\n\n\n\n

8. 6\/5 of (3 1\/3 \u2013 2 1\/2) \u00f7 (2 5\/21 \u2013 2)<\/strong><\/p>\n\n\n\n

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

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

6\/5 of (3 1\/3 \u2013 2 1\/2) \u00f7 (2 5\/21 \u2013 2)<\/p>\n\n\n\n

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

= 6\/5 of (10\/3 \u2013 5\/2) \u00f7 (47\/21 \u2013 2\/1)<\/p>\n\n\n\n

LCM of 3 and 2 is 6 and 1 and 21 is 21<\/p>\n\n\n\n

= 6\/5 of [(20 \u2013 15)\/ 6] \u00f7 [(47 \u2013 42)\/ 21]<\/p>\n\n\n\n

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

= 6\/5 of 5\/6 \u00f7 5\/21<\/p>\n\n\n\n

= 1 \u00f7 5\/21<\/p>\n\n\n\n

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

= 1 \u00d7 21\/5<\/p>\n\n\n\n

= 21\/5<\/p>\n\n\n\n

= 4 1\/5<\/p>\n\n\n\n

9. 10 1\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/strong><\/p>\n\n\n\n

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

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

10 1\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/p>\n\n\n\n

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

= 81\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/p>\n\n\n\n

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

= 81\/10 \u00f7 25\/63<\/p>\n\n\n\n

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

= 81\/10 \u00d7 63\/25<\/p>\n\n\n\n

= 5103\/250<\/p>\n\n\n\n

= 20 103\/250<\/p>\n\n\n\n

10. 5 3\/4 \u2013 3\/7 \u00d7 15 3\/4 + 2 2\/35 \u00f7 1 11\/25<\/strong><\/p>\n\n\n\n

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

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

5 3\/4 \u2013 3\/7 \u00d7 15 3\/4 + 2 2\/35 \u00f7 1 11\/25<\/p>\n\n\n\n

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

= 23\/4 \u2013 3\/7 \u00d7 63\/4 + 72\/35 \u00f7 36\/25<\/p>\n\n\n\n

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

= 23\/4 \u2013 3\/7 \u00d7 63\/4 + 72\/35 \u00d7 25\/36<\/p>\n\n\n\n

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

= 23\/4 \u2013 27\/4 + 10\/7<\/p>\n\n\n\n

LCM of 4 and 7 is 28<\/p>\n\n\n\n

= (161 \u2013 189 + 40)\/ 28<\/p>\n\n\n\n

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

= 3\/7<\/p>\n\n\n\n

11. 3\/4 of 7 3\/7 \u2013 5 3\/5 \u00f7 3 4\/15<\/strong><\/p>\n\n\n\n

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

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

3\/4 of 7 3\/7 \u2013 5 3\/5 \u00f7 3 4\/15<\/p>\n\n\n\n

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

= 3\/4 of 52\/7 \u2013 28\/5 \u00f7 49\/15<\/p>\n\n\n\n

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

= 39\/7 \u2013 28\/5 \u00f7 49\/15<\/p>\n\n\n\n

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

= 39\/7 \u2013 28\/5 \u00d7 15\/49<\/p>\n\n\n\n

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

= 39\/7 \u2013 12\/7<\/p>\n\n\n\n

= (39 \u2013 12)\/ 7<\/p>\n\n\n\n

= 27\/7<\/p>\n\n\n\n

= 3 6\/7<\/p>\n\n\n\n

Exercise 3E page: 52<\/h4>\n\n\n\n

1. A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is:<\/strong><\/p>\n\n\n\n

(i) the length of AB to that of CD?<\/strong><\/p>\n\n\n\n

(ii) \u00bd the length of AB to that of 1\/3 of CD?<\/strong><\/p>\n\n\n\n

(iii) 1\/5 of CD to that of AB?<\/strong><\/p>\n\n\n\n

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

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

Length of AB = 6 cm<\/p>\n\n\n\n

Length of CD = 15 cm<\/p>\n\n\n\n

(i) The length of AB to that of CD<\/p>\n\n\n\n

= 6\/15<\/p>\n\n\n\n

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

(ii) \u00bd the length of AB to that of 1\/3 of CD<\/p>\n\n\n\n

1\/2 of AB = 1\/2 \u00d7 6 = 3 cm<\/p>\n\n\n\n

1\/3 of CD = 1\/3 \u00d7 15 = 5 cm<\/p>\n\n\n\n

So \u00bd the length of AB to that of 1\/3 of CD = 3\/5<\/p>\n\n\n\n

(iii) 1\/5 of CD to that of AB<\/p>\n\n\n\n

1\/5 of CD = 1\/5 \u00d7 15 = 3 cm<\/p>\n\n\n\n

1\/5 of CD to that of AB = 3\/6 = 1\/2<\/p>\n\n\n\n

2. Subtract (2\/7 \u2013 5\/21) from the sum of 3\/4, 5\/7 and 7\/12.<\/strong><\/p>\n\n\n\n

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

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

(3\/4 + 5\/7 + 7\/12) \u2013 (2\/7 \u2013 5\/21)<\/p>\n\n\n\n

LCM of 4, 7 and 12 is 84 and 7 and 21 is 21<\/p>\n\n\n\n

= [(63 + 60 + 49)\/ 84] \u2013 [(6 \u2013 5)\/ 21]<\/p>\n\n\n\n

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

= 172\/84 \u2013 1\/21<\/p>\n\n\n\n

LCM of 21 and 84 is 84<\/p>\n\n\n\n

= (172 \u2013 4)\/ 84<\/p>\n\n\n\n

= 168\/ 84<\/p>\n\n\n\n

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

3. From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5 \u00bc kg, 9 \u00bd kg and 9 \u00be kg respectively.<\/strong><\/p>\n\n\n\n

(i) How many kg did he sell?<\/strong><\/p>\n\n\n\n

(ii) How many kg are still left in the sack?<\/strong><\/p>\n\n\n\n

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

Weight of potato = 120 kg<\/p>\n\n\n\n

(i) Potatoes sold by merchant = 6 kg + 5 \u00bc kg + 9 \u00bd kg + 9 \u00be kg<\/p>\n\n\n\n

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

= (6 + 21\/4 + 19\/2 + 39\/4) kg<\/p>\n\n\n\n

LCM of 1, 4 and 2 is 4<\/p>\n\n\n\n

= (24 + 21 + 38 + 39)\/ 4<\/p>\n\n\n\n

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

= 122\/4<\/p>\n\n\n\n

= 61\/2 kg<\/p>\n\n\n\n

= 30 \u00bd kg<\/p>\n\n\n\n

(ii) Potatoes left in the sack = 120 kg \u2013 30 \u00bd kg<\/p>\n\n\n\n

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

= (120\/ 1 \u2013 61\/2) kg<\/p>\n\n\n\n

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

= (240 \u2013 61)\/ 2<\/p>\n\n\n\n

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

= 179\/2 kg<\/p>\n\n\n\n

= 89 \u00bd kg<\/p>\n\n\n\n

4. If a boy works for six consecutive days for 8 hours, 7 \u00bd hours, 8 \u00bc hours, 6 \u00bc hours, 6 \u00be hours and 7 hours respectively, how much money will he earn at the rate of \u20b9 36 per hour?<\/strong><\/p>\n\n\n\n

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

Hours worked by a boy for six consecutive days = 8 hours + 7 \u00bd hours + 8 \u00bc hours + 6 \u00bc hours + 6 \u00be hours + 7 hours<\/p>\n\n\n\n

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

= (8 + 15\/2 + 33\/4 + 25\/4 + 27\/4 + 7) hours<\/p>\n\n\n\n

LCM of 2 and 4 is 4<\/p>\n\n\n\n

= (32 + 30 + 33 + 25 + 27 + 28)\/ 4 hours<\/p>\n\n\n\n

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

= 175\/4 hours<\/p>\n\n\n\n

= 43 \u00be hours<\/p>\n\n\n\n

He earned \u20b9 36 per hour<\/p>\n\n\n\n

So the total earnings = 175\/4 \u00d7 36<\/p>\n\n\n\n

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

= 175 \u00d7 9<\/p>\n\n\n\n

= \u20b9 1575<\/p>\n\n\n\n

5. A student bought 4 1\/3 m of yellow ribbon, 6 1\/6 m of red ribbon and 3 2\/9 m of blue ribbon decorating a room. How many metres of ribbon did he buy?<\/strong><\/p>\n\n\n\n

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

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

Length of yellow ribbon = 4 1\/3 m = 13\/3 m<\/p>\n\n\n\n

Length of red ribbon = 6 1\/6 m = 37\/6 m<\/p>\n\n\n\n

Length of blue ribbon = 3 2\/9 m = 29\/9 m<\/p>\n\n\n\n

So the total length = 13\/3 + 37\/6 + 29\/9<\/p>\n\n\n\n

LCM of 3, 6 and 9 is 18<\/p>\n\n\n\n

= (78 + 111 + 58)\/ 18<\/p>\n\n\n\n

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

= 247\/ 18<\/p>\n\n\n\n

= 13 13\/18 m<\/p>\n\n\n\n

6. In a business, Ram and Deepak invest 3\/5 and 2\/5 of the total investment. If \u20b9 40, 000 is the total investment, calculate the amount invested by each.<\/strong><\/p>\n\n\n\n

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

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

Total investment = \u20b9 40, 000<\/p>\n\n\n\n

Ram\u2019s investment = 3\/5 of \u20b9 40, 000<\/p>\n\n\n\n

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

= 3\/5 \u00d7 40, 000<\/p>\n\n\n\n

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

= 3 \u00d7 8000<\/p>\n\n\n\n

= \u20b9 24, 000<\/p>\n\n\n\n

Deepak\u2019s investment = 2\/5 of \u20b9 40, 000<\/p>\n\n\n\n

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

= 2\/5 \u00d7 40, 000<\/p>\n\n\n\n

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

= 2 \u00d7 8000<\/p>\n\n\n\n

= \u20b9 16, 000<\/p>\n\n\n\n

7. Geeta had 30 problems for home work. She worked out 2\/3 of them. How many problems were still left to be worked out by her?<\/strong><\/p>\n\n\n\n

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

Number of problems Geeta had for home work = 30<\/p>\n\n\n\n

Number of problems worked out by Geeta = 2\/3 of 30<\/p>\n\n\n\n

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

= 2\/3 \u00d7 30<\/p>\n\n\n\n

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

Number of problems still left to be worked out by her = 30 \u2013 20 = 10<\/p>\n\n\n\n

8. A picture was marked at \u20b9 90. It was sold at 3\/4 of its marked price. What was the sale price?<\/strong><\/p>\n\n\n\n

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

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

Marked price of picture = \u20b9 90<\/p>\n\n\n\n

Sale price of picture = \u00be of \u20b9 90<\/p>\n\n\n\n

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

= \u00be \u00d7 90<\/p>\n\n\n\n

= \u20b9 270\/4<\/p>\n\n\n\n

= \u20b9 67 \u00bd<\/p>\n\n\n\n

= \u20b9 67.50<\/p>\n\n\n\n

9. Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10 \u00bd kg?<\/strong><\/p>\n\n\n\n

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

Number of parcels = 15<\/p>\n\n\n\n

Weight of each parcel = 10 \u00bd kg = 21\/2 kg<\/p>\n\n\n\n

So the total weight of parcels = 15 of 21\/2 kg<\/p>\n\n\n\n

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

= 21\/2 \u00d7 15 kg<\/p>\n\n\n\n

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

= 315\/2<\/p>\n\n\n\n

= 157 \u00bd kg<\/p>\n\n\n\n

= 157.5 kg<\/p>\n\n\n\n

10. A rope is 25 \u00bd m long. How many pieces each of 1 \u00bd m length can be cut out from it?<\/strong><\/p>\n\n\n\n

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

Length of rope = 25 \u00bd m = 51\/2 m<\/p>\n\n\n\n

Length of each piece = 1 \u00bd m = 3\/2 m<\/p>\n\n\n\n

So the number of pieces = 51\/2 \u00f7 3\/2<\/p>\n\n\n\n

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

= 51\/2 \u00d7 2\/3<\/p>\n\n\n\n

= 17 pieces<\/p>\n\n\n\n

11. The heights of two vertical poles, above the earth\u2019s surface, are 14 \u00bc m and 22 1\/3 m respectively. How much higher is the second pole as compared with the height of the first pole?<\/strong><\/p>\n\n\n\n

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

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

Height of first pole above the earth\u2019s surface = 14 \u00bc m<\/p>\n\n\n\n

Height of second pole above the earth\u2019s surface = 22 1\/3 m<\/p>\n\n\n\n

So height of second pole when compared to first pole = 22 1\/3 \u2013 14 1\/4<\/p>\n\n\n\n

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

= 67\/3 \u2013 57\/4<\/p>\n\n\n\n

LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (268 \u2013 171)\/ 12<\/p>\n\n\n\n

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

= 97\/12 m<\/p>\n\n\n\n

= 8 1\/12 m<\/p>\n\n\n\n

12. Vijay weighed 65 \u00bd kg. He gained 1 2\/5 kg during the first week, 1 \u00bc kg during the second week, but lost 5\/16 kg during the third week. What was his weight after the third week?<\/strong><\/p>\n\n\n\n

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

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

Weight of Vijay = 65 \u00bd kg<\/p>\n\n\n\n

Weight gained during first week = 1 2\/5 kg<\/p>\n\n\n\n

Weight gained during second week = 1 \u00bc kg<\/p>\n\n\n\n

Weight lost during third week = 5\/16 kg<\/p>\n\n\n\n

So the weight of Vijay after third week = 65 \u00bd + 1 2\/5 + 1 \u00bc \u2013 5\/16<\/p>\n\n\n\n

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

= 131\/2 + 7\/5 + 5\/4 \u2013 5\/16<\/p>\n\n\n\n

LCM of 2, 5, 4 and 16 is 8-<\/p>\n\n\n\n

= (5240 + 112 + 100 \u2013 25)\/ 80<\/p>\n\n\n\n

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

= (5452 \u2013 25)\/ 80<\/p>\n\n\n\n

= 5427\/80 kg<\/p>\n\n\n\n

= 67 67\/80 kg<\/p>\n\n\n\n

13. A man spends 2\/5 of his salary on food and 3\/10 on house rent, electricity, etc. What fraction of his salary is still left with him?<\/strong><\/p>\n\n\n\n

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

Consider salary of man = \u20b9 1<\/p>\n\n\n\n

Salary spent on food = 2\/5 of \u20b9 1 = \u20b9 2\/5<\/p>\n\n\n\n

Salary spent on house rent = 3\/10 of \u20b9 1 = \u20b9 3\/10<\/p>\n\n\n\n

So the total salary spent = 2\/5 + 3\/10<\/p>\n\n\n\n

LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (4 + 3)\/ 10<\/p>\n\n\n\n

= 7\/10<\/p>\n\n\n\n

Salary still left with him = 1 \u2013 7\/10<\/p>\n\n\n\n

LCM of 1 and 10 is 10<\/p>\n\n\n\n

= (10 \u2013 7)\/ 10<\/p>\n\n\n\n

= 3\/10<\/p>\n\n\n\n

14. A man spends 2\/5 of his salary on food and 3\/10 of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?<\/strong><\/p>\n\n\n\n

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

Consider \u20b9 1 as the total salary<\/p>\n\n\n\n

Salary spent on food = 2\/5 of \u20b9 1 = \u20b9 2\/5<\/p>\n\n\n\n

So the remaining salary = 1 \u2013 2\/5<\/p>\n\n\n\n

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

= (5 \u2013 2)\/ 5<\/p>\n\n\n\n

= \u20b9 3\/5<\/p>\n\n\n\n

Salary spent on house rent = 3\/10 of 3\/5 = \u20b9 9\/50<\/p>\n\n\n\n

So the remaining salary = 3\/5 \u2013 9\/50<\/p>\n\n\n\n

LCM of 5 and 50 is 50<\/p>\n\n\n\n

= (30 \u2013 9)\/ 50<\/p>\n\n\n\n

= \u20b9 21\/50<\/p>\n\n\n\n

15. Shyam bought a refrigerator for \u20b9 5,000. He paid 1\/10 of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month?<\/strong><\/p>\n\n\n\n

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

Total cost of refrigerator = \u20b9 5000<\/p>\n\n\n\n

Cash paid = 1\/10 of \u20b9 5000<\/p>\n\n\n\n

= 1\/10 \u00d7 5000<\/p>\n\n\n\n

= \u20b9 500<\/p>\n\n\n\n

So the balance amount = 5000 \u2013 500 = \u20b9 4500<\/p>\n\n\n\n

Number of instalments = 12<\/p>\n\n\n\n

So the amount to be paid each month = 4500 \u00f7 12<\/p>\n\n\n\n

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

= 4500 \u00d7 1\/12<\/p>\n\n\n\n

= \u20b9 375<\/p>\n\n\n\n

Exercise 3B<\/h3>\n\n\n\n

1. For each pair, given below, state whether it forms like fractions or unlike fractions:<\/strong><\/p>\n\n\n\n

(i) 5\/8 and 7\/8<\/strong><\/p>\n\n\n\n

(ii) 8\/15 and 8\/21<\/strong><\/p>\n\n\n\n

(iii) 4\/9 and 9\/4<\/strong><\/p>\n\n\n\n

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

(i) 5\/8 and 7\/8 are like fractions.<\/p>\n\n\n\n

(ii) 8\/15 and 8\/21 are unlike fractions.<\/p>\n\n\n\n

(iii) 4\/9 and 9\/4 are unlike fractions.<\/p>\n\n\n\n

2. Convert given fractions into fractions with equal denominators:<\/strong><\/p>\n\n\n\n

(i) 5\/6 and 7\/9<\/strong><\/p>\n\n\n\n

(ii) 2\/3, 5\/6 and 7\/12<\/strong><\/p>\n\n\n\n

(iii) 4\/5, 17\/20, 23\/20 and 11\/16<\/strong><\/p>\n\n\n\n

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

(i) 5\/6 and 7\/9<\/p>\n\n\n\n

Here the LCM of 6 and 9 is 18<\/p>\n\n\n\n

5\/6 = (5 \u00d7 3)\/ (6 \u00d7 3) = 15\/18<\/p>\n\n\n\n

7\/9 = (7 \u00d7 2)\/ (9 \u00d7 2) = 14\/18<\/p>\n\n\n\n

Therefore, 15\/18 and 14\/18 are the required fractions.<\/p>\n\n\n\n

(ii) 2\/3, 5\/6 and 7\/12<\/p>\n\n\n\n

Here the LCM of 3, 6 and 12 is 12<\/p>\n\n\n\n

2\/3 = (2 \u00d7 4)\/ (3 \u00d7 4) = 8\/12<\/p>\n\n\n\n

5\/6 = (5 \u00d7 2)\/ (6 \u00d7 2) = 10\/12<\/p>\n\n\n\n

7\/12 = 7\/12<\/p>\n\n\n\n

Therefore, 8\/12, 10\/12 and 7\/12 are the required fractions.<\/p>\n\n\n\n

(iii) 4\/5, 17\/20, 23\/40 and 11\/16<\/p>\n\n\n\n

Here the LCM of 5, 20, 40 and 16 is 80<\/p>\n\n\n\n

4\/5 = (4 \u00d7 16)\/ (5 \u00d7 16) = 64\/80<\/p>\n\n\n\n

17\/20 = (17 \u00d7 4)\/ (20 \u00d7 4) = 68\/80<\/p>\n\n\n\n

23\/40 = (23 \u00d7 2)\/ (40 \u00d7 2) = 46\/80<\/p>\n\n\n\n

11\/16 = (11 \u00d7 5)\/ (16 \u00d7 5) = 55\/80<\/p>\n\n\n\n

Therefore, 64\/80, 68\/80, 46\/80 and 55\/80 are the required fractions.<\/p>\n\n\n\n

3. Convert given fractions into fractions with equal numerators:<\/strong><\/p>\n\n\n\n

(i) 8\/9 and 12\/17<\/strong><\/p>\n\n\n\n

(ii) 6\/13, 15\/23 and 12\/17<\/strong><\/p>\n\n\n\n

(iii) 15\/19, 25\/28, 9\/11 and 45\/47<\/strong><\/p>\n\n\n\n

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

(i) 8\/9 and 12\/17<\/p>\n\n\n\n

Here the LCM of 8 and 12 is 24<\/p>\n\n\n\n

8\/9 = (8 \u00d7 3)\/ (9 \u00d7 3) = 24\/27<\/p>\n\n\n\n

12\/17 = (12 \u00d7 2)\/ (17 \u00d7 2) = 24\/34<\/p>\n\n\n\n

Therfore, 24\/27 and 24\/34 are the required fractions.<\/p>\n\n\n\n

(ii) 6\/13, 15\/23 and 12\/17<\/p>\n\n\n\n

Here the LCM of 6, 15 and 12 is 60<\/p>\n\n\n\n

6\/13 = (6 \u00d7 10)\/ (13 \u00d7 10) = 60\/130<\/p>\n\n\n\n

15\/23 = (15 \u00d7 4)\/ (23 \u00d7 4) = 60\/92<\/p>\n\n\n\n

12\/17 = (12 \u00d7 5)\/ (17 \u00d7 5) = 60\/85<\/p>\n\n\n\n

Therefore, 60\/130, 60\/92 and 60\/85 are the required fractions.<\/p>\n\n\n\n

(iii) 15\/19, 25\/28, 9\/11 and 45\/47<\/p>\n\n\n\n

Here the LCM of 15, 25, 9 and 45 is 225<\/p>\n\n\n\n

15\/19 = (15 \u00d7 15)\/ (19 \u00d7 15) = 225\/285<\/p>\n\n\n\n

25\/28 = (25 \u00d7 9)\/ (28 \u00d7 9) = 225\/252<\/p>\n\n\n\n

9\/11 = (9 \u00d7 25)\/ (11 \u00d7 25) = 225\/275<\/p>\n\n\n\n

45\/47 = (45 \u00d7 5)\/ (47 \u00d7 5) = 225\/ 235<\/p>\n\n\n\n

Therefore, 225\/285, 225\/252, 225\/275 and 225\/235 are the required fractions.<\/p>\n\n\n\n

4. Put the given fractions in ascending order by making denominators equal:<\/strong><\/p>\n\n\n\n

(i) 1\/3, 2\/5, 3\/4 and 1\/6<\/strong><\/p>\n\n\n\n

(ii) 5\/6, 7\/8, 11\/12 and 3\/10<\/strong><\/p>\n\n\n\n

(iii) 5\/7, 3\/8, 9\/14 and 20\/21<\/strong><\/p>\n\n\n\n

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

(i) 1\/3, 2\/5, 3\/4 and 1\/6<\/p>\n\n\n\n

Here the LCM of 3, 5, 4 and 6 is 60<\/p>\n\n\n\n

1\/3 = (1 \u00d7 20)\/ (3 \u00d7 20) = 20\/60<\/p>\n\n\n\n

2\/5 = (2 \u00d7 12)\/ (5 \u00d7 12) = 24\/60<\/p>\n\n\n\n

3\/4 = (3 \u00d7 15)\/ (4 \u00d7 15) = 45\/60<\/p>\n\n\n\n

1\/6 = (1 \u00d7 10)\/ (6 \u00d7 10) = 10\/60<\/p>\n\n\n\n

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

10\/60 < 20\/60 < 24\/60 < 45\/60<\/p>\n\n\n\n

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

1\/6 < 1\/3 < 2\/5 < 3\/4<\/p>\n\n\n\n

Therefore, 1\/6, 1\/3, 2\/5 and 3\/4 are in ascending order.<\/p>\n\n\n\n

(ii) 5\/6, 7\/8, 11\/12 and 3\/10<\/p>\n\n\n\n

Here the LCM of 6, 8, 12 and 10 is 240<\/p>\n\n\n\n

5\/6 = (5 \u00d7 40)\/ (6 \u00d7 40) = 200\/240<\/p>\n\n\n\n

7\/8 = (7 \u00d7 30)\/ (8 \u00d7 30) = 210\/240<\/p>\n\n\n\n

11\/12 = (11 \u00d7 20)\/ (12 \u00d7 20) = 220\/240<\/p>\n\n\n\n

3\/10 = (3 \u00d7 24)\/ (10 \u00d7 24) = 72\/240<\/p>\n\n\n\n

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

72\/240 < 200\/240 < 210\/240 < 220\/240<\/p>\n\n\n\n

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

3\/10 < 5\/6 < 7\/8 < 11\/12<\/p>\n\n\n\n

Therefore, 3\/10, 5\/6, 7\/8 and 11\/12 are in ascending order.<\/p>\n\n\n\n

(iii) 5\/7, 3\/8, 9\/14 and 20\/21<\/p>\n\n\n\n

Here the LCM of 7, 8, 14 and 21 is 168<\/p>\n\n\n\n

5\/7 = (5 \u00d7 24)\/ (7 \u00d7 24) = 120\/168<\/p>\n\n\n\n

3\/8 = (3 \u00d7 21)\/ (8 \u00d7 21) = 63\/168<\/p>\n\n\n\n

9\/14 = (9 \u00d7 12)\/ (14 \u00d7 12) = 108\/168<\/p>\n\n\n\n

20\/21 = (20 \u00d7 8)\/ (21 \u00d7 8) = 160\/168<\/p>\n\n\n\n

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

63\/168 < 108\/ 168 < 120\/168 < 160\/168<\/p>\n\n\n\n

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

3\/8 < 9\/14 < 5\/7 < 20\/21<\/p>\n\n\n\n

Therefore, 3\/8, 9\/14, 5\/7 and 20\/21 are in ascending order.<\/p>\n\n\n\n

5. Arrange the given fractions in descending order by making numerators equal:<\/strong><\/p>\n\n\n\n

(i) 5\/6, 4\/15, 8\/9 and 1\/3<\/strong><\/p>\n\n\n\n

(ii) 3\/7, 4\/9, 5\/7 and 8\/11<\/strong><\/p>\n\n\n\n

(iii) 1\/10, 6\/11, 8\/11 and 3\/5<\/strong><\/p>\n\n\n\n

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

(i) 5\/6, 4\/15, 8\/9 and 1\/3<\/p>\n\n\n\n

Here the LCM of 5, 4, 8 and 1 is 40<\/p>\n\n\n\n

5\/6 = (5 \u00d7 8)\/ (6 \u00d7 8) = 40\/48<\/p>\n\n\n\n

4\/15 = (4 \u00d7 10)\/ (15 \u00d7 10) = 40\/150<\/p>\n\n\n\n

8\/9 = (8 \u00d7 5)\/ (9 \u00d7 5) = 40\/45<\/p>\n\n\n\n

1\/3 = (1 \u00d7 40)\/ (3 \u00d7 40) = 40\/120<\/p>\n\n\n\n

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

40\/45 > 40\/48 > 40\/120 > 40\/150<\/p>\n\n\n\n

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

8\/9 > 5\/6 > 1\/3 > 4\/15<\/p>\n\n\n\n

Therefore, 8\/9, 5\/6, 1\/3 and 4\/15 are in descending order.<\/p>\n\n\n\n

(ii) 3\/7, 4\/9, 5\/7 and 8\/11<\/p>\n\n\n\n

Here the LCM of 3, 4, 5 and 8 is 120<\/p>\n\n\n\n

3\/7 = (3 \u00d7 40)\/ (7 \u00d7 40) = 120\/280<\/p>\n\n\n\n

4\/9 = (4 \u00d7 30)\/ (9 \u00d7 30) = 120\/270<\/p>\n\n\n\n

5\/7 = (5 \u00d7 24)\/ (7 \u00d7 24) = 120\/168<\/p>\n\n\n\n

8\/11 = (8 \u00d7 15)\/ (11 \u00d7 15) = 120\/165<\/p>\n\n\n\n

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

120\/165 > 120\/168 > 120\/270 > 120\/280<\/p>\n\n\n\n

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

8\/11 > 5\/7 > 4\/9 > 3\/7<\/p>\n\n\n\n

Therefore, 8\/11, 5\/7, 4\/9 and 3\/7 are in descending order.<\/p>\n\n\n\n

(iii) 1\/10, 6\/11, 8\/11 and 3\/5<\/p>\n\n\n\n

Here the LCM of 1, 6, 8 and 3 is 24<\/p>\n\n\n\n

1\/10 = (1 \u00d7 24)\/ (10 \u00d7 24) = 24\/240<\/p>\n\n\n\n

6\/11 = (6 \u00d7 4)\/ (11 \u00d7 4) = 24\/44<\/p>\n\n\n\n

8\/11 = (8 \u00d7 3)\/ (11 \u00d7 3) = 24\/33<\/p>\n\n\n\n

3\/5 = (3 \u00d7 8)\/ (5 \u00d7 8) = 24\/40<\/p>\n\n\n\n

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

24\/33 > 24\/40 > 24\/44 > 24\/240<\/p>\n\n\n\n

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

8\/11 > 3\/5 > 6\/11 > 1\/10<\/p>\n\n\n\n

Therefore, 8\/11, 3\/5, 6\/11 and 1\/10 are in descending order.<\/p>\n\n\n\n

6. Find the greater fraction:<\/strong><\/p>\n\n\n\n

(i) 3\/5 and 11\/15<\/strong><\/p>\n\n\n\n

(ii) 4\/5 and 3\/10<\/strong><\/p>\n\n\n\n

(iii) 6\/7 and 5\/9<\/strong><\/p>\n\n\n\n

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

(i) 3\/5 and 11\/15<\/p>\n\n\n\n

Here the LCM of 5 and 15 is 15<\/p>\n\n\n\n

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

11\/15 = 11\/15<\/p>\n\n\n\n

So we get, 11\/15 > 9\/15<\/p>\n\n\n\n

Therefore, 11\/15 is greater.<\/p>\n\n\n\n

(ii) 4\/5 and 3\/10<\/p>\n\n\n\n

Here the LCM of 5 and 10 is 10<\/p>\n\n\n\n

4\/5 = (4 \u00d7 2)\/ (5 \u00d7 2) = 8\/10<\/p>\n\n\n\n

3\/10 = 3\/10<\/p>\n\n\n\n

So we get, 8\/10 > 3\/10<\/p>\n\n\n\n

4\/5 > 3\/10<\/p>\n\n\n\n

Therefore, 4\/5 is greater.<\/p>\n\n\n\n

(iii) 6\/7 and 5\/9<\/p>\n\n\n\n

Here LCM of 7 and 9 is 63<\/p>\n\n\n\n

6\/7 = (6 \u00d7 9)\/ (7 \u00d7 9) = 54\/63<\/p>\n\n\n\n

5\/9 = (5 \u00d7 7)\/ (9 \u00d7 7) = 35\/63<\/p>\n\n\n\n

So we get, 54\/63 > 35\/63<\/p>\n\n\n\n

6\/7 > 35\/63<\/p>\n\n\n\n

Therefore, 6\/7 is greater.<\/p>\n\n\n\n

7. Insert one fraction between:
(i) 3\/7 and 4\/9<\/strong><\/p>\n\n\n\n

(ii) 2 and 8\/3<\/strong><\/p>\n\n\n\n

(iii) 9\/17 and 6\/13<\/strong><\/p>\n\n\n\n

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

(i) 3\/7 and 4\/9<\/p>\n\n\n\n

So the fraction between 3\/7 and 4\/9<\/p>\n\n\n\n

= (3 + 4)\/ (7 + 9)<\/p>\n\n\n\n

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

(ii) 2 and 8\/3<\/p>\n\n\n\n

So the fraction between 2 and 8\/3<\/p>\n\n\n\n

= (2 + 8)\/ (1 + 3)<\/p>\n\n\n\n

= 10\/4<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

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

= 2 \u00bd<\/p>\n\n\n\n

(iii) 9\/17 and 6\/13<\/p>\n\n\n\n

So the fraction between 9\/17 and 6\/13<\/p>\n\n\n\n

= (9 + 6)\/ (17 + 13)<\/p>\n\n\n\n

= 15\/30<\/p>\n\n\n\n

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

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

8. Insert three fractions between:<\/strong><\/p>\n\n\n\n

(i) 2\/5 and 4\/9<\/strong><\/p>\n\n\n\n

(ii) 1\/2 and 5\/7<\/strong><\/p>\n\n\n\n

(iii) 3\/8 and 6\/11<\/strong><\/p>\n\n\n\n

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

(i) 2\/5 and 4\/9<\/p>\n\n\n\n

So the fraction between 2\/5 and 4\/9<\/p>\n\n\n\n

= (2 + 4)\/ (5 + 9)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/14<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

= 3\/7<\/p>\n\n\n\n

Fraction between 2\/5 and 3\/7<\/p>\n\n\n\n

= (2 + 3)\/ (5 + 7)<\/p>\n\n\n\n

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

Fraction between 3\/7 and 4\/9<\/p>\n\n\n\n

= (3 + 4)\/ (7 + 9)<\/p>\n\n\n\n

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

Therefore, three fractions between 2\/5 and 4\/9 will be 5\/12, 3\/7 and 7\/16.<\/p>\n\n\n\n

(ii) 1\/2 and 5\/7<\/p>\n\n\n\n

So the fraction between 1\/2 and 5\/7<\/p>\n\n\n\n

= (1 + 5)\/ (2 + 7)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/ 9<\/p>\n\n\n\n

Dividing by 3<\/p>\n\n\n\n

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

Fraction between 1\/2 and 2\/3<\/p>\n\n\n\n

= (1 + 2)\/ (2 + 3)<\/p>\n\n\n\n

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

Fraction between 2\/3 and 5\/7<\/p>\n\n\n\n

= (2 + 5)\/ (3 + 7)<\/p>\n\n\n\n

= 7\/ 10<\/p>\n\n\n\n

Therefore, three fractions between 1\/2 and 5\/7 will be 3\/5, 2\/3 and 7\/10.<\/p>\n\n\n\n

(iii) 3\/8 and 6\/11<\/p>\n\n\n\n

So the fraction between 3\/8 and 6\/11<\/p>\n\n\n\n

= (3 + 6)\/ (8 + 11)<\/p>\n\n\n\n

= 9\/ 19<\/p>\n\n\n\n

Fraction between 3\/8 and 9\/19<\/p>\n\n\n\n

= (3 + 9)\/ (8 + 19)<\/p>\n\n\n\n

= 12\/27<\/p>\n\n\n\n

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

Fraction between 9\/19 and 6\/11<\/p>\n\n\n\n

= (9 + 6)\/ (19 + 11)<\/p>\n\n\n\n

= 15\/30<\/p>\n\n\n\n

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

Therefore, three fractions between 3\/8 and 6\/11 will be 4\/9, 9\/19 and 1\/2.<\/p>\n\n\n\n

9. Insert two fractions between:<\/strong><\/p>\n\n\n\n

(i) 1 and 3\/11<\/strong><\/p>\n\n\n\n

(ii) 5\/9 and 1\/4<\/strong><\/p>\n\n\n\n

(iii) 5\/6 and 1 1\/5<\/strong><\/p>\n\n\n\n

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

(i) 1 and 3\/11<\/p>\n\n\n\n

Fraction between 1 and 3\/11<\/p>\n\n\n\n

= (1 + 3)\/ (1 + 11)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 4\/12<\/p>\n\n\n\n

= 1\/3<\/p>\n\n\n\n

Fraction between 1\/3 and 3\/11<\/p>\n\n\n\n

= (1 + 3)\/ (3 + 11)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 4\/14<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

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

Therefore, two fractions between 1 and 3\/11 will be 1\/3 and 2\/7.<\/p>\n\n\n\n

(ii) 5\/9 and 1\/4<\/p>\n\n\n\n

Fraction between 5\/9 and 1\/4<\/p>\n\n\n\n

= (5 + 1)\/ (9 + 4)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 6\/13<\/p>\n\n\n\n

Fraction between 6\/13 and 1\/4<\/p>\n\n\n\n

= (6 + 1)\/ (13 + 4)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 7\/17<\/p>\n\n\n\n

Therefore, two fractions between 5\/9 and 1\/4 will be 6\/13 and 7\/17.<\/p>\n\n\n\n

(iii) 5\/6 and 1 1\/5<\/p>\n\n\n\n

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

= 5\/6 + 6\/5<\/p>\n\n\n\n

Fraction between 5\/6 and 6\/5<\/p>\n\n\n\n

= (5 + 6)\/ (6 + 5)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

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

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

Fraction between 1 and 6\/5<\/p>\n\n\n\n

= (1 + 6)\/ (1 + 5)<\/p>\n\n\n\n

By addition<\/p>\n\n\n\n

= 7\/6<\/p>\n\n\n\n

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

Therefore, two fractions between 5\/6 and 1 1\/5 will be 1 and 1 1\/6.<\/p>\n\n\n\n

Exercise 3C<\/h3>\n\n\n\n

1. Reduce to a single fraction:<\/strong><\/p>\n\n\n\n

(i) 1\/2 + 2\/3<\/strong><\/p>\n\n\n\n

(ii) 3\/5 \u2013 1\/10<\/strong><\/p>\n\n\n\n

(iii) 2\/3 \u2013 1\/6<\/strong><\/p>\n\n\n\n

(iv) 1 1\/3 + 2 1\/4<\/strong><\/p>\n\n\n\n

(v) 1\/4 + 5\/6 \u2013 1\/12<\/strong><\/p>\n\n\n\n

(vi) 2\/3 \u2013 3\/5 + 3 \u2013 1\/5<\/strong><\/p>\n\n\n\n

(vii) 2\/3 \u2013 1\/5 + 1\/10<\/strong><\/p>\n\n\n\n

(viii) 2 1\/2 + 2 1\/3 \u2013 1 1\/4<\/strong><\/p>\n\n\n\n

(ix) 2 5\/8 \u2013 2 1\/6 + 4 3\/4<\/strong><\/p>\n\n\n\n

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

(i) 1\/2 + 2\/3<\/p>\n\n\n\n

Here the LCM of 2 and 3 is 6<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (2 \u00d7 3) + (2 \u00d7 2)\/ (3 \u00d7 2)<\/p>\n\n\n\n

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

= 3\/6 + 4\/6<\/p>\n\n\n\n

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

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

= 7\/6<\/p>\n\n\n\n

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

(ii) 3\/5 \u2013 1\/10<\/p>\n\n\n\n

Here the LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (3 \u00d7 2)\/ (5 \u00d7 2) \u2013 1\/10<\/p>\n\n\n\n

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

= 6\/10 \u2013 1\/10<\/p>\n\n\n\n

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

= (6 \u2013 1)\/ 10<\/p>\n\n\n\n

= 5\/10<\/p>\n\n\n\n

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

(iii) 2\/3 \u2013 1\/6<\/p>\n\n\n\n

Here the LCM of 3 and 6 is 6<\/p>\n\n\n\n

= (2 \u00d7 2)\/ (3 \u00d7 2) \u2013 1\/6<\/p>\n\n\n\n

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

= 4\/6 \u2013 1\/6<\/p>\n\n\n\n

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

= (4 \u2013 1)\/ 6<\/p>\n\n\n\n

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

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

(iv) 1 1\/3 + 2 1\/4<\/p>\n\n\n\n

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

= 4\/3 + 9\/4<\/p>\n\n\n\n

Here LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (4 \u00d7 4)\/ (3 \u00d7 4) + (9 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

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

= 16\/12 + 27\/12<\/p>\n\n\n\n

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

= (16 + 27)\/ 12<\/p>\n\n\n\n

= 43\/12<\/p>\n\n\n\n

= 3 7\/12<\/p>\n\n\n\n

(v) 1\/4 + 5\/6 \u2013 1\/12<\/p>\n\n\n\n

Here LCM of 4, 6 and 12 is 12<\/p>\n\n\n\n

= (1 \u00d7 3)\/ (4 \u00d7 3) + (5 \u00d7 2)\/ (6 \u00d7 2) \u2013 1\/12<\/p>\n\n\n\n

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

= 3\/12 + 10\/12 \u2013 1\/12<\/p>\n\n\n\n

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

= (3 + 10 \u2013 1)\/ 12<\/p>\n\n\n\n

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

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

(vi) 2\/3 \u2013 3\/5 + 3 \u2013 1\/5<\/p>\n\n\n\n

Here LCM of 3 and 5 is 15<\/p>\n\n\n\n

= (2 \u00d7 5)\/ (3 \u00d7 5) \u2013 (3 \u00d7 3)\/ (5 \u00d7 3) + (3 \u00d7 15)\/ 15 \u2013 (1 \u00d7 3)\/ (5 \u00d7 3)<\/p>\n\n\n\n

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

= 10\/15 \u2013 9\/15 + 45\/15 \u2013 3\/15<\/p>\n\n\n\n

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

= (10 \u2013 9 + 45 \u2013 3)\/ 15<\/p>\n\n\n\n

= (55 \u2013 12)\/ 15<\/p>\n\n\n\n

= 43\/15<\/p>\n\n\n\n

= 2 13\/15<\/p>\n\n\n\n

(vii) 2\/3 \u2013 1\/5 + 1\/10<\/p>\n\n\n\n

Here the LCM of 3, 5 and 10 is 30<\/p>\n\n\n\n

= (2 \u00d7 10)\/ (3 \u00d7 10) \u2013 (1 \u00d7 6)\/ (5 \u00d7 6) + (1 \u00d7 3)\/ (10 \u00d7 3)<\/p>\n\n\n\n

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

= 20\/30 \u2013 6\/30 + 3\/30<\/p>\n\n\n\n

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

= (20 \u2013 6 + 3)\/ 30<\/p>\n\n\n\n

= (23 \u2013 6)\/ 30<\/p>\n\n\n\n

= 17\/30<\/p>\n\n\n\n

(viii) 2 1\/2 + 2 1\/3 \u2013 1 \u00bc<\/p>\n\n\n\n

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

= 5\/2 + 7\/3 \u2013 5\/4<\/p>\n\n\n\n

Here the LCM of 2, 3 and 4 is 12<\/p>\n\n\n\n

= (5 \u00d7 6)\/ (2 \u00d7 6) + (7 \u00d7 4)\/ (3 \u00d7 4) \u2013 (5 \u00d7 3)\/ (4 \u00d7 3)<\/p>\n\n\n\n

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

= 30\/12 + 28\/12 \u2013 15\/12<\/p>\n\n\n\n

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

= (30 + 28 \u2013 15)\/ 12<\/p>\n\n\n\n

= (58 \u2013 15)\/ 12<\/p>\n\n\n\n

= 43\/12<\/p>\n\n\n\n

= 3 7\/12<\/p>\n\n\n\n

(ix) 2 5\/8 \u2013 2 1\/6 + 4 3\/4<\/p>\n\n\n\n

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

= 21\/8 \u2013 13\/6 + 19\/4<\/p>\n\n\n\n

Here the LCM of 8, 6 and 4 is 24<\/p>\n\n\n\n

= (21 \u00d7 3)\/ (8 \u00d7 3) \u2013 (13 \u00d7 4)\/ (6 \u00d7 4) + (19 \u00d7 6)\/ (4 \u00d7 6)<\/p>\n\n\n\n

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

= 63\/24 \u2013 52\/24 + 114\/24<\/p>\n\n\n\n

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

= (63 \u2013 52 + 114)\/ 24<\/p>\n\n\n\n

= (177 \u2013 52)\/ 24<\/p>\n\n\n\n

= 125\/24<\/p>\n\n\n\n

= 5 5\/24<\/p>\n\n\n\n

2. Simplify:<\/strong><\/p>\n\n\n\n

(i) 3\/4 \u00d7 6<\/strong><\/p>\n\n\n\n

(ii) 2\/3 \u00d7 15<\/strong><\/p>\n\n\n\n

(iii) 3\/4 \u00d7 1\/2<\/strong><\/p>\n\n\n\n

(iv) 9\/12 \u00d7 4\/7<\/strong><\/p>\n\n\n\n

(v) 45 \u00d7 2 1\/3<\/strong><\/p>\n\n\n\n

(vi) 36 \u00d7 3 1\/4<\/strong><\/p>\n\n\n\n

(vii) 2 \u00f7 1\/3<\/strong><\/p>\n\n\n\n

(viii) 3 \u00f7 2\/5<\/strong><\/p>\n\n\n\n

(ix) 1 \u00f7 3\/5<\/strong><\/p>\n\n\n\n

(x) 1\/3 \u00f7 1\/4<\/strong><\/p>\n\n\n\n

(xi) -5\/8 \u00f7 3\/4<\/strong><\/p>\n\n\n\n

(xii) 3 3\/7 \u00f7 1 1\/14<\/strong><\/p>\n\n\n\n

(xiii) 3 \u00be \u00d7 1 1\/5 \u00d7 20\/21<\/strong><\/p>\n\n\n\n

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

(i) 3\/4 \u00d7 6<\/p>\n\n\n\n

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

= 3\/4 \u00d7 6\/1<\/p>\n\n\n\n

= (3 \u00d7 6)\/ (4 \u00d7 1)<\/p>\n\n\n\n

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

= 18\/4<\/p>\n\n\n\n

HCF of 18 and 4 is 2<\/p>\n\n\n\n

Dividing both numerator and denominator by 2<\/p>\n\n\n\n

= (18 \u00f7 2)\/ (4 \u00f7 2)<\/p>\n\n\n\n

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

= 4 1\/2<\/p>\n\n\n\n

(ii) 2\/3 \u00d7 15<\/p>\n\n\n\n

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

= 2\/3 \u00d7 15\/1<\/p>\n\n\n\n

= (2 \u00d7 15)\/ (3 \u00d7 1)<\/p>\n\n\n\n

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

= 30\/3<\/p>\n\n\n\n

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

(iii) 3\/4 \u00d7 1\/2<\/p>\n\n\n\n

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

= (3 \u00d7 1)\/ (4 \u00d7 2)<\/p>\n\n\n\n

= 3\/8<\/p>\n\n\n\n

(iv) 9\/12 \u00d7 4\/7<\/p>\n\n\n\n

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

= (9 \u00d7 4)\/ (12 \u00d7 7)<\/p>\n\n\n\n

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

= 36\/84<\/p>\n\n\n\n

Here the HCF of 36 and 84 is 12<\/p>\n\n\n\n

Dividing both numerator and denominator by 12<\/p>\n\n\n\n

= (36 \u00f7 12)\/ (84 \u00f7 12)<\/p>\n\n\n\n

= 3\/7<\/p>\n\n\n\n

(v) 45 \u00d7 2 1\/3<\/p>\n\n\n\n

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

= 45\/1 \u00d7 7\/3<\/p>\n\n\n\n

= (45 \u00d7 7)\/ (1 \u00d7 3)<\/p>\n\n\n\n

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

= 315\/3<\/p>\n\n\n\n

= 105<\/p>\n\n\n\n

(vi) 36 \u00d7 3 1\/4<\/p>\n\n\n\n

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

= 36\/1 \u00d7 13\/4<\/p>\n\n\n\n

= (36 \u00d7 13)\/ (4 \u00d7 1)<\/p>\n\n\n\n

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

= 468\/4<\/p>\n\n\n\n

= 117<\/p>\n\n\n\n

(vii) 2 \u00f7 1\/3<\/p>\n\n\n\n

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

= 2\/1 \u00d7 3\/1<\/p>\n\n\n\n

= (2 \u00d7 3)\/ (1 \u00d7 1)<\/p>\n\n\n\n

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

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

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

(viii) 3 \u00f7 2\/5<\/p>\n\n\n\n

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

= 3\/1 \u00d7 5\/2<\/p>\n\n\n\n

= (3 \u00d7 5)\/ (1 \u00d7 2)<\/p>\n\n\n\n

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

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

= 7 \u00bd<\/p>\n\n\n\n

(ix) 1 \u00f7 3\/5<\/p>\n\n\n\n

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

= 1 \u00d7 5\/3<\/p>\n\n\n\n

= (1 \u00d7 5)\/ 3<\/p>\n\n\n\n

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

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

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

(x) 1\/3 \u00f7 1\/4<\/p>\n\n\n\n

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

= 1\/3 \u00d7 4\/1<\/p>\n\n\n\n

= (1 \u00d7 4)\/ (3 \u00d7 1)<\/p>\n\n\n\n

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

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

= 1 1\/3<\/p>\n\n\n\n

(xi) -5\/8 \u00f7 3\/4<\/p>\n\n\n\n

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

= -5\/8 \u00d7 4\/3<\/p>\n\n\n\n

= (-5 \u00d7 4)\/ (8 \u00d7 3)<\/p>\n\n\n\n

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

= \u2013 20\/24<\/p>\n\n\n\n

Here the HCF of 20 and 24 is 4<\/p>\n\n\n\n

So by dividing both numerator and denominator by 4<\/p>\n\n\n\n

= (-20 \u00f7 4)\/ (24 \u00f7 4)<\/p>\n\n\n\n

= -5\/6<\/p>\n\n\n\n

(xii) 3 3\/7 \u00f7 1 1\/14<\/p>\n\n\n\n

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

= 24\/7 \u00d7 14\/15<\/p>\n\n\n\n

= (24 \u00d7 14)\/ (7 \u00d7 15)<\/p>\n\n\n\n

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

= 336\/105<\/p>\n\n\n\n

Here the HCF of 336 and 105 is 21<\/p>\n\n\n\n

So by dividing both numerator and denominator by 21<\/p>\n\n\n\n

= (336 \u00f7 21)\/ (105 \u00f7 21)<\/p>\n\n\n\n

= 16\/5<\/p>\n\n\n\n

= 3 1\/5<\/p>\n\n\n\n

(xiii) 3 \u00be \u00d7 1 1\/5 \u00d7 20\/21<\/p>\n\n\n\n

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

= 15\/4 \u00d7 6\/5 \u00d7 20\/21<\/p>\n\n\n\n

= (15 \u00d7 6 \u00d7 20)\/ (4 \u00d7 5 \u00d7 21)<\/p>\n\n\n\n

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

= 1800\/420<\/p>\n\n\n\n

Here the HCF of 1800 and 420 is 60<\/p>\n\n\n\n

So by dividing both numerator and denominator by 60<\/p>\n\n\n\n

= (1800 \u00f7 60)\/ (420 \u00f7 60)<\/p>\n\n\n\n

= 30\/7<\/p>\n\n\n\n

= 4 2\/7<\/p>\n\n\n\n

3. Subtract:<\/strong><\/p>\n\n\n\n

(i) 2 from 2\/3<\/strong><\/p>\n\n\n\n

(ii) 1\/8 from 5\/8<\/strong><\/p>\n\n\n\n

(iii) -2\/5 and 2\/5<\/strong><\/p>\n\n\n\n

(iv) \u2013 3\/7 from 3\/7<\/strong><\/p>\n\n\n\n

(v) 0 from \u2013 4\/5<\/strong><\/p>\n\n\n\n

(vi) 2\/9 from 4\/5<\/strong><\/p>\n\n\n\n

(vii) \u2013 4\/7 from -6\/11<\/strong><\/p>\n\n\n\n

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

(i) 2 from 2\/3<\/p>\n\n\n\n

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

= 2\/3 \u2013 2\/1<\/p>\n\n\n\n

LCM of 3 and 1 is 3<\/p>\n\n\n\n

= 2\/3 \u2013 (2 \u00d7 3)\/ 3<\/p>\n\n\n\n

= 2\/3 \u2013 6\/3<\/p>\n\n\n\n

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

= (2 \u2013 6)\/ 3<\/p>\n\n\n\n

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

= \u2013 1 1\/3<\/p>\n\n\n\n

(ii) 1\/8 from 5\/8<\/p>\n\n\n\n

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

= 5\/8 \u2013 1\/8<\/p>\n\n\n\n

= (5 \u2013 1)\/ 8<\/p>\n\n\n\n

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

= 4\/8<\/p>\n\n\n\n

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

(iii) -2\/5 and 2\/5<\/p>\n\n\n\n

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

= 2\/5 \u2013 (-2\/5)<\/p>\n\n\n\n

= 2\/5 + 2\/5<\/p>\n\n\n\n

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

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

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

(iv) \u2013 3\/7 from 3\/7<\/p>\n\n\n\n

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

= 3\/7 \u2013 (-3\/7)<\/p>\n\n\n\n

= 3\/7 + 3\/7<\/p>\n\n\n\n

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

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

= 6\/7<\/p>\n\n\n\n

(v) 0 from \u2013 4\/5<\/p>\n\n\n\n

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

= -4\/5 \u2013 0<\/p>\n\n\n\n

= \u2013 4\/5<\/p>\n\n\n\n

(vi) 2\/9 from 4\/5<\/p>\n\n\n\n

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

= 4\/5 \u2013 2\/9<\/p>\n\n\n\n

LCM of 5 and 9 is 45<\/p>\n\n\n\n

= (4 \u00d7 9)\/ (5 \u00d7 9) \u2013 (2 \u00d7 5)\/ (9 \u00d7 5)<\/p>\n\n\n\n

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

= 36\/45 \u2013 10\/45<\/p>\n\n\n\n

= (36 \u2013 10)\/ 45<\/p>\n\n\n\n

= 26\/45<\/p>\n\n\n\n

(vii) \u2013 4\/7 from -6\/11<\/p>\n\n\n\n

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

= -6\/11 \u2013 (-4\/7)<\/p>\n\n\n\n

= \u2013 6\/11 + 4\/7<\/p>\n\n\n\n

Here LCM of 7 and 11 is 77<\/p>\n\n\n\n

= (- 6 \u00d7 7)\/ (11 \u00d7 7) + (4 \u00d7 11)\/ (7 \u00d7 11)<\/p>\n\n\n\n

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

= \u2013 42\/77 + 44\/77<\/p>\n\n\n\n

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

= (-42 + 44)\/ 77<\/p>\n\n\n\n

= 2\/77<\/p>\n\n\n\n

4. Find the value of:<\/strong><\/p>\n\n\n\n

(i) \u00bd and 10 kg<\/strong><\/p>\n\n\n\n

(ii) 3\/5 of 1 hour<\/strong><\/p>\n\n\n\n

(iii) 4\/7 of 2 1\/3 kg<\/strong><\/p>\n\n\n\n

(iv) 3 \u00bd times of 2 metre<\/strong><\/p>\n\n\n\n

(v) 1\/2 of 2 2\/3<\/strong><\/p>\n\n\n\n

(vi) 5\/11 of 4\/5 of 22 kg<\/strong><\/p>\n\n\n\n

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

(i) \u00bd and 10 kg<\/p>\n\n\n\n

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

= (1\/2 \u00d7 10) kg<\/p>\n\n\n\n

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

(ii) 3\/5 of 1 hour<\/p>\n\n\n\n

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

= (3\/5 \u00d7 60) minutes<\/p>\n\n\n\n

= 3 \u00d7 12<\/p>\n\n\n\n

= 36 minutes<\/p>\n\n\n\n

(iii) 4\/7 of 2 1\/3 kg<\/p>\n\n\n\n

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

= (4\/7 \u00d7 7\/3) kg<\/p>\n\n\n\n

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

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

= 1 1\/3 kg<\/p>\n\n\n\n

(iv) 3 \u00bd times of 2 metre<\/p>\n\n\n\n

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

= (7\/2 \u00d7 2) metres<\/p>\n\n\n\n

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

(v) 1\/2 of 2 2\/3<\/p>\n\n\n\n

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

= 1\/2 \u00d7 8\/3<\/p>\n\n\n\n

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

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

= 1 1\/3<\/p>\n\n\n\n

(vi) 5\/11 of 4\/5 of 22 kg<\/p>\n\n\n\n

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

= (5\/11 \u00d7 4\/5 \u00d7 22\/1) kg<\/p>\n\n\n\n

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

= 4 \u00d7 2<\/p>\n\n\n\n

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

5. Simplify and reduce to a simple fraction:<\/strong><\/p>\n\n\n\n

(i) 3\/ 3 \u00be<\/strong><\/p>\n\n\n\n

(ii) 3\/5\/ 7<\/strong><\/p>\n\n\n\n

(iii) 3\/ 5\/7<\/strong><\/p>\n\n\n\n

(iv) 2 1\/5\/ 1 1\/10<\/strong><\/p>\n\n\n\n

(v) 2\/5 of 6\/11 \u00d7 1 \u00bc<\/strong><\/p>\n\n\n\n

(vi) 2 \u00bc \u00f7 1\/7 \u00d7 1\/3<\/strong><\/p>\n\n\n\n

(vii) 1\/3 \u00d7 4 2\/3 \u00f7 3 \u00bd \u00d7 1\/2<\/strong><\/p>\n\n\n\n

(viii) 2\/3 \u00d7 1 \u00bc \u00f7 3\/7 of 2 5\/8<\/strong><\/p>\n\n\n\n

(ix) 0 \u00f7 8\/11<\/strong><\/p>\n\n\n\n

(x) 4\/5 \u00f7 7\/15 of 8\/9<\/strong><\/p>\n\n\n\n

(xi) 4\/5 \u00f7 7\/15 \u00d7 8\/9<\/strong><\/p>\n\n\n\n

(xii) 4\/5 of 7\/15 \u00f7 8\/9<\/strong><\/p>\n\n\n\n

(xiii) 1\/2 of 3\/4 \u00d7 1\/2 \u00f7 2\/3<\/strong><\/p>\n\n\n\n

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

(i) 3\/ 3 \u00be<\/p>\n\n\n\n

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

= 3 \/ 15\/4<\/p>\n\n\n\n

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

= (3 \u00d7 4)\/ 15<\/p>\n\n\n\n

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

(ii) 3\/5\/ 7<\/p>\n\n\n\n

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

= 3\/5 \u00d7 1\/7<\/p>\n\n\n\n

= 3\/35<\/p>\n\n\n\n

(iii) 3\/ 5\/7<\/p>\n\n\n\n

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

= 3 \u00d7 7\/5<\/p>\n\n\n\n

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

= 21\/5<\/p>\n\n\n\n

= 4 1\/5<\/p>\n\n\n\n

(iv) 2 1\/5\/ 1 1\/10<\/p>\n\n\n\n

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

= 11\/5 \/ 11\/10<\/p>\n\n\n\n

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

= 11\/5 \u00d7 10\/11<\/p>\n\n\n\n

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

(v) 2\/5 of 6\/11 \u00d7 1 \u00bc<\/p>\n\n\n\n

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

= 2\/5 of 6\/11 \u00d7 5\/4<\/p>\n\n\n\n

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

= 12\/55 \u00d7 5\/4<\/p>\n\n\n\n

= 3\/11<\/p>\n\n\n\n

(vi) 2 \u00bc \u00f7 1\/7 \u00d7 1\/3<\/p>\n\n\n\n

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

= 9\/4 \u00f7 1\/7 \u00d7 1\/3<\/p>\n\n\n\n

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

= 9\/4 \u00d7 7\/1 \u00d7 1\/3<\/p>\n\n\n\n

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

= 5 \u00bc<\/p>\n\n\n\n

(vii) 1\/3 \u00d7 4 2\/3 \u00f7 3 \u00bd \u00d7 1\/2<\/p>\n\n\n\n

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

= 1\/3 \u00d7 14\/3 \u00f7 7\/2 \u00d7 1\/2<\/p>\n\n\n\n

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

= 1\/3 \u00d7 14\/3 \u00d7 2\/7 \u00d7 1\/2<\/p>\n\n\n\n

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

(viii) 2\/3 \u00d7 1 \u00bc \u00f7 3\/7 of 2 5\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00f7 3\/7 of 21\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00f7 9\/8<\/p>\n\n\n\n

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

= 2\/3 \u00d7 5\/4 \u00d7 8\/9<\/p>\n\n\n\n

= 20\/27<\/p>\n\n\n\n

(ix) 0 \u00f7 8\/11<\/p>\n\n\n\n

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

= 0 \u00d7 11\/8<\/p>\n\n\n\n

= 0<\/p>\n\n\n\n

(x) 4\/5 \u00f7 7\/15 of 8\/9<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 4\/5 \u00f7 156\/135<\/p>\n\n\n\n

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

= 4\/5 \u00d7 135\/56<\/p>\n\n\n\n

= 27\/14<\/p>\n\n\n\n

= 1 13\/14<\/p>\n\n\n\n

(xi) 4\/5 \u00f7 7\/15 \u00d7 8\/9<\/p>\n\n\n\n

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

= 4\/5 \u00d7 15\/7 \u00d7 8\/9<\/p>\n\n\n\n

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

= 32\/21<\/p>\n\n\n\n

= 1 11\/21<\/p>\n\n\n\n

(xii) 4\/5 of 7\/15 \u00f7 8\/9<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 28\/75 \u00f7 8\/9<\/p>\n\n\n\n

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

= 28\/75 \u00d7 9\/8<\/p>\n\n\n\n

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

= (7 \u00d7 3)\/ (25 \u00d7 2)<\/p>\n\n\n\n

= 21\/50<\/p>\n\n\n\n

(xiii) 1\/2 of 3\/4 \u00d7 1\/2 \u00f7 2\/3<\/p>\n\n\n\n

From BODMAS rule<\/p>\n\n\n\n

= 3\/8 \u00d7 1\/2 \u00f7 2\/3<\/p>\n\n\n\n

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

= 3\/8 \u00d7 1\/2 \u00d7 3\/2<\/p>\n\n\n\n

= 9\/32<\/p>\n\n\n\n

6. A bought 3 \u00be kg of wheat and 2 \u00bd kg of rice. Find the total weight wheat and rice bought.<\/strong><\/p>\n\n\n\n

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

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

Weight of wheat = 3 \u00be kg = 15\/4 kg<\/p>\n\n\n\n

Weight of rice = 2 \u00bd kg = 5\/2 kg<\/p>\n\n\n\n

So the total weight of wheat and rice = 15\/4 + 5\/2<\/p>\n\n\n\n

Here the LCM of 4 and 2 is 4<\/p>\n\n\n\n

= (15 \u00d7 1)\/ (4 \u00d7 1) + (5 \u00d7 2)\/ (2 \u00d7 2)<\/p>\n\n\n\n

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

= (15 + 10)\/ 4<\/p>\n\n\n\n

= 25\/4 kg<\/p>\n\n\n\n

= 6 \u00bc kg<\/p>\n\n\n\n

7. Which is greater, 3\/5 or 7\/10 and by how much?<\/strong><\/p>\n\n\n\n

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

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

3 \u00d7 10 = 30 and 7 \u00d7 5 = 35<\/p>\n\n\n\n

Here 30 is smaller than 35<\/p>\n\n\n\n

So 3\/5 < 7\/10<\/p>\n\n\n\n

We know that difference between 7\/10 and 3\/5<\/p>\n\n\n\n

= 7\/10 \u2013 3\/5<\/p>\n\n\n\n

LCM of 10 and 5 is 10<\/p>\n\n\n\n

= (7 \u00d7 1)\/ (10 \u00d7 1) \u2013 (3 \u00d7 2)\/ (5 \u00d7 2)<\/p>\n\n\n\n

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

= (7 \u2013 6)\/ 10<\/p>\n\n\n\n

= 1\/10<\/p>\n\n\n\n

Therefore, 7\/10 is greater than 3\/5 by 1\/10.<\/p>\n\n\n\n

8. What number should be added to 8 2\/3 to get 12 5\/6?<\/strong><\/p>\n\n\n\n

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

To find the fraction we must subtract 8 2\/3 from 12 5\/6<\/p>\n\n\n\n

So the required number = 12 5\/6 \u2013 8 2\/3<\/p>\n\n\n\n

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

= 77\/6 \u2013 26\/3<\/p>\n\n\n\n

Here the LCM of 3 and 6 is 6<\/p>\n\n\n\n

= (77 \u00d7 1)\/ (6 \u00d7 1) \u2013 (26 \u00d7 2)\/ (3 \u00d7 2)<\/p>\n\n\n\n

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

= (77 \u2013 52)\/ 6<\/p>\n\n\n\n

= 25\/6<\/p>\n\n\n\n

= 4 1\/6<\/p>\n\n\n\n

9. What should be subtracted from 8 \u00be to get 2 2\/3?<\/strong><\/p>\n\n\n\n

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

Required number = 8 \u00be \u2013 2 2\/3<\/p>\n\n\n\n

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

= 35\/4 \u2013 8\/3<\/p>\n\n\n\n

LCM of 4 and 3 is 12<\/p>\n\n\n\n

= (35 \u00d7 3)\/ (4 \u00d7 3) \u2013 (8 \u00d7 4)\/ (3 \u00d7 4)<\/p>\n\n\n\n

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

= (105 \u2013 32)\/ 12<\/p>\n\n\n\n

= 73\/12<\/p>\n\n\n\n

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

10. A rectangular field is 16 \u00bd m long and 12 2\/5 m wide. Find the perimeter of the field.<\/strong><\/p>\n\n\n\n

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

Dimensions of rectangular field are<\/p>\n\n\n\n

Length = 16 \u00bd m<\/p>\n\n\n\n

Breadth = 12 2\/5 m<\/p>\n\n\n\n

So the perimeter = 2 (l + b)<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 2 \u00d7 (16 \u00bd + 12 2\/5)<\/p>\n\n\n\n

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

= 2 \u00d7 (33\/2 + 62\/5)<\/p>\n\n\n\n

LCM of 2 and 5 is 10<\/p>\n\n\n\n

= 2 \u00d7 [(33 \u00d7 5)\/ (2 \u00d7 5) + (62 \u00d7 2)\/ (5 \u00d7 2)]<\/p>\n\n\n\n

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

= 2 \u00d7 [(165 + 124)\/ 10)]
So we get<\/p>\n\n\n\n

= 2 \u00d7 289\/10<\/p>\n\n\n\n

= 289\/5 m<\/p>\n\n\n\n

= 57 4\/5 m<\/p>\n\n\n\n

11. Sugar costs \u20b9 37 \u00bd per kg. Find the cost of 8 \u00be kg sugar.<\/strong><\/p>\n\n\n\n

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

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

Cost of 1 kg sugar = \u20b9 37 1\/2<\/p>\n\n\n\n

So the cost of 8 \u00be kg sugar = 37 \u00bd \u00d7 8 \u00be<\/p>\n\n\n\n

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

= 75\/2 \u00d7 35\/4<\/p>\n\n\n\n

= 2625\/8<\/p>\n\n\n\n

= \u20b9 328 1\/8<\/p>\n\n\n\n

12. A motor cycle runs 31 \u00bc km consuming 1 litre of petrol. How much distance will it run consuming 1 3\/5 litre of petrol?<\/strong><\/p>\n\n\n\n

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

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

Distance covered consuming 1 litre petrol = 31 \u00bc km = 125\/4 km<\/p>\n\n\n\n

So the distance covered consuming 1 3\/5 litre petrol = 125\/4 \u00d7 8\/5<\/p>\n\n\n\n

= 1000\/20<\/p>\n\n\n\n

= 50 km<\/p>\n\n\n\n

13. A rectangular park has length = 23 2\/5 m and breadth = 16 2\/3 m. Find the area of the park.<\/strong><\/p>\n\n\n\n

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

Dimensions of rectangular park are<\/p>\n\n\n\n

Length = 23 2\/5 m = 117\/5 m<\/p>\n\n\n\n

Breadth = 16 2\/3 m = 50\/3 m<\/p>\n\n\n\n

So the area = l \u00d7 b<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 117\/5 \u00d7 50\/3<\/p>\n\n\n\n

= 39 \u00d7 10<\/p>\n\n\n\n

= 390 m2<\/sup><\/p>\n\n\n\n

14. Each of 40 identical boxes weighs 4 4\/5 kg. Find the total weight of all the boxes.<\/strong><\/p>\n\n\n\n

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

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

Weight of one box = 4 4\/5 kg = 24\/5 kg<\/p>\n\n\n\n

So the weight of 40 boxes = 40 \u00d7 24\/5<\/p>\n\n\n\n

= 8 \u00d7 24<\/p>\n\n\n\n

= 192 kg<\/p>\n\n\n\n

15. Out of 24 kg of wheat, 5\/6th<\/sup> of wheat is consumed. Find, how much wheat is still left?<\/strong><\/p>\n\n\n\n

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

Wheat available = 24 kg<\/p>\n\n\n\n

Wheat consumed = 5\/6th<\/sup> of 24 kg<\/p>\n\n\n\n

= 5\/6 \u00d7 24<\/p>\n\n\n\n

= 20kg<\/p>\n\n\n\n

So the remaining wheat = 24 \u2013 20 kg = 4 kg<\/p>\n\n\n\n

16. A rod of length 2 2\/5 metre is divided into five equal parts. Find the length of each part so obtained.<\/strong><\/p>\n\n\n\n

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

Length of rod = 2\/5 m<\/p>\n\n\n\n

It is given that the length of rod should be divided into 5 equal parts<\/p>\n\n\n\n

So the length of each part of rod = 2 2\/5 \u00f7 5<\/p>\n\n\n\n

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

= 12\/5 \u00d7 1\/5<\/p>\n\n\n\n

= 12\/25 m<\/p>\n\n\n\n

17. If A = 3 3\/8 and B = 6 5\/8, find: (i) A \u00f7 B (ii) B \u00f7 A.<\/strong><\/p>\n\n\n\n

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

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

A = 3 3\/8 = 27\/8<\/p>\n\n\n\n

B = 6 5\/8 = 53\/8<\/p>\n\n\n\n

(i) A \u00f7 B<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 27\/8 \u00f7 53\/8<\/p>\n\n\n\n

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

= 27\/8 \u00d7 8\/53<\/p>\n\n\n\n

= 27\/53<\/p>\n\n\n\n

(ii) B \u00f7 A<\/p>\n\n\n\n

Substituting the values<\/p>\n\n\n\n

= 53\/8 \u00f7 27\/8<\/p>\n\n\n\n

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

= 53\/8 \u00d7 8\/27<\/p>\n\n\n\n

= 53\/27<\/p>\n\n\n\n

= 1 26\/27<\/p>\n\n\n\n

18. Cost of 3 5\/7 litres of oil is \u20b9 83 \u00bd. Find the cost of one litre oil.<\/strong><\/p>\n\n\n\n

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

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

Cost of 3 5\/7 litres of oil = \u20b9 83 \u00bd<\/p>\n\n\n\n

So the cost of one litre oil = \u20b9 83 \u00bd \u00f7 3 5\/7<\/p>\n\n\n\n

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

= \u20b9 167\/2 \u00f7 26\/7<\/p>\n\n\n\n

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

= \u20b9 167\/2 \u00d7 7\/26<\/p>\n\n\n\n

= \u20b9 1169\/52<\/p>\n\n\n\n

= \u20b9 22 25\/52<\/p>\n\n\n\n

19. The product of two numbers is 20 5\/7. If one of these numbers is 6 2\/3, find the other.<\/strong><\/p>\n\n\n\n

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

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

Product of two numbers = 20 5\/7 = 145\/7<\/p>\n\n\n\n

One number = 6 2\/3 = 20\/3<\/p>\n\n\n\n

So the other number = 145\/7 \u00f7 20\/3<\/p>\n\n\n\n

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

= 145\/7 \u00d7 3\/20<\/p>\n\n\n\n

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

= 87\/28<\/p>\n\n\n\n

= 3 3\/28<\/p>\n\n\n\n

20. By what number should 5 5\/6 be multiplied to get 3 1\/3?<\/strong><\/p>\n\n\n\n

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

Here the required number = 3 1\/3 \u00f7 5 5\/6<\/p>\n\n\n\n

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

= 10\/3 \u00f7 35\/6<\/p>\n\n\n\n

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

= 10\/3 \u00d7 6\/35<\/p>\n\n\n\n

= 4\/7<\/p>\n\n\n\n

Exercise 3D<\/h3>\n\n\n\n

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

1. 6 + {4\/3 + (3\/4 \u2013 1\/3)}<\/strong><\/p>\n\n\n\n

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

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

6 + {4\/3 + (3\/4 \u2013 1\/3)}<\/p>\n\n\n\n

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

= 6 + {4\/3 + 3\/4 \u2013 1\/3}<\/p>\n\n\n\n

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

= 6\/1 + 4\/3 + 3\/4 \u2013 1\/3<\/p>\n\n\n\n

Here the LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (72 + 16 + 9 \u2013 4)\/ 12<\/p>\n\n\n\n

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

= (97 \u2013 4)\/ 12<\/p>\n\n\n\n

= 93\/12<\/p>\n\n\n\n

= 31\/4<\/p>\n\n\n\n

= 7 \u00be<\/p>\n\n\n\n

2. 8 \u2013 {3\/2 + (3\/5 \u2013 1\/2)}<\/strong><\/p>\n\n\n\n

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

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

8 \u2013 {3\/2 + (3\/5 \u2013 1\/2)}<\/p>\n\n\n\n

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

= 8 \u2013 {3\/2 + 3\/5 \u2013 1\/2}<\/p>\n\n\n\n

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

= 8\/1 \u2013 3\/2 \u2013 3\/5 + 1\/2<\/p>\n\n\n\n

Here the LCM of 2 and 5 is 10<\/p>\n\n\n\n

= (80 \u2013 15 \u2013 6 + 5)\/ 10<\/p>\n\n\n\n

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

= (85 \u2013 21)\/ 10<\/p>\n\n\n\n

= 64\/10<\/p>\n\n\n\n

Dividing by 2<\/p>\n\n\n\n

= 32\/5<\/p>\n\n\n\n

= 6 2\/5<\/p>\n\n\n\n

3. 1\/4 (1\/4 + 1\/3) \u2013 2\/5<\/strong><\/p>\n\n\n\n

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

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

1\/4 (1\/4 + 1\/3) \u2013 2\/5<\/p>\n\n\n\n

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

= 1\/4 [(3 + 4)\/ 12] \u2013 2\/5<\/p>\n\n\n\n

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

= 1\/4 \u00d7 7\/12 \u2013 2\/5<\/p>\n\n\n\n

= 7\/48 \u2013 2\/5<\/p>\n\n\n\n

Here the LCM of 48 and 5 is 240<\/p>\n\n\n\n

= (35 \u2013 96)\/ 240<\/p>\n\n\n\n

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

= -61\/240<\/p>\n\n\n\n

4. 2 3\/4 \u2013 [3 1\/8 \u00f7 {5 \u2013 (4 2\/3 \u2013 11\/12)}]<\/strong><\/p>\n\n\n\n

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

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

2 3\/4 \u2013 [3 1\/8 \u00f7 {5 \u2013 (4 2\/3 \u2013 11\/12)}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 (14\/3 \u2013 11\/12)}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 (56 \u2013 11)\/12}]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00f7 {5 \u2013 45\/12}]<\/p>\n\n\n\n

LCM of 12 and 1 is 12<\/p>\n\n\n\n

= 11\/4 \u2013 [25\/8 \u00f7 {(60 \u2013 45)\/12}]<\/p>\n\n\n\n

By subtraction<\/p>\n\n\n\n

= 11\/4 \u2013 [25\/8 \u00f7 15\/12]<\/p>\n\n\n\n

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

= 11\/4 \u2013 [25\/8 \u00d7 12\/15]<\/p>\n\n\n\n

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

= 11\/4 \u2013 5\/2<\/p>\n\n\n\n

LCM of 4 and 2 is 4<\/p>\n\n\n\n

= (11 \u2013 10)\/ 4<\/p>\n\n\n\n

= 1\/4<\/p>\n\n\n\n

5. 12 1\/2 \u2013 [8 1\/2 + {9 \u2013 (5 \u2013 )<\/strong><\/p>\n\n\n\n

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

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

12 1\/2 \u2013 [8 1\/2 + {9 \u2013 (5 \u2013
\"Selina)<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + {9 \u2013 (5 \u2013 1)}]<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + {9 \u2013 4}]<\/p>\n\n\n\n

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

= 25\/2 \u2013 [17\/2 + 5]<\/p>\n\n\n\n

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

= 25\/2 \u2013 17\/2 \u2013 5\/1<\/p>\n\n\n\n

LCM of 2 and 1 is 2<\/p>\n\n\n\n

= (25 \u2013 17 \u2013 10)\/ 2<\/p>\n\n\n\n

= (25 \u2013 27)\/ 2<\/p>\n\n\n\n

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

= \u2013 2\/2<\/p>\n\n\n\n

= -1<\/p>\n\n\n\n

6. 1 1\/5 \u00f7 {2 1\/3 \u2013 (5 + )} \u2013 3 \u00bd<\/strong><\/p>\n\n\n\n

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

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

1 1\/5 \u00f7 {2 1\/3 \u2013 (5 +
\"Selina)} \u2013 3 \u00bd<\/p>\n\n\n\n

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

= 6\/5 \u00f7 {7\/3 \u2013 (5 \u2013 1)} \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00f7 {7\/3 \u2013 4} \u2013 7\/2<\/p>\n\n\n\n

LCM of 3 and 1 is 3<\/p>\n\n\n\n

= 6\/5 \u00f7 {(7 \u2013 12)\/ 3} \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00f7 -5\/3 \u2013 7\/2<\/p>\n\n\n\n

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

= 6\/5 \u00d7 3\/-5 \u2013 7\/2<\/p>\n\n\n\n

= \u2013 18\/25 \u2013 7\/2<\/p>\n\n\n\n

LCM of 25 and 2 is 50<\/p>\n\n\n\n

= (- 36 \u2013 175)\/ 50<\/p>\n\n\n\n

= \u2013 211\/50<\/p>\n\n\n\n

= \u2013 4 11\/50<\/p>\n\n\n\n

7. (1\/2 + 2\/3) \u00f7 (3\/4 \u2013 2\/9)<\/strong><\/p>\n\n\n\n

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

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

(1\/2 + 2\/3) \u00f7 (3\/4 \u2013 2\/9)<\/p>\n\n\n\n

LCM of 2 and 3 is 6 and 4 and 9 is 36<\/p>\n\n\n\n

= (3 + 4)\/ 6 \u00f7 (27 \u2013 8)\/ 36<\/p>\n\n\n\n

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

= 7\/6 \u00f7 19\/36<\/p>\n\n\n\n

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

= 7\/6 \u00d7 36\/19<\/p>\n\n\n\n

= 42\/19<\/p>\n\n\n\n

= 2 4\/19<\/p>\n\n\n\n

8. 6\/5 of (3 1\/3 \u2013 2 1\/2) \u00f7 (2 5\/21 \u2013 2)<\/strong><\/p>\n\n\n\n

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

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

6\/5 of (3 1\/3 \u2013 2 1\/2) \u00f7 (2 5\/21 \u2013 2)<\/p>\n\n\n\n

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

= 6\/5 of (10\/3 \u2013 5\/2) \u00f7 (47\/21 \u2013 2\/1)<\/p>\n\n\n\n

LCM of 3 and 2 is 6 and 1 and 21 is 21<\/p>\n\n\n\n

= 6\/5 of [(20 \u2013 15)\/ 6] \u00f7 [(47 \u2013 42)\/ 21]<\/p>\n\n\n\n

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

= 6\/5 of 5\/6 \u00f7 5\/21<\/p>\n\n\n\n

= 1 \u00f7 5\/21<\/p>\n\n\n\n

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

= 1 \u00d7 21\/5<\/p>\n\n\n\n

= 21\/5<\/p>\n\n\n\n

= 4 1\/5<\/p>\n\n\n\n

9. 10 1\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/strong><\/p>\n\n\n\n

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

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

10 1\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/p>\n\n\n\n

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

= 81\/8 of 4\/5 \u00f7 35\/36 of 20\/49<\/p>\n\n\n\n

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

= 81\/10 \u00f7 25\/63<\/p>\n\n\n\n

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

= 81\/10 \u00d7 63\/25<\/p>\n\n\n\n

= 5103\/250<\/p>\n\n\n\n

= 20 103\/250<\/p>\n\n\n\n

10. 5 3\/4 \u2013 3\/7 \u00d7 15 3\/4 + 2 2\/35 \u00f7 1 11\/25<\/strong><\/p>\n\n\n\n

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

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

5 3\/4 \u2013 3\/7 \u00d7 15 3\/4 + 2 2\/35 \u00f7 1 11\/25<\/p>\n\n\n\n

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

= 23\/4 \u2013 3\/7 \u00d7 63\/4 + 72\/35 \u00f7 36\/25<\/p>\n\n\n\n

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

= 23\/4 \u2013 3\/7 \u00d7 63\/4 + 72\/35 \u00d7 25\/36<\/p>\n\n\n\n

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

= 23\/4 \u2013 27\/4 + 10\/7<\/p>\n\n\n\n

LCM of 4 and 7 is 28<\/p>\n\n\n\n

= (161 \u2013 189 + 40)\/ 28<\/p>\n\n\n\n

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

= 3\/7<\/p>\n\n\n\n

11. 3\/4 of 7 3\/7 \u2013 5 3\/5 \u00f7 3 4\/15<\/strong><\/p>\n\n\n\n

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

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

3\/4 of 7 3\/7 \u2013 5 3\/5 \u00f7 3 4\/15<\/p>\n\n\n\n

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

= 3\/4 of 52\/7 \u2013 28\/5 \u00f7 49\/15<\/p>\n\n\n\n

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

= 39\/7 \u2013 28\/5 \u00f7 49\/15<\/p>\n\n\n\n

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

= 39\/7 \u2013 28\/5 \u00d7 15\/49<\/p>\n\n\n\n

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

= 39\/7 \u2013 12\/7<\/p>\n\n\n\n

= (39 \u2013 12)\/ 7<\/p>\n\n\n\n

= 27\/7<\/p>\n\n\n\n

= 3 6\/7<\/p>\n\n\n\n

Exercise 3E<\/h3>\n\n\n\n

1. A line AB is of length 6 cm. Another line CD is of length 15 cm. What fraction is:<\/strong><\/p>\n\n\n\n

(i) the length of AB to that of CD?<\/strong><\/p>\n\n\n\n

(ii) \u00bd the length of AB to that of 1\/3 of CD?<\/strong><\/p>\n\n\n\n

(iii) 1\/5 of CD to that of AB?<\/strong><\/p>\n\n\n\n

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

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

Length of AB = 6 cm<\/p>\n\n\n\n

Length of CD = 15 cm<\/p>\n\n\n\n

(i) The length of AB to that of CD<\/p>\n\n\n\n

= 6\/15<\/p>\n\n\n\n

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

(ii) \u00bd the length of AB to that of 1\/3 of CD<\/p>\n\n\n\n

1\/2 of AB = 1\/2 \u00d7 6 = 3 cm<\/p>\n\n\n\n

1\/3 of CD = 1\/3 \u00d7 15 = 5 cm<\/p>\n\n\n\n

So \u00bd the length of AB to that of 1\/3 of CD = 3\/5<\/p>\n\n\n\n

(iii) 1\/5 of CD to that of AB<\/p>\n\n\n\n

1\/5 of CD = 1\/5 \u00d7 15 = 3 cm<\/p>\n\n\n\n

1\/5 of CD to that of AB = 3\/6 = 1\/2<\/p>\n\n\n\n

2. Subtract (2\/7 \u2013 5\/21) from the sum of 3\/4, 5\/7 and 7\/12.<\/strong><\/p>\n\n\n\n

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

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

(3\/4 + 5\/7 + 7\/12) \u2013 (2\/7 \u2013 5\/21)<\/p>\n\n\n\n

LCM of 4, 7 and 12 is 84 and 7 and 21 is 21<\/p>\n\n\n\n

= [(63 + 60 + 49)\/ 84] \u2013 [(6 \u2013 5)\/ 21]<\/p>\n\n\n\n

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

= 172\/84 \u2013 1\/21<\/p>\n\n\n\n

LCM of 21 and 84 is 84<\/p>\n\n\n\n

= (172 \u2013 4)\/ 84<\/p>\n\n\n\n

= 168\/ 84<\/p>\n\n\n\n

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

3. From a sack of potatoes weighing 120 kg, a merchant sells portions weighing 6 kg, 5 \u00bc kg, 9 \u00bd kg and 9 \u00be kg respectively.<\/strong><\/p>\n\n\n\n

(i) How many kg did he sell?<\/strong><\/p>\n\n\n\n

(ii) How many kg are still left in the sack?<\/strong><\/p>\n\n\n\n

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

Weight of potato = 120 kg<\/p>\n\n\n\n

(i) Potatoes sold by merchant = 6 kg + 5 \u00bc kg + 9 \u00bd kg + 9 \u00be kg<\/p>\n\n\n\n

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

= (6 + 21\/4 + 19\/2 + 39\/4) kg<\/p>\n\n\n\n

LCM of 1, 4 and 2 is 4<\/p>\n\n\n\n

= (24 + 21 + 38 + 39)\/ 4<\/p>\n\n\n\n

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

= 122\/4<\/p>\n\n\n\n

= 61\/2 kg<\/p>\n\n\n\n

= 30 \u00bd kg<\/p>\n\n\n\n

(ii) Potatoes left in the sack = 120 kg \u2013 30 \u00bd kg<\/p>\n\n\n\n

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

= (120\/ 1 \u2013 61\/2) kg<\/p>\n\n\n\n

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

= (240 \u2013 61)\/ 2<\/p>\n\n\n\n

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

= 179\/2 kg<\/p>\n\n\n\n

= 89 \u00bd kg<\/p>\n\n\n\n

4. If a boy works for six consecutive days for 8 hours, 7 \u00bd hours, 8 \u00bc hours, 6 \u00bc hours, 6 \u00be hours and 7 hours respectively, how much money will he earn at the rate of \u20b9 36 per hour?<\/strong><\/p>\n\n\n\n

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

Hours worked by a boy for six consecutive days = 8 hours + 7 \u00bd hours + 8 \u00bc hours + 6 \u00bc hours + 6 \u00be hours + 7 hours<\/p>\n\n\n\n

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

= (8 + 15\/2 + 33\/4 + 25\/4 + 27\/4 + 7) hours<\/p>\n\n\n\n

LCM of 2 and 4 is 4<\/p>\n\n\n\n

= (32 + 30 + 33 + 25 + 27 + 28)\/ 4 hours<\/p>\n\n\n\n

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

= 175\/4 hours<\/p>\n\n\n\n

= 43 \u00be hours<\/p>\n\n\n\n

He earned \u20b9 36 per hour<\/p>\n\n\n\n

So the total earnings = 175\/4 \u00d7 36<\/p>\n\n\n\n

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

= 175 \u00d7 9<\/p>\n\n\n\n

= \u20b9 1575<\/p>\n\n\n\n

5. A student bought 4 1\/3 m of yellow ribbon, 6 1\/6 m of red ribbon and 3 2\/9 m of blue ribbon decorating a room. How many metres of ribbon did he buy?<\/strong><\/p>\n\n\n\n

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

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

Length of yellow ribbon = 4 1\/3 m = 13\/3 m<\/p>\n\n\n\n

Length of red ribbon = 6 1\/6 m = 37\/6 m<\/p>\n\n\n\n

Length of blue ribbon = 3 2\/9 m = 29\/9 m<\/p>\n\n\n\n

So the total length = 13\/3 + 37\/6 + 29\/9<\/p>\n\n\n\n

LCM of 3, 6 and 9 is 18<\/p>\n\n\n\n

= (78 + 111 + 58)\/ 18<\/p>\n\n\n\n

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

= 247\/ 18<\/p>\n\n\n\n

= 13 13\/18 m<\/p>\n\n\n\n

6. In a business, Ram and Deepak invest 3\/5 and 2\/5 of the total investment. If \u20b9 40, 000 is the total investment, calculate the amount invested by each.<\/strong><\/p>\n\n\n\n

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

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

Total investment = \u20b9 40, 000<\/p>\n\n\n\n

Ram\u2019s investment = 3\/5 of \u20b9 40, 000<\/p>\n\n\n\n

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

= 3\/5 \u00d7 40, 000<\/p>\n\n\n\n

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

= 3 \u00d7 8000<\/p>\n\n\n\n

= \u20b9 24, 000<\/p>\n\n\n\n

Deepak\u2019s investment = 2\/5 of \u20b9 40, 000<\/p>\n\n\n\n

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

= 2\/5 \u00d7 40, 000<\/p>\n\n\n\n

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

= 2 \u00d7 8000<\/p>\n\n\n\n

= \u20b9 16, 000<\/p>\n\n\n\n

7. Geeta had 30 problems for home work. She worked out 2\/3 of them. How many problems were still left to be worked out by her?<\/strong><\/p>\n\n\n\n

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

Number of problems Geeta had for home work = 30<\/p>\n\n\n\n

Number of problems worked out by Geeta = 2\/3 of 30<\/p>\n\n\n\n

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

= 2\/3 \u00d7 30<\/p>\n\n\n\n

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

Number of problems still left to be worked out by her = 30 \u2013 20 = 10<\/p>\n\n\n\n

8. A picture was marked at \u20b9 90. It was sold at 3\/4 of its marked price. What was the sale price?<\/strong><\/p>\n\n\n\n

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

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

Marked price of picture = \u20b9 90<\/p>\n\n\n\n

Sale price of picture = \u00be of \u20b9 90<\/p>\n\n\n\n

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

= \u00be \u00d7 90<\/p>\n\n\n\n

= \u20b9 270\/4<\/p>\n\n\n\n

= \u20b9 67 \u00bd<\/p>\n\n\n\n

= \u20b9 67.50<\/p>\n\n\n\n

9. Mani had sent fifteen parcels of oranges. What was the total weight of the parcels, if each weighed 10 \u00bd kg?<\/strong><\/p>\n\n\n\n

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

Number of parcels = 15<\/p>\n\n\n\n

Weight of each parcel = 10 \u00bd kg = 21\/2 kg<\/p>\n\n\n\n

So the total weight of parcels = 15 of 21\/2 kg<\/p>\n\n\n\n

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

= 21\/2 \u00d7 15 kg<\/p>\n\n\n\n

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

= 315\/2<\/p>\n\n\n\n

= 157 \u00bd kg<\/p>\n\n\n\n

= 157.5 kg<\/p>\n\n\n\n

10. A rope is 25 \u00bd m long. How many pieces each of 1 \u00bd m length can be cut out from it?<\/strong><\/p>\n\n\n\n

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

Length of rope = 25 \u00bd m = 51\/2 m<\/p>\n\n\n\n

Length of each piece = 1 \u00bd m = 3\/2 m<\/p>\n\n\n\n

So the number of pieces = 51\/2 \u00f7 3\/2<\/p>\n\n\n\n

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

= 51\/2 \u00d7 2\/3<\/p>\n\n\n\n

= 17 pieces<\/p>\n\n\n\n

11. The heights of two vertical poles, above the earth\u2019s surface, are 14 \u00bc m and 22 1\/3 m respectively. How much higher is the second pole as compared with the height of the first pole?<\/strong><\/p>\n\n\n\n

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

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

Height of first pole above the earth\u2019s surface = 14 \u00bc m<\/p>\n\n\n\n

Height of second pole above the earth\u2019s surface = 22 1\/3 m<\/p>\n\n\n\n

So height of second pole when compared to first pole = 22 1\/3 \u2013 14 1\/4<\/p>\n\n\n\n

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

= 67\/3 \u2013 57\/4<\/p>\n\n\n\n

LCM of 3 and 4 is 12<\/p>\n\n\n\n

= (268 \u2013 171)\/ 12<\/p>\n\n\n\n

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

= 97\/12 m<\/p>\n\n\n\n

= 8 1\/12 m<\/p>\n\n\n\n

12. Vijay weighed 65 \u00bd kg. He gained 1 2\/5 kg during the first week, 1 \u00bc kg during the second week, but lost 5\/16 kg during the third week. What was his weight after the third week?<\/strong><\/p>\n\n\n\n

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

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

Weight of Vijay = 65 \u00bd kg<\/p>\n\n\n\n

Weight gained during first week = 1 2\/5 kg<\/p>\n\n\n\n

Weight gained during second week = 1 \u00bc kg<\/p>\n\n\n\n

Weight lost during third week = 5\/16 kg<\/p>\n\n\n\n

So the weight of Vijay after third week = 65 \u00bd + 1 2\/5 + 1 \u00bc \u2013 5\/16<\/p>\n\n\n\n

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

= 131\/2 + 7\/5 + 5\/4 \u2013 5\/16<\/p>\n\n\n\n

LCM of 2, 5, 4 and 16 is 8-<\/p>\n\n\n\n

= (5240 + 112 + 100 \u2013 25)\/ 80<\/p>\n\n\n\n

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

= (5452 \u2013 25)\/ 80<\/p>\n\n\n\n

= 5427\/80 kg<\/p>\n\n\n\n

= 67 67\/80 kg<\/p>\n\n\n\n

13. A man spends 2\/5 of his salary on food and 3\/10 on house rent, electricity, etc. What fraction of his salary is still left with him?<\/strong><\/p>\n\n\n\n

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

Consider salary of man = \u20b9 1<\/p>\n\n\n\n

Salary spent on food = 2\/5 of \u20b9 1 = \u20b9 2\/5<\/p>\n\n\n\n

Salary spent on house rent = 3\/10 of \u20b9 1 = \u20b9 3\/10<\/p>\n\n\n\n

So the total salary spent = 2\/5 + 3\/10<\/p>\n\n\n\n

LCM of 5 and 10 is 10<\/p>\n\n\n\n

= (4 + 3)\/ 10<\/p>\n\n\n\n

= 7\/10<\/p>\n\n\n\n

Salary still left with him = 1 \u2013 7\/10<\/p>\n\n\n\n

LCM of 1 and 10 is 10<\/p>\n\n\n\n

= (10 \u2013 7)\/ 10<\/p>\n\n\n\n

= 3\/10<\/p>\n\n\n\n

14. A man spends 2\/5 of his salary on food and 3\/10 of the remaining on house rent, electricity, etc. What fraction of his salary is still left with him?<\/strong><\/p>\n\n\n\n

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

Consider \u20b9 1 as the total salary<\/p>\n\n\n\n

Salary spent on food = 2\/5 of \u20b9 1 = \u20b9 2\/5<\/p>\n\n\n\n

So the remaining salary = 1 \u2013 2\/5<\/p>\n\n\n\n

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

= (5 \u2013 2)\/ 5<\/p>\n\n\n\n

= \u20b9 3\/5<\/p>\n\n\n\n

Salary spent on house rent = 3\/10 of 3\/5 = \u20b9 9\/50<\/p>\n\n\n\n

So the remaining salary = 3\/5 \u2013 9\/50<\/p>\n\n\n\n

LCM of 5 and 50 is 50<\/p>\n\n\n\n

= (30 \u2013 9)\/ 50<\/p>\n\n\n\n

= \u20b9 21\/50<\/p>\n\n\n\n

15. Shyam bought a refrigerator for \u20b9 5,000. He paid 1\/10 of the price in cash and the rest in 12 equal monthly instalments. How much had he to pay each month?<\/strong><\/p>\n\n\n\n

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

Total cost of refrigerator = \u20b9 5000<\/p>\n\n\n\n

Cash paid = 1\/10 of \u20b9 5000<\/p>\n\n\n\n

= 1\/10 \u00d7 5000<\/p>\n\n\n\n

= \u20b9 500<\/p>\n\n\n\n

So the balance amount = 5000 \u2013 500 = \u20b9 4500<\/p>\n\n\n\n

Number of instalments = 12<\/p>\n\n\n\n

So the amount to be paid each month = 4500 \u00f7 12<\/p>\n\n\n\n

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

= 4500 \u00d7 1\/12<\/p>\n\n\n\n

= \u20b9 375<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Selina Class 7 ICSE Solutions Mathematics : Chapter 3- Fraction (Including Problems)<\/p>\n\n\n\n

Download PDF<\/strong>: Selina Class 7 ICSE Solutions Mathematics : Chapter 3-\u00a0Fraction (Including Problems) PDF<\/a><\/p>\n\n\n\n

Chapterwise Selina Publishers ICSE Solutions for Class 7 Mathematics :<\/strong><\/h2>\n\n\n\n