{"id":546000,"date":"2021-10-05T10:59:36","date_gmt":"2021-10-05T10:59:36","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546000"},"modified":"2021-10-07T04:32:38","modified_gmt":"2021-10-07T04:32:38","slug":"rd-sharma-solutions-for-class-8-maths-chapter-1-rational-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-1-rational-numbers\/","title":{"rendered":"RD Sharma Solutions for Class 8 Maths Chapter 1\u2013Rational Numbers"},"content":{"rendered":"\n

Class 8: Maths Chapter 1 solutions. Complete Class 8 Maths Chapter 1 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 1\u2013Rational Numbers<\/h2>\n\n\n\n

RD Sharma 8th Maths Chapter 1, Class 8 Maths Chapter 1 solutions<\/p>\n\n\n\n

EXERCISE 1.1 PAGE NO: 1.5<\/h4>\n\n\n\n

1. Add the following rational numbers:<\/strong><\/p>\n\n\n\n

(i) -5\/7 and 3\/7<\/strong><\/p>\n\n\n\n

(ii) -15\/4 and 7\/4<\/strong><\/p>\n\n\n\n

(iii) -8\/11 and -4\/11<\/strong><\/p>\n\n\n\n

(iv) 6\/13 and -9\/13<\/strong><\/p>\n\n\n\n

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

Since the denominators are of same positive numbers we can add them directly<\/p>\n\n\n\n

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

(ii) -15\/4 + 7\/4 = (-15+7)\/4 = -8\/4<\/p>\n\n\n\n

Further dividing by 4 we get,<\/p>\n\n\n\n

-8\/4 = -2<\/p>\n\n\n\n

(iii) -8\/11 + -4\/11 = (-8 + (-4))\/11 = (-8-4)\/11 = -12\/11<\/p>\n\n\n\n

(iv) 6\/13 + -9\/13 = (6 + (-9))\/13 = (6-9)\/13 = -3\/13<\/p>\n\n\n\n

2. Add the following rational numbers:<\/strong><\/p>\n\n\n\n

(i) 3\/4 and -5\/8<\/strong><\/p>\n\n\n\n

Solution: <\/strong>The denominators are 4 and 8<\/p>\n\n\n\n

By taking LCM for 4 and 8 is 8<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

3\/4 = (3\u00d72) \/ (4\u00d72) = 6\/8 and<\/p>\n\n\n\n

-5\/8 = (-5\u00d71) \/ (8\u00d71) = -5\/8<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

6\/8 + -5\/8 = (6 + (-5))\/8 = (6-5)\/8 = 1\/8<\/p>\n\n\n\n

(ii) 5\/-9 and 7\/3<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

5\/-9 = (5 \u00d7 -1)\/ (-9 \u00d7 -1) = -5\/9<\/p>\n\n\n\n

The denominators are 9 and 3<\/p>\n\n\n\n

By taking LCM for 9 and 3 is 9<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-5\/9 = (-5\u00d71) \/ (9\u00d71) = -5\/9 and<\/p>\n\n\n\n

7\/3 = (7\u00d73) \/ (3\u00d73) = 21\/9<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-5\/9 + 21\/9 = (-5+21)\/9 = 16\/9<\/p>\n\n\n\n

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

Solution: <\/strong>The denominators are 1 and 5<\/p>\n\n\n\n

By taking LCM for 1 and 5 is 5<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-3\/1 = (-3\u00d75) \/ (1\u00d75) = -15\/5 and<\/p>\n\n\n\n

3\/5 = (3\u00d71) \/ (5\u00d71) = 3\/5<\/p>\n\n\n\n

Now, the denominators are same we can add them directly<\/p>\n\n\n\n

-15\/5 + 3\/5 = (-15+3)\/5 = -12\/5<\/p>\n\n\n\n

(iv) -7\/27 and 11\/18<\/strong><\/p>\n\n\n\n

Solution:<\/strong> The denominators are 27 and 18<\/p>\n\n\n\n

By taking LCM for 27 and 18 is 54<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-7\/27 = (-7\u00d72) \/ (27\u00d72) = -14\/54 and<\/p>\n\n\n\n

11\/18 = (11\u00d73) \/ (18\u00d73) = 33\/54<\/p>\n\n\n\n

Now, the denominators are same we can add them directly<\/p>\n\n\n\n

-14\/54 + 33\/54 = (-14+33)\/54 = 19\/54<\/p>\n\n\n\n

(v) 31\/-4 and -5\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

31\/-4 = (31 \u00d7 -1)\/ (-4 \u00d7 -1) = -31\/4<\/p>\n\n\n\n

The denominators are 4 and 8<\/p>\n\n\n\n

By taking LCM for 4 and 8 is 8<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-31\/4 = (-31\u00d72) \/ (4\u00d72) = -62\/8 and<\/p>\n\n\n\n

-5\/8 = (-5\u00d71) \/ (8\u00d71) = -5\/8<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-62\/8 + (-5)\/8 = (-62 + (-5))\/8 = (-62-5)\/8 = -67\/8<\/p>\n\n\n\n

(vi) 5\/36 and -7\/12<\/strong><\/p>\n\n\n\n

Solution:<\/strong> The denominators are 36 and 12<\/p>\n\n\n\n

By taking LCM for 36 and 12 is 36<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

5\/36 = (5\u00d71) \/ (36\u00d71) = 5\/36 and<\/p>\n\n\n\n

-7\/12 = (-7\u00d73) \/ (12\u00d73) = -21\/36<\/p>\n\n\n\n

Now, the denominators are same we can add them directly<\/p>\n\n\n\n

5\/36 + -21\/36 = (5 + (-21))\/36 = 5-21\/36 = -16\/36 = -4\/9<\/p>\n\n\n\n

(vii) -5\/16 and 7\/24<\/strong><\/p>\n\n\n\n

Solution:<\/strong> The denominators are 16 and 24<\/p>\n\n\n\n

By taking LCM for 16 and 24 is 48<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-5\/16 = (-5\u00d73) \/ (16\u00d73) = -15\/48 and<\/p>\n\n\n\n

7\/24 = (7\u00d72) \/ (24\u00d72) = 14\/48<\/p>\n\n\n\n

Now, the denominators are same we can add them directly<\/p>\n\n\n\n

-15\/48 + 14\/48 = (-15 + 14)\/48 = -1\/48<\/p>\n\n\n\n

(viii) 7\/-18 and 8\/27<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

7\/-18 = (7 \u00d7 -1)\/ (-18 \u00d7 -1) = -7\/18<\/p>\n\n\n\n

The denominators are 18 and 27<\/p>\n\n\n\n

By taking LCM for 18 and 27 is 54<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

-7\/18 = (-7\u00d73) \/ (18\u00d73) = -21\/54 and<\/p>\n\n\n\n

8\/27 = (8\u00d72) \/ (27\u00d72) = 16\/54<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-21\/54 + 16\/54 = (-21 + 16)\/54 = -5\/54<\/p>\n\n\n\n

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

(i) 8\/9 + -11\/6<\/strong><\/p>\n\n\n\n

Solution: <\/strong>let us take the LCM for 9 and 6 which is 18<\/p>\n\n\n\n

(8\u00d72)\/(9\u00d72) + (-11\u00d73)\/(6\u00d73)<\/p>\n\n\n\n

16\/18 + -33\/18<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(16-33)\/18 = -17\/18<\/p>\n\n\n\n

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

Solution: <\/strong>Firstly convert the denominator to positive number<\/p>\n\n\n\n

5\/-7 = (5\u00d7-1)\/(-7\u00d7-1) = -5\/7<\/p>\n\n\n\n

3\/1 + -5\/7<\/p>\n\n\n\n

Now let us take the LCM for 1 and 7 which is 7<\/p>\n\n\n\n

(3\u00d77)\/(1\u00d77) + (-5\u00d71)\/(7\u00d71)<\/p>\n\n\n\n

21\/7 + -5\/7<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(21-5)\/7 = 16\/7<\/p>\n\n\n\n

(iii) 1\/-12 + 2\/-15<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly convert the denominator to positive number<\/p>\n\n\n\n

1\/-12 = (1\u00d7-1)\/(-12\u00d7-1) = -1\/12<\/p>\n\n\n\n

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

-1\/12 + -2\/15<\/p>\n\n\n\n

Now let us take the LCM for 12 and 15 which is 60<\/p>\n\n\n\n

(-1\u00d75)\/(12\u00d75) + (-2\u00d74)\/(15\u00d74)<\/p>\n\n\n\n

-5\/60 + -8\/60<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-5-8)\/60 = -13\/60<\/p>\n\n\n\n

(iv) -8\/19 + -4\/57<\/strong><\/p>\n\n\n\n

Solution: <\/strong>let us take the LCM for 19 and 57 which is 57<\/p>\n\n\n\n

(-8\u00d73)\/(19\u00d73) + (-4\u00d71)\/(57\u00d71)<\/p>\n\n\n\n

-24\/57 + -4\/57<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-24-4)\/57 = -28\/57<\/p>\n\n\n\n

(v) 7\/9 + 3\/-4<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly convert the denominator to positive number<\/p>\n\n\n\n

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

7\/9 + -3\/4<\/p>\n\n\n\n

Now let us take the LCM for 9 and 4 which is 36<\/p>\n\n\n\n

(7\u00d74)\/(9\u00d74) + (-3\u00d79)\/(4\u00d79)<\/p>\n\n\n\n

28\/36 + -27\/36<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(28-27)\/36 = 1\/36<\/p>\n\n\n\n

(vi) 5\/26 + 11\/-39<\/strong><\/p>\n\n\n\n

Solution<\/strong>: Firstly convert the denominator to positive number<\/p>\n\n\n\n

11\/-39 = (11\u00d7-1)\/(-39\u00d7-1) = -11\/39<\/p>\n\n\n\n

5\/26 + -11\/39<\/p>\n\n\n\n

Now let us take the LCM for 26 and 39 which is 78<\/p>\n\n\n\n

(5\u00d73)\/(26\u00d73) + (-11\u00d72)\/(39\u00d72)<\/p>\n\n\n\n

15\/78 + -22\/78<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(15-22)\/78 = -7\/78<\/p>\n\n\n\n

(vii) -16\/9 + -5\/12<\/strong><\/p>\n\n\n\n

Solution:<\/strong> let us take the LCM for 9 and 12 which is 108<\/p>\n\n\n\n

(-16\u00d712)\/(9\u00d712) + (-5\u00d79)\/(12\u00d79)<\/p>\n\n\n\n

-192\/108 + -45\/108<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-192-45)\/108 = -237\/108<\/p>\n\n\n\n

Further divide the fraction by 3 we get,<\/p>\n\n\n\n

-237\/108 = -79\/36<\/p>\n\n\n\n

(viii) -13\/8 + 5\/36<\/strong><\/p>\n\n\n\n

Solution: <\/strong>let us take the LCM for 8 and 36 which is 72<\/p>\n\n\n\n

(-13\u00d79)\/(8\u00d79) + (5\u00d72)\/(36\u00d72)<\/p>\n\n\n\n

-117\/72 + 10\/72<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-117+10)\/72 = -107\/72<\/p>\n\n\n\n

(ix) 0 + -3\/5<\/strong><\/p>\n\n\n\n

Solution: <\/strong>We know that anything added to 0 results in the same.<\/p>\n\n\n\n

0 + -3\/5 = -3\/5<\/p>\n\n\n\n

(x) 1 + -4\/5<\/strong><\/p>\n\n\n\n

Solution:<\/strong> let us take the LCM for 1 and 5 which is 5<\/p>\n\n\n\n

(1\u00d75)\/(1\u00d75) + (-4\u00d71)\/(5\u00d71)<\/p>\n\n\n\n

5\/5 + -4\/5<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

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

4. Add and express the sum as a mixed fraction:<\/strong><\/p>\n\n\n\n

(i) -12\/5 and 43\/10<\/strong><\/p>\n\n\n\n

Solution: <\/strong>let us add the given fraction<\/p>\n\n\n\n

-12\/5 + 43\/10<\/p>\n\n\n\n

let us take the LCM for 5 and 10 which is 10<\/p>\n\n\n\n

(-12\u00d72)\/(5\u00d72) + (43\u00d71)\/(10\u00d71)<\/p>\n\n\n\n

-24\/10 + 43\/10<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-24+43)\/10 = 19\/10<\/p>\n\n\n\n

19\/10 can be written as 19101910 in mixed fraction.<\/p>\n\n\n\n

(ii) 24\/7 and -11\/4<\/strong><\/p>\n\n\n\n

Solution: <\/strong>let us add the given fraction<\/p>\n\n\n\n

24\/7 + -11\/4<\/p>\n\n\n\n

let us take the LCM for 7 and 4 which is 28<\/p>\n\n\n\n

(24\u00d74)\/(7\u00d74) + (-11\u00d77)\/(4\u00d77)<\/p>\n\n\n\n

96\/28 + -77\/28<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(96-77)\/28 = 19\/28<\/p>\n\n\n\n

(iii) -31\/6 and -27\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong> let us add the given fraction<\/p>\n\n\n\n

-31\/6 + -27\/8<\/p>\n\n\n\n

let us take the LCM for 6 and 8 which is 24<\/p>\n\n\n\n

(-31\u00d74)\/(6\u00d74) + (-27\u00d73)\/(8\u00d73)<\/p>\n\n\n\n

-124\/24 + -81\/24<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(-124-81)\/24 = -205\/24<\/p>\n\n\n\n

-205\/24 can be written as \u221281324\u221281324 in mixed fraction.<\/p>\n\n\n\n

(iv) 101\/6 and 7\/8<\/strong><\/p>\n\n\n\n

Solution:<\/strong> let us add the given fraction<\/p>\n\n\n\n

101\/6 + 7\/8<\/p>\n\n\n\n

let us take the LCM for 6 and 8 which is 24<\/p>\n\n\n\n

(101\u00d74)\/(6\u00d74) + (7\u00d73)\/(8\u00d73)<\/p>\n\n\n\n

404\/24 + 21\/24<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

(404+21)\/24 = 425\/24<\/p>\n\n\n\n

425\/24 can be written as 171724171724 in mixed fraction.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.2 PAGE NO: 1.14<\/h4>\n\n\n\n

1. Verify commutativity of addition of rational numbers for each of the following pairs of rational numbers:<\/strong><\/p>\n\n\n\n

(i) -11\/5 and 4\/7<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By using the commutativity law, the addition of rational numbers is commutative \u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

-11\/5 and 4\/7 as<\/p>\n\n\n\n

-11\/5 + 4\/7 and 4\/7 + -11\/5<\/p>\n\n\n\n

The denominators are 5 and 7<\/p>\n\n\n\n

By taking LCM for 5 and 7 is 35<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -11\/5 = (-11 \u00d7 7) \/ (5 \u00d77) = -77\/35<\/p>\n\n\n\n

4\/7 = (4 \u00d75) \/ (7 \u00d75) = 20\/35<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-77\/35 + 20\/35 = (-77+20)\/35 = -57\/35<\/p>\n\n\n\n

4\/7 + -11\/5<\/p>\n\n\n\n

The denominators are 7 and 5<\/p>\n\n\n\n

By taking LCM for 7 and 5 is 35<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, 4\/7 = (4 \u00d7 5) \/ (7 \u00d75) = 20\/35<\/p>\n\n\n\n

-11\/5 = (-11 \u00d77) \/ (5 \u00d77) = -77\/35<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

20\/35 + -77\/35 = (20 + (-77))\/35 = (20-77)\/35 = -57\/35<\/p>\n\n\n\n

\u2234 -11\/5 + 4\/7 = 4\/7 + -11\/5 is satisfied.<\/p>\n\n\n\n

(ii) 4\/9 and 7\/-12<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

7\/-12 = (7 \u00d7 -1)\/ (-12 \u00d7 -1) = -7\/12<\/p>\n\n\n\n

By using the commutativity law, the addition of rational numbers is commutative.<\/p>\n\n\n\n

\u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

4\/9 and -7\/12 as<\/p>\n\n\n\n

4\/9 + -7\/12 and -7\/12 + 4\/9<\/p>\n\n\n\n

The denominators are 9 and 12<\/p>\n\n\n\n

By taking LCM for 9 and 12 is 36<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, 4\/9 = (4 \u00d7 4) \/ (9 \u00d74) = 16\/36<\/p>\n\n\n\n

-7\/12 = (-7 \u00d73) \/ (12 \u00d73) = -21\/36<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

16\/36 + (-21)\/36 = (16 + (-21))\/36 = (16-21)\/36 = -5\/36<\/p>\n\n\n\n

-7\/12 + 4\/9<\/p>\n\n\n\n

The denominators are 12 and 9<\/p>\n\n\n\n

By taking LCM for 12 and 9 is 36<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -7\/12 = (-7 \u00d73) \/ (12 \u00d73) = -21\/36<\/p>\n\n\n\n

4\/9 = (4 \u00d7 4) \/ (9 \u00d74) = 16\/36<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-21\/36 + 16\/36 = (-21 + 16)\/36 = -5\/36<\/p>\n\n\n\n

\u2234 4\/9 + -7\/12 = -7\/12 + 4\/9 is satisfied.<\/p>\n\n\n\n

(iii) -3\/5 and -2\/-15<\/strong><\/p>\n\n\n\n

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

By using the commutativity law, the addition of rational numbers is commutative.<\/p>\n\n\n\n

\u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

-3\/5 and -2\/-15 as<\/p>\n\n\n\n

-3\/5 + -2\/-15 and -2\/-15 + -3\/5<\/p>\n\n\n\n

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

The denominators are 5 and 15<\/p>\n\n\n\n

By taking LCM for 5 and 15 is 15<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -3\/5 = (-3 \u00d7 3) \/ (5\u00d73) = -9\/15<\/p>\n\n\n\n

2\/15 = (2 \u00d71) \/ (15 \u00d71) = 2\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-9\/15 + 2\/15 = (-9 + 2)\/15 = -7\/15<\/p>\n\n\n\n

-2\/-15 + -3\/5<\/p>\n\n\n\n

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

The denominators are 15 and 5<\/p>\n\n\n\n

By taking LCM for 15 and 5 is 15<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, 2\/15 = (2 \u00d71) \/ (15 \u00d71) = 2\/15<\/p>\n\n\n\n

-3\/5 = (-3 \u00d7 3) \/ (5\u00d73) = -9\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

2\/15 + -9\/15 = (2 + (-9))\/15 = (2-9)\/15 = -7\/15<\/p>\n\n\n\n

\u2234 -3\/5 + -2\/-15 = -2\/-15 + -3\/5 is satisfied.<\/p>\n\n\n\n

(iv) 2\/-7 and 12\/-35<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

2\/-7 = (2 \u00d7 -1)\/ (-7 \u00d7 -1) = -2\/7<\/p>\n\n\n\n

12\/-35 = (12 \u00d7 -1)\/ (-35 \u00d7 -1) = -12\/35<\/p>\n\n\n\n

By using the commutativity law, the addition of rational numbers is commutative.<\/p>\n\n\n\n

\u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

-2\/7 and -12\/35 as<\/p>\n\n\n\n

-2\/7 + -12\/35 and -12\/35 + -2\/7<\/p>\n\n\n\n

The denominators are 7 and 35<\/p>\n\n\n\n

By taking LCM for 7 and 35 is 35<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -2\/7 = (-2 \u00d7 5) \/ (7 \u00d75) = -10\/35<\/p>\n\n\n\n

-12\/35 = (-12 \u00d71) \/ (35 \u00d71) = -12\/35<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-10\/35 + (-12)\/35 = (-10 + (-12))\/35 = (-10-12)\/35 = -22\/35<\/p>\n\n\n\n

-12\/35 + -2\/7<\/p>\n\n\n\n

The denominators are 35 and 7<\/p>\n\n\n\n

By taking LCM for 35 and 7 is 35<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -12\/35 = (-12 \u00d71) \/ (35 \u00d71) = -12\/35<\/p>\n\n\n\n

-2\/7 = (-2 \u00d7 5) \/ (7 \u00d75) = -10\/35<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-12\/35 + -10\/35 = (-12 + (-10))\/35 = (-12-10)\/35 = -22\/35<\/p>\n\n\n\n

\u2234 -2\/7 + -12\/35 = -12\/35 + -2\/7 is satisfied.<\/p>\n\n\n\n

(v) 4 and -3\/5<\/strong><\/p>\n\n\n\n

Solution:<\/strong> By using the commutativity law, the addition of rational numbers is commutative.<\/p>\n\n\n\n

\u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

4\/1 and -3\/5 as<\/p>\n\n\n\n

4\/1 + -3\/5 and -3\/5 + 4\/1<\/p>\n\n\n\n

The denominators are 1 and 5<\/p>\n\n\n\n

By taking LCM for 1 and 5 is 5<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, 4\/1 = (4 \u00d7 5) \/ (1\u00d75) = 20\/5<\/p>\n\n\n\n

-3\/5 = (-3 \u00d71) \/ (5 \u00d71) = -3\/5<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

20\/5 + -3\/5 = (20 + (-3))\/5 = (20-3)\/5 = 17\/5<\/p>\n\n\n\n

-3\/5 + 4\/1<\/p>\n\n\n\n

The denominators are 5 and 1<\/p>\n\n\n\n

By taking LCM for 5 and 1 is 5<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -3\/5 = (-3 \u00d71) \/ (5 \u00d71) = -3\/5<\/p>\n\n\n\n

4\/1 = (4 \u00d7 5) \/ (1\u00d75) = 20\/5<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-3\/5 + 20\/5 = (-3 + 20)\/5 = 17\/5<\/p>\n\n\n\n

\u2234 4\/1 + -3\/5 = -3\/5 + 4\/1 is satisfied.<\/p>\n\n\n\n

(vi) -4 and 4\/-7<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly we need to convert the denominators to positive numbers.<\/p>\n\n\n\n

4\/-7 = (4 \u00d7 -1)\/ (-7 \u00d7 -1) = -4\/7<\/p>\n\n\n\n

By using the commutativity law, the addition of rational numbers is commutative.<\/p>\n\n\n\n

\u2234 a\/b + c\/d = c\/d + a\/b<\/p>\n\n\n\n

In order to verify the above property let us consider the given fraction<\/p>\n\n\n\n

-4\/1 and -4\/7 as<\/p>\n\n\n\n

-4\/1 + -4\/7 and -4\/7 + -4\/1<\/p>\n\n\n\n

The denominators are 1 and 7<\/p>\n\n\n\n

By taking LCM for 1 and 7 is 7<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -4\/1 = (-4 \u00d7 7) \/ (1\u00d77) = -28\/7<\/p>\n\n\n\n

-4\/7 = (-4 \u00d71) \/ (7 \u00d71) = -4\/7<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-28\/7 + -4\/7 = (-28 + (-4))\/7 = (-28-4)\/7 = -32\/7<\/p>\n\n\n\n

-4\/7 + -4\/1<\/p>\n\n\n\n

The denominators are 7 and 1<\/p>\n\n\n\n

By taking LCM for 7 and 1 is 7<\/p>\n\n\n\n

We rewrite the given fraction in order to get the same denominator<\/p>\n\n\n\n

Now, -4\/7 = (-4 \u00d71) \/ (7 \u00d71) = -4\/7<\/p>\n\n\n\n

-4\/1 = (-4 \u00d7 7) \/ (1\u00d77) = -28\/7<\/p>\n\n\n\n

Since the denominators are same we can add them directly<\/p>\n\n\n\n

-4\/7 + -28\/7 = (-4 + (-28))\/7 = (-4-28)\/7 = -32\/7<\/p>\n\n\n\n

\u2234 -4\/1 + -4\/7 = -4\/7 + -4\/1 is satisfied.<\/p>\n\n\n\n

2. Verify associativity of addition of rational numbers i.e., (x + y) + z = x + (y + z), when:<\/strong><\/p>\n\n\n\n

(i) x = \u00bd, y = 2\/3, z = -1\/5<\/strong><\/p>\n\n\n\n

Solution: <\/strong>As the property states (x + y) + z = x + (y + z)<\/strong><\/p>\n\n\n\n

Use the values as such,<\/p>\n\n\n\n

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

Let us consider LHS (1\/2 + 2\/3) + (-1\/5)<\/p>\n\n\n\n

Taking LCM for 2 and 3 is 6<\/p>\n\n\n\n

(1\u00d7 3)\/(2\u00d73) + (2\u00d72)\/(3\u00d72)<\/p>\n\n\n\n

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

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

3\/6 + 4\/6 = 7\/6<\/p>\n\n\n\n

7\/6 + (-1\/5)<\/p>\n\n\n\n

Taking LCM for 6 and 5 is 30<\/p>\n\n\n\n

(7\u00d75)\/(6\u00d75) + (-1\u00d76)\/(5\u00d76)<\/p>\n\n\n\n

35\/30 + (-6)\/30<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

(35+(-6))\/30 = (35-6)\/30 = 29\/30<\/p>\n\n\n\n

Let us consider RHS 1\/2 + (2\/3 + (-1\/5))<\/p>\n\n\n\n

Taking LCM for 3 and 5 is 15<\/p>\n\n\n\n

(2\/3 + (-1\/5)) = (2\u00d75)\/(3\u00d75) + (-1\u00d73)\/(5\u00d73)<\/p>\n\n\n\n

= 10\/15 + (-3)\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

10\/15 + (-3)\/15 = (10-3)\/15 = 7\/15<\/p>\n\n\n\n

1\/2 + 7\/15<\/p>\n\n\n\n

Taking LCM for 2 and 15 is 30<\/p>\n\n\n\n

1\/2 + 7\/15 = (1\u00d715)\/(2\u00d715) + (7\u00d72)\/(15\u00d72)<\/p>\n\n\n\n

= 15\/30 + 14\/30<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

= (15 + 14)\/30 = 29\/30<\/p>\n\n\n\n

\u2234 LHS = RHS associativity of addition of rational numbers is verified.<\/p>\n\n\n\n

(ii) x = -2\/5, y = 4\/3, z = -7\/10<\/strong><\/p>\n\n\n\n

Solution: <\/strong>As the property states (x + y) + z = x + (y + z)<\/strong><\/p>\n\n\n\n

Use the values as such,<\/p>\n\n\n\n

(-2\/5 + 4\/3) + (-7\/10) = -2\/5 + (4\/3 + (-7\/10))<\/p>\n\n\n\n

Let us consider LHS (-2\/5 + 4\/3) + (-7\/10)<\/p>\n\n\n\n

Taking LCM for 5 and 3 is 15<\/p>\n\n\n\n

(-2\u00d7 3)\/(5\u00d73) + (4\u00d75)\/(3\u00d75)<\/p>\n\n\n\n

-6\/15 + 20\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

-6\/15 + 20\/15= (-6+20)\/15 = 14\/15<\/p>\n\n\n\n

14\/15 + (-7\/10)<\/p>\n\n\n\n

Taking LCM for 15 and 10 is 30<\/p>\n\n\n\n

(14\u00d72)\/(15\u00d72) + (-7\u00d73)\/(10\u00d73)<\/p>\n\n\n\n

28\/30 + (-21)\/30<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

(28+(-21))\/30 = (28-21)\/30 = 7\/30<\/p>\n\n\n\n

Let us consider RHS -2\/5 + (4\/3 + (-7\/10))<\/p>\n\n\n\n

Taking LCM for 3 and 10 is 30<\/p>\n\n\n\n

(4\/3 + (-7\/10)) = (4\u00d710)\/(3\u00d710) + (-7\u00d73)\/(10\u00d73)<\/p>\n\n\n\n

= 40\/30 + (-21)\/30<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

40\/30 + (-21)\/30 = (40-21)\/30 = 19\/30<\/p>\n\n\n\n

-2\/5 + 19\/30<\/p>\n\n\n\n

Taking LCM for 5 and 30 is 30<\/p>\n\n\n\n

-2\/5 + 19\/30 = (-2\u00d76)\/(5\u00d76) + (19\u00d71)\/(30\u00d71)<\/p>\n\n\n\n

= -12\/30 + 19\/30<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

= (-12 + 19)\/30 = 7\/30<\/p>\n\n\n\n

\u2234 LHS = RHS associativity of addition of rational numbers is verified.<\/p>\n\n\n\n

(iii) x = -7\/11, y = 2\/-5, z = -3\/22<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly convert the denominators to positive numbers<\/p>\n\n\n\n

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

As the property states (x + y) + z = x + (y + z)<\/strong><\/p>\n\n\n\n

Use the values as such,<\/p>\n\n\n\n

(-7\/11 + -2\/5) + (-3\/22) = -7\/11 + (-2\/5 + (-3\/22))<\/p>\n\n\n\n

Let us consider LHS (-7\/11 + -2\/5) + (-3\/22)<\/p>\n\n\n\n

Taking LCM for 11 and 5 is 55<\/p>\n\n\n\n

(-7\u00d75)\/(11\u00d75) + (-2\u00d711)\/(5\u00d711)<\/p>\n\n\n\n

-35\/55 + -22\/55<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

-35\/55 + -22\/55 = (-35-22)\/55 = -57\/55<\/p>\n\n\n\n

-57\/55 + (-3\/22)<\/p>\n\n\n\n

Taking LCM for 55 and 22 is 110<\/p>\n\n\n\n

(-57\u00d72)\/(55\u00d72) + (-3\u00d75)\/(22\u00d75)<\/p>\n\n\n\n

-114\/110 + (-15)\/110<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

(-114+(-15))\/110 = (-114-15)\/110 = -129\/110<\/p>\n\n\n\n

Let us consider RHS -7\/11 + (-2\/5 + (-3\/22))<\/p>\n\n\n\n

Taking LCM for 5 and 22 is 110<\/p>\n\n\n\n

(-2\/5 + (-3\/22))= (-2\u00d722)\/(5\u00d722) + (-3\u00d75)\/(22\u00d75)<\/p>\n\n\n\n

= -44\/110 + (-15)\/110<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

-44\/110 + (-15)\/110 = (-44-15)\/110 = -59\/110<\/p>\n\n\n\n

-7\/11 + -59\/110<\/p>\n\n\n\n

Taking LCM for 11 and 110 is 110<\/p>\n\n\n\n

-7\/11 + -59\/110 = (-7\u00d710)\/(11\u00d710) + (-59\u00d71)\/(110\u00d71)<\/p>\n\n\n\n

= -70\/110 + -59\/110<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

= (-70 -59)\/110 = -129\/110<\/p>\n\n\n\n

\u2234 LHS = RHS associativity of addition of rational numbers is verified.<\/p>\n\n\n\n

(iv) x = -2, y = 3\/5, z = -4\/3<\/strong><\/p>\n\n\n\n

Solution: <\/strong>As the property states (x + y) + z = x + (y + z)<\/strong><\/p>\n\n\n\n

Use the values as such,<\/p>\n\n\n\n

(-2\/1 + 3\/5) + (-4\/3) = -2\/1 + (3\/5 + (-4\/3))<\/p>\n\n\n\n

Let us consider LHS (-2\/1 + 3\/5) + (-4\/3)<\/p>\n\n\n\n

Taking LCM for 1 and 5 is 5<\/p>\n\n\n\n

(-2\u00d75)\/(1\u00d75) + (3\u00d71)\/(5\u00d71)<\/p>\n\n\n\n

-10\/5 + 3\/5<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

-10\/5 + 3\/5= (-10+3)\/5 = -7\/5<\/p>\n\n\n\n

-7\/5 + (-4\/3)<\/p>\n\n\n\n

Taking LCM for 5 and 3 is 15<\/p>\n\n\n\n

(-7\u00d73)\/(5\u00d73) + (-4\u00d75)\/(3\u00d75)<\/p>\n\n\n\n

-21\/15 + (-20)\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

(-21+(-20))\/15 = (-21-20)\/15 = -41\/15<\/p>\n\n\n\n

Let us consider RHS -2\/1 + (3\/5 + (-4\/3))<\/p>\n\n\n\n

Taking LCM for 5 and 3 is 15<\/p>\n\n\n\n

(3\/5 + (-4\/3)) = (3\u00d73)\/(5\u00d73) + (-4\u00d75)\/(3\u00d75)<\/p>\n\n\n\n

= 9\/15 + (-20)\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

9\/15 + (-20)\/15 = (9-20)\/15 = -11\/15<\/p>\n\n\n\n

-2\/1 + -11\/15<\/p>\n\n\n\n

Taking LCM for 1 and 15 is 15<\/p>\n\n\n\n

-2\/1 + -11\/15 = (-2\u00d715)\/(1\u00d715) + (-11\u00d71)\/(15\u00d71)<\/p>\n\n\n\n

= -30\/15 + -11\/15<\/p>\n\n\n\n

Since the denominators are same we can add them directly,<\/p>\n\n\n\n

= (-30 -11)\/15 = -41\/15<\/p>\n\n\n\n

\u2234 LHS = RHS associativity of addition of rational numbers is verified.<\/p>\n\n\n\n

3. Write the additive of each of the following rational numbers:<\/strong><\/p>\n\n\n\n

(i) -2\/17<\/strong><\/p>\n\n\n\n

(ii) 3\/-11<\/strong><\/p>\n\n\n\n

(iii) -17\/5<\/strong><\/p>\n\n\n\n

(iv) -11\/-25<\/strong><\/p>\n\n\n\n

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

(i) The additive inverse of -2\/17 is 2\/17<\/p>\n\n\n\n

(ii) The additive inverse of 3\/-11 is 3\/11<\/p>\n\n\n\n

(iii) The additive inverse of -17\/5 is 17\/5<\/p>\n\n\n\n

(iv) The additive inverse of -11\/-25 is -11\/25<\/p>\n\n\n\n

4. Write the negative(additive) inverse of each of the following:<\/strong><\/p>\n\n\n\n

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

(ii) 7\/-9<\/strong><\/p>\n\n\n\n

(iii) -16\/13<\/strong><\/p>\n\n\n\n

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

(v) 0<\/strong><\/p>\n\n\n\n

(vi) 1<\/strong><\/p>\n\n\n\n

(vii) \u2013 1<\/strong><\/p>\n\n\n\n

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

(i) The negative (additive) inverse of -2\/5 is 2\/5<\/p>\n\n\n\n

(ii) The negative (additive) inverse of 7\/-9 is 7\/9<\/p>\n\n\n\n

(iii) The negative (additive) inverse of -16\/13 is 16\/13<\/p>\n\n\n\n

(iv) The negative (additive) inverse of -5\/1 is 5<\/p>\n\n\n\n

(v) The negative (additive) inverse of 0 is 0<\/p>\n\n\n\n

(vi) The negative (additive) inverse of 1 is -1<\/p>\n\n\n\n

(vii) The negative (additive) inverse of -1 is 1<\/p>\n\n\n\n

5. Using commutativity and associativity of addition of rational numbers, express each of the following as a rational number:<\/strong><\/p>\n\n\n\n

(i) 2\/5 + 7\/3 + -4\/5 + -1\/3<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

2\/5 + -4\/5 + 7\/3 + -1\/3<\/p>\n\n\n\n

Now the denominators which are same can be added directly.<\/p>\n\n\n\n

(2+(-4))\/5 + (7+(-1))\/3<\/p>\n\n\n\n

(2-4)\/5 + (7-1)\/3<\/p>\n\n\n\n

-2\/5 + 6\/3<\/p>\n\n\n\n

By taking LCM for 5 and 3 we get, 15<\/p>\n\n\n\n

(-2\u00d73)\/(5\u00d73) + (6\u00d75)\/(3\u00d75)<\/p>\n\n\n\n

-6\/15 + 30\/15<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(-6+30)\/15 = 24\/15<\/p>\n\n\n\n

Further can be divided by 3 we get,<\/p>\n\n\n\n

24\/15 = 8\/5<\/p>\n\n\n\n

(ii) 3\/7 + -4\/9 + -11\/7 + 7\/9<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

3\/7 + -11\/7 + -4\/9 + 7\/9<\/p>\n\n\n\n

Now the denominators which are same can be added directly.<\/p>\n\n\n\n

(3+ (-11))\/7 + (-4+ 7)\/9<\/p>\n\n\n\n

(3-11)\/7 + (-4+7)\/9<\/p>\n\n\n\n

-8\/7 + 3\/9<\/p>\n\n\n\n

-8\/7 + 1\/3<\/p>\n\n\n\n

By taking LCM for 7 and 3 we get, 21<\/p>\n\n\n\n

(-8\u00d73)\/ (7\u00d73) + (1\u00d77)\/ (3\u00d77)<\/p>\n\n\n\n

-24\/21 + 7\/21<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(-24+7)\/21 = -17\/21<\/p>\n\n\n\n

(iii) 2\/5 + 8\/3 + -11\/15 + 4\/5 + -2\/3<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

2\/5 + 4\/5 + 8\/3 + -2\/3 + -11\/15<\/p>\n\n\n\n

Now the denominators which are same can be added directly.<\/p>\n\n\n\n

(2 + 4)\/5 + (8 + (-2))\/3 + -11\/15<\/p>\n\n\n\n

6\/5 + (8-2)\/3 + -11\/15<\/p>\n\n\n\n

6\/5 + 6\/3 + -11\/15<\/p>\n\n\n\n

6\/5 + 2\/1 + -11\/15<\/p>\n\n\n\n

By taking LCM for 5, 1 and 15 we get, 15<\/p>\n\n\n\n

(6\u00d73)\/ (5\u00d73) + (2\u00d715)\/ (1\u00d715) + (-11\u00d71)\/ (15\u00d71)<\/p>\n\n\n\n

18\/15 + 30\/15 + -11\/15<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(18+30+ (-11))\/15 = (18+30-11)\/15 = 37\/15<\/p>\n\n\n\n

(iv) 4\/7 + 0 + -8\/9 + -13\/7 + 17\/21<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

4\/7 + -13\/7 + -8\/9 + 17\/21<\/p>\n\n\n\n

Now the denominators which are same can be added directly.<\/p>\n\n\n\n

(4 + (-13))\/7 + -8\/9 + 17\/21<\/p>\n\n\n\n

(4-13)\/7 + -8\/9 + 17\/21<\/p>\n\n\n\n

-9\/7 + -8\/9 + 17\/21<\/p>\n\n\n\n

By taking LCM for 7, 9 and 21 we get, 63<\/p>\n\n\n\n

(-9\u00d79)\/ (7\u00d79) + (-8\u00d77)\/ (9\u00d77) + (17\u00d73)\/ (21\u00d73)<\/p>\n\n\n\n

-81\/63 + -56\/63 + 51\/63<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(-81+(-56)+ 51)\/63 = (-81-56+51)\/63 = -86\/63<\/p>\n\n\n\n

6. Re-arrange suitably and find the sum in each of the following:<\/strong><\/p>\n\n\n\n

(i) 11\/12 + -17\/3 + 11\/2 + -25\/2<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

11\/12 + -17\/3 + (11-25)\/2<\/p>\n\n\n\n

11\/12 + -17\/3 + -14\/2<\/p>\n\n\n\n

By taking LCM for 12, 3 and 2 we get, 12<\/p>\n\n\n\n

(11\u00d71)\/(12\u00d71) + (-17\u00d74)\/(3\u00d74) + (-14\u00d76)\/(2\u00d76)<\/p>\n\n\n\n

11\/12 + -68\/12 + -84\/12<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(11-68-84)\/12 = -141\/12<\/p>\n\n\n\n

(ii)-6\/7 + -5\/6 + -4\/9 + -15\/7<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

-6\/7 + -15\/7 + -5\/6 + -4\/9<\/p>\n\n\n\n

(-6 -15)\/7 + -5\/6 + -4\/9<\/p>\n\n\n\n

-21\/7 + -5\/6 + -4\/9<\/p>\n\n\n\n

-3\/1 + -5\/6 + -4\/9<\/p>\n\n\n\n

By taking LCM for 1, 6 and 9 we get, 18<\/p>\n\n\n\n

(-3\u00d718)\/(1\u00d718) + (-5\u00d73)\/(6\u00d73) + (-4\u00d72)\/(9\u00d72)<\/p>\n\n\n\n

-54\/18 + -15\/18 + -8\/18<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(-54-15-8)\/18 = -77\/18<\/p>\n\n\n\n

(iii) 3\/5 + 7\/3 + 9\/ 5+ -13\/15 + -7\/3<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

3\/5 + 9\/5 + 7\/3 + -7\/3 + -13\/15<\/p>\n\n\n\n

(3+9)\/5 + -13\/15<\/p>\n\n\n\n

12\/5 + -13\/15<\/p>\n\n\n\n

By taking LCM for 5 and 15 we get, 15<\/p>\n\n\n\n

(12\u00d73)\/(5\u00d73) + (-13\u00d71)\/(15\u00d71)<\/p>\n\n\n\n

36\/15 + -13\/15<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(36-13)\/15 = 23\/15<\/p>\n\n\n\n

(iv) 4\/13 + -5\/8 + -8\/13 + 9\/13<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

4\/13 + -8\/13 + 9\/13 + -5\/8<\/p>\n\n\n\n

(4-8+9)\/13 + -5\/8<\/p>\n\n\n\n

5\/13 + -5\/8<\/p>\n\n\n\n

By taking LCM for 13 and 8 we get, 104<\/p>\n\n\n\n

(5\u00d78)\/(13\u00d78) + (-5\u00d713)\/(8\u00d713)<\/p>\n\n\n\n

40\/104 + -65\/104<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(40-65)\/104 = -25\/104<\/p>\n\n\n\n

(v) 2\/3 + -4\/5 + 1\/3 + 2\/5<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

2\/3 + 1\/3 + -4\/5 + 2\/5<\/p>\n\n\n\n

(2+1)\/3 + (-4+2)\/5<\/p>\n\n\n\n

3\/3 + -2\/5<\/p>\n\n\n\n

1\/1 + -2\/5<\/p>\n\n\n\n

By taking LCM for 1 and 5 we get, 5<\/p>\n\n\n\n

(1\u00d75)\/(1\u00d75) + (-2\u00d71)\/(5\u00d71)<\/p>\n\n\n\n

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

Since the denominators are same can be added directly<\/p>\n\n\n\n

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

(vi) 1\/8 + 5\/12 + 2\/7 + 7\/12 + 9\/7 + -5\/16<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Firstly group the rational numbers with same denominators<\/p>\n\n\n\n

1\/8 + 5\/12 + 7\/12 + 2\/7 + 9\/7 + -5\/16<\/p>\n\n\n\n

1\/8 + (5+7)\/12 + (2+9)\/7 + -5\/16<\/p>\n\n\n\n

1\/8 + 12\/12 + 11\/7 + -5\/16<\/p>\n\n\n\n

1\/8 + 1\/1 + 11\/7 + -5\/16<\/p>\n\n\n\n

By taking LCM for 8, 1, 7 and 16 we get, 112<\/p>\n\n\n\n

(1\u00d714)\/(8\u00d714) + (1\u00d7112)\/(1\u00d7112) + (11\u00d716)\/(7\u00d716) + (-5\u00d77)\/(16\u00d77)<\/p>\n\n\n\n

14\/112 + 112\/112 + 176\/112 + -35\/112<\/p>\n\n\n\n

Since the denominators are same can be added directly<\/p>\n\n\n\n

(14+112+176-35)\/112 = 267\/112<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.3 PAGE NO: 1.18<\/h4>\n\n\n\n

1. Subtract the first rational number from the second in each of the following:<\/strong><\/p>\n\n\n\n

(i) 3\/8, 5\/8<\/strong><\/p>\n\n\n\n

(ii) -7\/9, 4\/9<\/strong><\/p>\n\n\n\n

(iii) -2\/11, -9\/11<\/strong><\/p>\n\n\n\n

(iv) 11\/13, -4\/13<\/strong><\/p>\n\n\n\n

(v) \u00bc, -3\/8<\/strong><\/p>\n\n\n\n

(vi) -2\/3, 5\/6<\/strong><\/p>\n\n\n\n

(vii) -6\/7, -13\/14<\/strong><\/p>\n\n\n\n

(viii) -8\/33, -7\/22<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> let us subtract<\/p>\n\n\n\n

5\/8 \u2013 3\/8<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

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

Further we can divide by 2 we get,<\/p>\n\n\n\n

2\/8 = 1\/4<\/p>\n\n\n\n

(ii)<\/strong> let us subtract<\/p>\n\n\n\n

4\/9 \u2013 -7\/9<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(4+7)\/9 = 11\/9<\/p>\n\n\n\n

(iii)<\/strong> let us subtract<\/p>\n\n\n\n

-9\/11 \u2013 -2\/11<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(-9+2)\/11 = -7\/11<\/p>\n\n\n\n

(iv)<\/strong> let us subtract<\/p>\n\n\n\n

-4\/13 \u2013 11\/13<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(-4-11)\/13 = -15\/13<\/p>\n\n\n\n

(v)<\/strong> let us subtract<\/p>\n\n\n\n

-3\/8 \u2013 1\/4<\/p>\n\n\n\n

By taking LCM for 8 and 4 which is 8<\/p>\n\n\n\n

-3\/8 \u2013 1\/4 = (-3\u00d71)\/(8\u00d71) \u2013 (1\u00d72)\/(4\u00d72) = -3\/8 \u2013 2\/8<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

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

(vi)<\/strong> let us subtract<\/p>\n\n\n\n

5\/6 \u2013 -2\/3<\/p>\n\n\n\n

By taking LCM for 6 and 3 which is 6<\/p>\n\n\n\n

5\/6 \u2013 -2\/3 = (5\u00d71)\/(6\u00d71) \u2013 (-2\u00d72)\/(3\u00d72) = 5\/6 \u2013 -4\/6<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(5+4)\/6 = 9\/6<\/p>\n\n\n\n

Further we can divide by 3 we get,<\/p>\n\n\n\n

9\/6 = 3\/2<\/p>\n\n\n\n

(vii)<\/strong> let us subtract<\/p>\n\n\n\n

-13\/14 \u2013 -6\/7<\/p>\n\n\n\n

By taking LCM for 14 and 7 which is 14<\/p>\n\n\n\n

-13\/14 \u2013 -6\/7 = (-13\u00d71)\/(14\u00d71) \u2013 (-6\u00d72)\/(7\u00d72) = -13\/14 \u2013 -12\/14<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(-13+12)\/14 = -1\/14<\/p>\n\n\n\n

(viii)<\/strong> let us subtract<\/p>\n\n\n\n

-7\/22 \u2013 -8\/33<\/p>\n\n\n\n

By taking LCM for 22 and 33 which is 66<\/p>\n\n\n\n

-7\/22 \u2013 -8\/33 = (-7\u00d73)\/(22\u00d73) \u2013 (-8\u00d72)\/(33\u00d72) = -21\/66 \u2013 -16\/66<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(-21+16)\/66 = -5\/66<\/p>\n\n\n\n

2. Evaluate each of the following:<\/strong><\/p>\n\n\n\n

(i) 2\/3 \u2013 3\/5<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By taking LCM for 3 and 5 which is 15<\/p>\n\n\n\n

2\/3 \u2013 3\/5 = (2\u00d75 \u2013 3\u00d73)\/15<\/p>\n\n\n\n

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

(ii) -4\/7 \u2013 2\/-3<\/strong><\/p>\n\n\n\n

Solution:<\/strong> convert the denominator to positive number by multiplying by -1<\/p>\n\n\n\n

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

-4\/7 \u2013 -2\/3<\/p>\n\n\n\n

By taking LCM for 7 and 3 which is 21<\/p>\n\n\n\n

-4\/7 \u2013 -2\/3 = (-4\u00d73 \u2013 -2\u00d77)\/21<\/p>\n\n\n\n

= (-12+14)\/21<\/p>\n\n\n\n

= 2\/21<\/p>\n\n\n\n

(iii) 4\/7 \u2013 -5\/-7<\/strong><\/p>\n\n\n\n

Solution:<\/strong> convert the denominator to positive number by multiplying by -1<\/p>\n\n\n\n

-5\/-7 = 5\/7<\/p>\n\n\n\n

4\/7 \u2013 5\/7<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

(4-5)\/7 = -1\/7<\/p>\n\n\n\n

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

Solution: <\/strong>By taking LCM for 1 and 9 which is 9<\/p>\n\n\n\n

-2\/1 \u2013 5\/9 = (-2\u00d79 \u2013 5\u00d71)\/9<\/p>\n\n\n\n

= (-18 \u2013 5)\/9<\/p>\n\n\n\n

= -23\/9<\/p>\n\n\n\n

(v) -3\/-8 \u2013 -2\/7<\/strong><\/p>\n\n\n\n

Solution: <\/strong>convert the denominator to positive number by multiplying by -1<\/p>\n\n\n\n

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

3\/8 \u2013 -2\/7<\/p>\n\n\n\n

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

3\/8 \u2013 -2\/7 = (3\u00d77 \u2013 -2\u00d78)\/56<\/p>\n\n\n\n

= (21 + 16)\/56<\/p>\n\n\n\n

= 37\/56<\/p>\n\n\n\n

(vi) -4\/13 \u2013 -5\/26<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By taking LCM for 13 and 26 which is 26<\/p>\n\n\n\n

-4\/13 \u2013 -5\/26 = (-4\u00d72 \u2013 -5\u00d71)\/26<\/p>\n\n\n\n

= (-8 + 5)\/26<\/p>\n\n\n\n

= -3\/26<\/p>\n\n\n\n

(vii) -5\/14 \u2013 -2\/7<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By taking LCM for 14 and 7 which is 14<\/p>\n\n\n\n

-5\/14 \u2013 -2\/7 = (-5\u00d71 \u2013 -2\u00d72)\/14<\/p>\n\n\n\n

= (-5 + 4)\/14<\/p>\n\n\n\n

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

(viii) 13\/15 \u2013 12\/25<\/strong><\/p>\n\n\n\n

Solution:<\/strong> By taking LCM for 15 and 25 which is 75<\/p>\n\n\n\n

13\/15 \u2013 12\/25 = (13\u00d75 \u2013 12\u00d73)\/75<\/p>\n\n\n\n

= (65 \u2013 36)\/75<\/p>\n\n\n\n

= 29\/75<\/p>\n\n\n\n

(ix) -6\/13 \u2013 -7\/13<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Since the denominators are same we can subtract directly<\/p>\n\n\n\n

-6\/13 \u2013 -7\/13 = (-6+7)\/13<\/p>\n\n\n\n

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

(x) 7\/24 \u2013 19\/36<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By taking LCM for 24 and 36 which is 72<\/p>\n\n\n\n

7\/24 \u2013 19\/36 = (7\u00d73 \u2013 19\u00d72)\/72<\/p>\n\n\n\n

= (21 \u2013 38)\/72<\/p>\n\n\n\n

= -17\/72<\/p>\n\n\n\n

(xi) 5\/63 \u2013 -8\/21<\/strong><\/p>\n\n\n\n

Solution: <\/strong>By taking LCM for 63 and 21 which is 63<\/p>\n\n\n\n

5\/63 \u2013 -8\/21 = (5\u00d71 \u2013 -8\u00d73)\/63<\/p>\n\n\n\n

= (5 + 24)\/63<\/p>\n\n\n\n

= 29\/63<\/p>\n\n\n\n

3. The sum of the two numbers is 5\/9. If one of the numbers is 1\/3, find the other.<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us note down the given details<\/p>\n\n\n\n

Sum of two numbers = 5\/9<\/p>\n\n\n\n

One of the number = 1\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Other number = sum of number \u2013 given number<\/p>\n\n\n\n

= 5\/9 \u2013 1\/3<\/p>\n\n\n\n

By taking LCM for 9 and 3 which is 9<\/p>\n\n\n\n

5\/9 \u2013 1\/3 = (5\u00d71 \u2013 1\u00d73)\/9<\/p>\n\n\n\n

= (5 \u2013 3)\/9<\/p>\n\n\n\n

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

\u2234 the other number is 2\/9<\/p>\n\n\n\n

4. The sum of the two numbers is -1\/3. If one of the numbers is -12\/3, find the other.<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us note down the given details<\/p>\n\n\n\n

Sum of two numbers = -1\/3<\/p>\n\n\n\n

One of the number = -12\/3<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Other number = sum of number \u2013 given number<\/p>\n\n\n\n

= -1\/3 \u2013 -12\/3<\/p>\n\n\n\n

Since the denominators are same we can subtract directly<\/p>\n\n\n\n

= (-1+12)\/3 = 11\/3<\/p>\n\n\n\n

\u2234 the other number is 11\/3<\/p>\n\n\n\n

5. The sum of the two numbers is -4\/3. If one of the numbers is -5, find the other.<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us note down the given details<\/p>\n\n\n\n

Sum of two numbers = -4\/3<\/p>\n\n\n\n

One of the number = -5\/1<\/p>\n\n\n\n

By using the formula,<\/p>\n\n\n\n

Other number = sum of number \u2013 given number<\/p>\n\n\n\n

= -4\/3 \u2013 -5\/1<\/p>\n\n\n\n

By taking LCM for 3 and 1 which is 3<\/p>\n\n\n\n

-4\/3 \u2013 -5\/1 = (-4\u00d71 \u2013 -5\u00d73)\/3<\/p>\n\n\n\n

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

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

\u2234 the other number is 11\/3<\/p>\n\n\n\n

6. The sum of the two rational numbers is -8. If one of the numbers is -15\/7, find the other.<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us note down the given details<\/p>\n\n\n\n

Sum of two rational numbers = -8\/1<\/p>\n\n\n\n

One of the number = -15\/7<\/p>\n\n\n\n

Let us consider the other number as x<\/p>\n\n\n\n

x + -15\/7 = -8<\/p>\n\n\n\n

(7x -15)\/7 = -8<\/p>\n\n\n\n

7x -15 = -8\u00d77<\/p>\n\n\n\n

7x \u2013 15 = -56<\/p>\n\n\n\n

7x = -56+15<\/p>\n\n\n\n

x = -41\/7<\/p>\n\n\n\n

\u2234 the other number is -41\/7<\/p>\n\n\n\n

7. What should be added to -7\/8 so as to get 5\/9?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be added to -7\/8 to get 5\/9<\/p>\n\n\n\n

So, -7\/8 + x = 5\/9<\/p>\n\n\n\n

(-7 + 8x)\/8 = 5\/9<\/p>\n\n\n\n

(-7 + 8x) \u00d7 9 = 5 \u00d7 8<\/p>\n\n\n\n

-63 + 72x = 40<\/p>\n\n\n\n

72x = 40 + 63<\/p>\n\n\n\n

x = 103\/72<\/p>\n\n\n\n

\u2234 the required number is 103\/72<\/p>\n\n\n\n

8. What number should be added to -5\/11 so as to get 26\/33?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be added to -5\/11 to get 26\/33<\/p>\n\n\n\n

So, -5\/11 + x = 26\/33<\/p>\n\n\n\n

x = 26\/33 + 5\/11<\/p>\n\n\n\n

let us take LCM for 33 and 11 which is 33<\/p>\n\n\n\n

x = (26\u00d71 + 5\u00d73)\/33<\/p>\n\n\n\n

= (26 + 15)\/33<\/p>\n\n\n\n

= 41\/33<\/p>\n\n\n\n

\u2234 the required number is 41\/33<\/p>\n\n\n\n

9. What number should be added to -5\/7 to get -2\/3?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be added to -5\/7 to get -2\/3<\/p>\n\n\n\n

So, -5\/7 + x = -2\/3<\/p>\n\n\n\n

x = -2\/3 + 5\/7<\/p>\n\n\n\n

let us take LCM for 3 and 7 which is 21<\/p>\n\n\n\n

x = (-2\u00d77 + 5\u00d73)\/21<\/p>\n\n\n\n

= (-14 + 15)\/21<\/p>\n\n\n\n

= 1\/21<\/p>\n\n\n\n

\u2234 the required number is 1\/21<\/p>\n\n\n\n

10. What number should be subtracted from -5\/3 to get 5\/6?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be subtracted from -5\/3 to get 5\/6<\/p>\n\n\n\n

So, -5\/3 \u2013 x = 5\/6<\/p>\n\n\n\n

x = -5\/3 \u2013 5\/6<\/p>\n\n\n\n

let us take LCM for 3 and 6 which is 6<\/p>\n\n\n\n

x = (-5\u00d72 \u2013 5\u00d71)\/6<\/p>\n\n\n\n

= (-10 \u2013 5)\/6<\/p>\n\n\n\n

= -15\/6<\/p>\n\n\n\n

Further we can divide by 3 we get,<\/p>\n\n\n\n

-15\/6 = -5\/2<\/p>\n\n\n\n

\u2234 the required number is -5\/2<\/p>\n\n\n\n

11. What number should be subtracted from 3\/7 to get 5\/4?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be subtracted from 3\/7 to get 5\/4<\/p>\n\n\n\n

So, 3\/7 \u2013 x = 5\/4<\/p>\n\n\n\n

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

let us take LCM for 7 and 4 which is 28<\/p>\n\n\n\n

x = (3\u00d74 \u2013 5\u00d77)\/28<\/p>\n\n\n\n

= (12 \u2013 35)\/28<\/p>\n\n\n\n

= -23\/28<\/p>\n\n\n\n

\u2234 the required number is -23\/28<\/p>\n\n\n\n

12. What should be added to (2\/3 + 3\/5) to get -2\/15?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be added to (2\/3 + 3\/5) to get -2\/15<\/p>\n\n\n\n

x + (2\/3 + 3\/5) = -2\/15<\/p>\n\n\n\n

By taking LCM of 3 and 5 which is 15 we get,<\/p>\n\n\n\n

(15x + 2\u00d75 + 3\u00d73)15 = -2\/15<\/p>\n\n\n\n

15x + 10 + 9 = -2<\/p>\n\n\n\n

15x = -2-19<\/p>\n\n\n\n

x = -21\/15<\/p>\n\n\n\n

Further we can divide by 3 we get,<\/p>\n\n\n\n

-21\/15 = -7\/5<\/p>\n\n\n\n

\u2234 the required number is -7\/5<\/p>\n\n\n\n

13. What should be added to (1\/2 + 1\/3 + 1\/5) to get 3?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be added to (1\/2 + 1\/3 + 1\/5) to get 3<\/p>\n\n\n\n

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

By taking LCM of 2, 3 and 5 which is 30 we get,<\/p>\n\n\n\n

(30x + 1\u00d715 + 1\u00d710 + 1\u00d76 )30 = 3<\/p>\n\n\n\n

30x + 15 + 10 + 6 = 3 \u00d7 30<\/p>\n\n\n\n

30x + 31 = 90<\/p>\n\n\n\n

30x = 90-31<\/p>\n\n\n\n

x = 59\/30<\/p>\n\n\n\n

\u2234 the required number is 59\/30<\/p>\n\n\n\n

14. What number should be subtracted from (3\/4 \u2013 2\/3) to get -1\/6?<\/strong><\/p>\n\n\n\n

Solution: <\/strong>Let us consider a number as x to be subtracted from (3\/4 \u2013 2\/3) to get -1\/6<\/p>\n\n\n\n

So, (3\/4 \u2013 2\/3) \u2013 x = -1\/6<\/p>\n\n\n\n

x = 3\/4 \u2013 2\/3 + 1\/6<\/p>\n\n\n\n

Let us take LCM for 4 and 3 which is 12<\/p>\n\n\n\n

x = (3\u00d73 \u2013 2\u00d74)\/12 + 1\/6<\/p>\n\n\n\n

= (9 \u2013 8)\/12 + 1\/6<\/p>\n\n\n\n

= 1\/12 + 1\/6<\/p>\n\n\n\n

Let us take LCM for 12 and 6 which is 12<\/p>\n\n\n\n

= (1\u00d71 + 1\u00d72)\/12<\/p>\n\n\n\n

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

Further we can divide by 3 we get,<\/p>\n\n\n\n

3\/12 = 1\/4 \u2234 the required number is \u00bc<\/p>\n\n\n\n

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

(i) -4\/13 \u2013 -3\/26 = \u2026.<\/strong><\/p>\n\n\n\n

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

-4\/13 \u2013 -3\/26<\/p>\n\n\n\n

Let us take LCM for 13 and 26 which is 26<\/p>\n\n\n\n

(-4\u00d72 + 3\u00d71)\/26<\/p>\n\n\n\n

(-8+3)\/26 = -5\/26<\/p>\n\n\n\n

(ii) -9\/14 + \u2026. = -1<\/strong><\/p>\n\n\n\n

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

Let us consider the number to be added as x<\/p>\n\n\n\n

-9\/14 + x = -1<\/p>\n\n\n\n

x = -1 + 9\/14<\/p>\n\n\n\n

By taking LCM as 14 we get,<\/p>\n\n\n\n

x = (-1\u00d714 + 9)\/14<\/p>\n\n\n\n

= (-14+9)\/14<\/p>\n\n\n\n

= -5\/14<\/p>\n\n\n\n

(iii) -7\/9 + \u2026. =3<\/strong><\/p>\n\n\n\n

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

Let us consider the number to be added as x<\/p>\n\n\n\n

-7\/9 + x = 3<\/p>\n\n\n\n

x = 3 + 7\/9<\/p>\n\n\n\n

By taking LCM as 9 we get,<\/p>\n\n\n\n

x = (3\u00d79 + 7)\/9<\/p>\n\n\n\n

= (27 + 7)\/9<\/p>\n\n\n\n

= 34\/9<\/p>\n\n\n\n

(iv) \u2026 + 15\/23 = 4<\/strong><\/p>\n\n\n\n

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

Let us consider the number to be added as x<\/p>\n\n\n\n

x + 15\/23 = 4<\/p>\n\n\n\n

x = 4 \u2013 15\/23<\/p>\n\n\n\n

By taking LCM as 23 we get,<\/p>\n\n\n\n

x = (4\u00d723 \u2013 15)\/23<\/p>\n\n\n\n

= (92 \u2013 15)\/23<\/p>\n\n\n\n

= 77\/23<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.4 PAGE NO: 1.22<\/h4>\n\n\n\n

1. Simplify each of the following and write as a rational number of the form p\/q:<\/strong><\/p>\n\n\n\n

(i) 3\/4 + 5\/6 + -7\/8<\/strong><\/p>\n\n\n\n

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

3\/4 + 5\/6 -7\/8<\/p>\n\n\n\n

By taking LCM for 4, 6 and 8 which is 24<\/p>\n\n\n\n

((3\u00d76) + (5\u00d74) \u2013 (7\u00d73))\/24<\/p>\n\n\n\n

(18 + 20 \u2013 21)\/24<\/p>\n\n\n\n

(38-21)\/24<\/p>\n\n\n\n

17\/24<\/p>\n\n\n\n

(ii) 2\/3 + -5\/6 + -7\/9<\/strong><\/p>\n\n\n\n

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

2\/3 + -5\/6 + -7\/9<\/p>\n\n\n\n

By taking LCM for 3, 6 and 9 which is 18<\/p>\n\n\n\n

((2\u00d76) + (-5\u00d73) + (-7\u00d72))\/18<\/p>\n\n\n\n

(12 \u2013 15 \u2013 14)\/18<\/p>\n\n\n\n

-17\/18<\/p>\n\n\n\n

(iii) -11\/2 + 7\/6 + -5\/8<\/strong><\/p>\n\n\n\n

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

-11\/2 + 7\/6 + -5\/8<\/p>\n\n\n\n

By taking LCM for 2, 6 and 8 which is 24<\/p>\n\n\n\n

((-11\u00d712) + (7\u00d74) + (-5\u00d73))\/24<\/p>\n\n\n\n

(-132 + 28 \u2013 15)\/24<\/p>\n\n\n\n

-119\/24<\/p>\n\n\n\n

(iv) -4\/5 + -7\/10 + -8\/15<\/strong><\/p>\n\n\n\n

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

-4\/5 + -7\/10 + -8\/15<\/p>\n\n\n\n

By taking LCM for 5, 10 and 15 which is 30<\/p>\n\n\n\n

((-4\u00d76) + (-7\u00d73) + (-8\u00d72))\/30<\/p>\n\n\n\n

(-24 \u2013 21 \u2013 16)\/30<\/p>\n\n\n\n

-61\/30<\/p>\n\n\n\n

(v) -9\/10 + 22\/15 + 13\/-20<\/strong><\/p>\n\n\n\n

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

-9\/10 + 22\/15 + 13\/-20<\/p>\n\n\n\n

By taking LCM for 10, 15 and 20 which is 60<\/p>\n\n\n\n

((-9\u00d76) + (22\u00d74) + (-13\u00d73))\/60<\/p>\n\n\n\n

(-54 + 88 \u2013 39)\/60<\/p>\n\n\n\n

-5\/60 = -1\/12<\/p>\n\n\n\n

(vi) 5\/3 + 3\/-2 + -7\/3 +3<\/strong><\/p>\n\n\n\n

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

5\/3 + 3\/-2 + -7\/3 +3<\/p>\n\n\n\n

By taking LCM for 3, 2, 3 and 1 which is 6<\/p>\n\n\n\n

((5\u00d72) + (-3\u00d73) + (-7\u00d72) + (3\u00d76))\/6<\/p>\n\n\n\n

(10 \u2013 9 \u2013 14 + 18)\/6<\/p>\n\n\n\n

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

2. Express each of the following as a rational number of the form p\/q:<\/strong><\/p>\n\n\n\n

(i) -8\/3 + -1\/4 + -11\/6 + 3\/8 \u2013 3<\/strong><\/p>\n\n\n\n

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

-8\/3 + -1\/4 + -11\/6 + 3\/8 \u2013 3<\/p>\n\n\n\n

By taking LCM for 3, 4, 6, 8 and 1 which is 24<\/p>\n\n\n\n

((-8\u00d78) + (-1\u00d76) + (-11\u00d74) + (3\u00d73) \u2013 (3\u00d724))\/24<\/p>\n\n\n\n

(-64 \u2013 6 \u2013 44 + 9 \u2013 72)\/24<\/p>\n\n\n\n

-177\/24<\/p>\n\n\n\n

Further divide by 3 we get,<\/p>\n\n\n\n

-177\/24 = -59\/8<\/p>\n\n\n\n

(ii) 6\/7 + 1 + -7\/9 + 19\/21 + -12\/7<\/strong><\/p>\n\n\n\n

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

6\/7 + 1 + -7\/9 + 19\/21 + -12\/7<\/p>\n\n\n\n

By taking LCM for 7, 1, 9, 21 and 7 which is 63<\/p>\n\n\n\n

((6\u00d79) + (1\u00d763) + (-7\u00d77) + (19\u00d73) + (-12\u00d79))\/63<\/p>\n\n\n\n

(54 + 63 \u2013 49 + 57 \u2013 108)\/63<\/p>\n\n\n\n

17\/63<\/p>\n\n\n\n

(iii) 15\/2 + 9\/8 + -11\/3 + 6 + -7\/6<\/strong><\/p>\n\n\n\n

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

15\/2 + 9\/8 + -11\/3 + 6 + -7\/6<\/p>\n\n\n\n

By taking LCM for 2, 8, 3, 1 and 6 which is 24<\/p>\n\n\n\n

((15\u00d712) + (9\u00d73) + (-11\u00d78) + (6\u00d724) + (-7\u00d74))\/24<\/p>\n\n\n\n

(180 + 27 \u2013 88 + 144 \u2013 28)\/24<\/p>\n\n\n\n

235\/24<\/p>\n\n\n\n

(iv) -7\/4 +0 + -9\/5 + 19\/10 + 11\/14<\/strong><\/p>\n\n\n\n

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

-7\/4 +0 + -9\/5 + 19\/10 + 11\/14<\/p>\n\n\n\n

By taking LCM for 4, 5, 10 and 14 which is 140<\/p>\n\n\n\n

((-7\u00d735) + (-9\u00d728) + (19\u00d714) + (11\u00d710))\/140<\/p>\n\n\n\n

(-245 \u2013 252 + 266 + 110)\/140<\/p>\n\n\n\n

-121\/140<\/p>\n\n\n\n

(v) -7\/4 +5\/3 + -1\/2 + -5\/6 + 2<\/strong><\/p>\n\n\n\n

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

-7\/4 +5\/3 + -1\/2 + -5\/6 + 2<\/p>\n\n\n\n

By taking LCM for 4, 3, 2, 6 and 1 which is 12<\/p>\n\n\n\n

((-7\u00d73) + (5\u00d74) + (-1\u00d76) + (-5\u00d72) + (2\u00d712))\/12<\/p>\n\n\n\n

(-21 + 20 \u2013 6 \u2013 10 + 24)\/12<\/p>\n\n\n\n

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

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

(i) -3\/2 + 5\/4 \u2013 7\/4<\/strong><\/p>\n\n\n\n

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

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

By taking LCM for 2 and 4 which is 4<\/p>\n\n\n\n

((-3\u00d72) + (5\u00d71) \u2013 (7\u00d71))\/4<\/p>\n\n\n\n

(-6 + 5 \u2013 7)\/4<\/p>\n\n\n\n

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

Further divide by 2 we get,<\/p>\n\n\n\n

-8\/2 = -2<\/p>\n\n\n\n

(ii) 5\/3 \u2013 7\/6 + -2\/3<\/strong><\/p>\n\n\n\n

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

5\/3 \u2013 7\/6 + -2\/3<\/p>\n\n\n\n

By taking LCM for 3 and 6 which is 6<\/p>\n\n\n\n

((5\u00d72) \u2013 (7\u00d71) + (-2\u00d72))\/6<\/p>\n\n\n\n

(10 \u2013 7 \u2013 4)\/6<\/p>\n\n\n\n

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

(iii) 5\/4 \u2013 7\/6 \u2013 -2\/3<\/strong><\/p>\n\n\n\n

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

5\/4 \u2013 7\/6 \u2013 -2\/3<\/p>\n\n\n\n

By taking LCM for 4, 6 and 3 which is 12<\/p>\n\n\n\n

((5\u00d73) \u2013 (7\u00d72) \u2013 (-2\u00d74))\/12<\/p>\n\n\n\n

(15 \u2013 14 + 8)\/12<\/p>\n\n\n\n

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

Further can divide by 3 we get,<\/p>\n\n\n\n

9\/12 = 3\/4<\/p>\n\n\n\n

(iv) -2\/5 \u2013 -3\/10 \u2013 -4\/7<\/strong><\/p>\n\n\n\n

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

-2\/5 \u2013 -3\/10 \u2013 -4\/7<\/p>\n\n\n\n

By taking LCM for 5, 10 and 7 which is 70<\/p>\n\n\n\n

((-2\u00d714) \u2013 (-3\u00d77) \u2013 (-4\u00d710))\/70<\/p>\n\n\n\n

(-28 + 21 + 40)\/70<\/p>\n\n\n\n

33\/70<\/p>\n\n\n\n

(v) 5\/6 + -2\/5 \u2013 -2\/15<\/strong><\/p>\n\n\n\n

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

5\/6 + -2\/5 \u2013 -2\/15<\/p>\n\n\n\n

By taking LCM for 6, 5 and 15 which is 30<\/p>\n\n\n\n

((5\u00d75) + (-2\u00d76) \u2013 (-2\u00d72))\/30<\/p>\n\n\n\n

(25 \u2013 12 + 4)\/30<\/p>\n\n\n\n

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

(vi) 3\/8 \u2013 -2\/9 + -5\/36<\/strong><\/p>\n\n\n\n

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

3\/8 \u2013 -2\/9 + -5\/36<\/p>\n\n\n\n

By taking LCM for 8, 9 and 36 which is 72<\/p>\n\n\n\n

((3\u00d79) \u2013 (-2\u00d78) + (-5\u00d72))\/72<\/p>\n\n\n\n

(27 + 16 \u2013 10)\/72<\/p>\n\n\n\n

33\/72<\/p>\n\n\n\n

Further can divide by 3 we get,<\/p>\n\n\n\n

33\/72 = 11\/24<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.5 PAGE NO: 1.25<\/h3>\n\n\n\n

1. Multiply:<\/strong><\/p>\n\n\n\n

(i) 7\/11 by 5\/4<\/strong><\/p>\n\n\n\n

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

7\/11 by 5\/4<\/p>\n\n\n\n

(7\/11) \u00d7 (5\/4) = (7\u00d75)\/(11\u00d74)<\/p>\n\n\n\n

= 35\/44<\/p>\n\n\n\n

(ii) 5\/7 by -3\/4<\/strong><\/p>\n\n\n\n

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

5\/7 by -3\/4<\/p>\n\n\n\n

(5\/7) \u00d7 (-3\/4) = (5\u00d7-3)\/(7\u00d74)<\/p>\n\n\n\n

= -15\/28<\/p>\n\n\n\n

(iii) -2\/9 by 5\/11<\/strong><\/p>\n\n\n\n

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

-2\/9 by 5\/11<\/p>\n\n\n\n

(-2\/9) \u00d7 (5\/11) = (-2\u00d75)\/(9\u00d711)<\/p>\n\n\n\n

= -10\/99<\/p>\n\n\n\n

(iv) -3\/17 by -5\/-4<\/strong><\/p>\n\n\n\n

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

-3\/17 by -5\/-4<\/p>\n\n\n\n

(-3\/17) \u00d7 (-5\/-4) = (-3\u00d7-5)\/(17\u00d7-4)<\/p>\n\n\n\n

= 15\/-68<\/p>\n\n\n\n

= -15\/68<\/p>\n\n\n\n

(v) 9\/-7 by 36\/-11<\/strong><\/p>\n\n\n\n

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

9\/-7 by 36\/-11<\/p>\n\n\n\n

(9\/-7) \u00d7 (36\/-11) = (9\u00d736)\/(-7\u00d7-11)<\/p>\n\n\n\n

= 324\/77<\/p>\n\n\n\n

(vi) -11\/13 by -21\/7<\/strong><\/p>\n\n\n\n

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

-11\/13 by -21\/7<\/p>\n\n\n\n

(-11\/13) \u00d7 (-21\/7) = (-11\u00d7-21)\/(13\u00d77)<\/p>\n\n\n\n

= 231\/91 = 33\/13<\/p>\n\n\n\n

(vii) -3\/5 by -4\/7<\/strong><\/p>\n\n\n\n

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

-3\/5 by -4\/7<\/p>\n\n\n\n

(-3\/5) \u00d7 (-4\/7) = (-3\u00d7-4)\/(5\u00d77)<\/p>\n\n\n\n

= 12\/35<\/p>\n\n\n\n

(viii) -15\/11 by 7<\/strong><\/p>\n\n\n\n

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

-15\/11 by 7<\/p>\n\n\n\n

(-15\/11) \u00d7 7 = (-15\u00d77)\/11<\/p>\n\n\n\n

= -105\/11<\/p>\n\n\n\n

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

(i) -5\/17 by 51\/-60<\/strong><\/p>\n\n\n\n

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

-5\/17 by 51\/-60<\/p>\n\n\n\n

(-5\/17) \u00d7 (51\/-60) = (-5\u00d751)\/(17\u00d7-60)<\/p>\n\n\n\n

= -255\/-1020<\/p>\n\n\n\n

Further can divide by 255 we get,<\/p>\n\n\n\n

-255\/-1020 = 1\/4<\/p>\n\n\n\n

(ii) -6\/11 by -55\/36<\/strong><\/p>\n\n\n\n

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

-6\/11 by -55\/36<\/p>\n\n\n\n

(-6\/11) \u00d7 (-55\/36) = (-6\u00d7-55)\/(11\u00d736)<\/p>\n\n\n\n

= 330\/396<\/p>\n\n\n\n

Further can divide by 66 we get,<\/p>\n\n\n\n

330\/396 = 5\/6<\/p>\n\n\n\n

(iii) -8\/25 by -5\/16<\/strong><\/p>\n\n\n\n

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

-8\/25 by -5\/16<\/p>\n\n\n\n

(-8\/25) \u00d7 (-5\/16) = (-8\u00d7-5)\/(25\u00d716)<\/p>\n\n\n\n

= 40\/400<\/p>\n\n\n\n

Further can divide by 40 we get,<\/p>\n\n\n\n

40\/400 = 1\/10<\/p>\n\n\n\n

(iv) 6\/7 by -49\/36<\/strong><\/p>\n\n\n\n

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

6\/7 by -49\/36<\/p>\n\n\n\n

(6\/7) \u00d7 (-49\/36) = (6\u00d7-49)\/(7\u00d736)<\/p>\n\n\n\n

= 294\/252<\/p>\n\n\n\n

Further can divide by 42 we get,<\/p>\n\n\n\n

294\/252 = -7\/6<\/p>\n\n\n\n

(v) 8\/-9 by -7\/-16<\/strong><\/p>\n\n\n\n

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

8\/-9 by -7\/-16<\/p>\n\n\n\n

(8\/-9) \u00d7 (-7\/-16) = (8\u00d7-7)\/(-9\u00d7-16)<\/p>\n\n\n\n

= -56\/144<\/p>\n\n\n\n

Further can divide by 8 we get,<\/p>\n\n\n\n

-56\/144 = -7\/18<\/p>\n\n\n\n

(vi) -8\/9 by 3\/64<\/strong><\/p>\n\n\n\n

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

-8\/9 by 3\/64<\/p>\n\n\n\n

(-8\/9) \u00d7 (3\/64) = (-8\u00d73)\/(9\u00d764)<\/p>\n\n\n\n

= -24\/576<\/p>\n\n\n\n

Further can divide by 24 we get,<\/p>\n\n\n\n

-24\/576 = -1\/24<\/p>\n\n\n\n

3. Simplify each of the following and express the result as a rational number in standard form:<\/strong><\/p>\n\n\n\n

(i) (-16\/21) \u00d7 (14\/5)<\/strong><\/p>\n\n\n\n

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

(-16\/21) \u00d7 (14\/5) = (-16\/3) \u00d7 (2\/5) (divisible by 7)<\/p>\n\n\n\n

= (-16\u00d72)\/(3\u00d75)<\/p>\n\n\n\n

= -32\/15<\/p>\n\n\n\n

(ii) (7\/6) \u00d7 (-3\/28)<\/strong><\/p>\n\n\n\n

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

(7\/6) \u00d7 (-3\/28) = (1\/2) \u00d7 (-1\/4) (divisible by 7 and 3)<\/p>\n\n\n\n

= -1\/8<\/p>\n\n\n\n

(iii) (-19\/36) \u00d7 16<\/strong><\/p>\n\n\n\n

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

-19\/36 \u00d7 16 = (-19\/9) \u00d7 4 (divisible by 4)<\/p>\n\n\n\n

= (-19\u00d74)\/9 = -76\/9<\/p>\n\n\n\n

(iv) (-13\/9) \u00d7 (27\/-26)<\/strong><\/p>\n\n\n\n

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

(-13\/9) \u00d7 (27\/-26) = (-1\/1) \u00d7 (3\/-2) (divisible by 13 and 9)<\/p>\n\n\n\n

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

(v) (-9\/16) \u00d7 (-64\/-27)<\/strong><\/p>\n\n\n\n

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

(-9\/16) \u00d7 (-64\/-27) = (-1\/1) \u00d7 (-4\/-3) (divisible by 9 and 16)<\/p>\n\n\n\n

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

(vi) (-50\/7) \u00d7 (14\/3)<\/strong><\/p>\n\n\n\n

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

(-50\/7) \u00d7 (14\/3) = (-50\/1) \u00d7 (2\/3) (divisible by 7)<\/p>\n\n\n\n

= (-50\u00d72)\/(1\u00d73)<\/p>\n\n\n\n

= -100\/3<\/p>\n\n\n\n

(vii) (-11\/9) \u00d7 (-81\/-88)<\/strong><\/p>\n\n\n\n

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

(-11\/9) \u00d7 (-81\/-88) = (-1\/1) \u00d7 (-9\/-8) (divisible by 11 and 9)<\/p>\n\n\n\n

= (-1\u00d7-9)\/(1\u00d7-8)<\/p>\n\n\n\n

= 9\/-8 = -9\/8<\/p>\n\n\n\n

(viii) (-5\/9) \u00d7 (72\/-25)<\/strong><\/p>\n\n\n\n

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

(-5\/9) \u00d7 (72\/-25) = (-1\/1) \u00d7 (8\/-5) (divisible by 5 and 9)<\/p>\n\n\n\n

= (-1\u00d78)\/(1\u00d7-5)<\/p>\n\n\n\n

= -8\/-5 = 8\/5<\/p>\n\n\n\n

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

(i) ((25\/8) \u00d7 (2\/5)) \u2013 ((3\/5) \u00d7 (-10\/9))<\/strong><\/p>\n\n\n\n

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

((25\/8) \u00d7 (2\/5)) \u2013 ((3\/5) \u00d7 (-10\/9)) = (25\u00d72)\/(8\u00d75) \u2013 (3\u00d7-10)\/(5\u00d79)<\/p>\n\n\n\n

= 50\/40 \u2013 -30\/45<\/p>\n\n\n\n

= 5\/4 + 2\/3 (divisible by 5 and 3)<\/p>\n\n\n\n

By taking LCM for 4 and 3 which is 12<\/p>\n\n\n\n

= ((5\u00d73) + (2\u00d74))\/12<\/p>\n\n\n\n

= (15+8)\/12<\/p>\n\n\n\n

= 23\/12<\/p>\n\n\n\n

(ii) ((1\/2) \u00d7 (1\/4)) + ((1\/2) \u00d7 6)<\/strong><\/p>\n\n\n\n

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

((1\/2) \u00d7 (1\/4)) + ((1\/2) \u00d7 6) <\/strong>= (1\u00d71)\/(2\u00d74) + (1\u00d73) (divisible by 2)<\/p>\n\n\n\n

= 1\/8 +3<\/p>\n\n\n\n

By taking LCM for 8 and 1 which is 8<\/p>\n\n\n\n

= ((1\u00d71) + (3\u00d78))\/8<\/p>\n\n\n\n

= (1+24)\/8<\/p>\n\n\n\n

= 25\/8<\/p>\n\n\n\n

(iii) (-5 \u00d7 (2\/15)) \u2013 (-6 \u00d7 (2\/9))<\/strong><\/p>\n\n\n\n

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

(-5 \u00d7 (2\/15)) \u2013 (-6 \u00d7 (2\/9)) = (-1 \u00d7 (2\/3)) \u2013 (-2 \u00d7 (2\/3)) (divisible by 5 and 3)<\/p>\n\n\n\n

= (-2\/3) + (4\/3)<\/p>\n\n\n\n

Since the denominators are same we can add directly<\/p>\n\n\n\n

= (-2+4)\/3<\/p>\n\n\n\n

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

(iv) ((-9\/4) \u00d7 (5\/3)) + ((13\/2) \u00d7 (5\/6))<\/strong><\/p>\n\n\n\n

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

((-9\/4) \u00d7 (5\/3)) + ((13\/2) \u00d7 (5\/6)) = (-9\u00d75)\/(4\u00d73) + (13\u00d75)\/(2\u00d76)<\/p>\n\n\n\n

= -45\/12 + 65\/12<\/p>\n\n\n\n

Since the denominators are same we can add directly<\/p>\n\n\n\n

= (-45+65)\/12<\/p>\n\n\n\n

= 20\/12 (divisible by 2)<\/p>\n\n\n\n

= 10\/6 (divisible by 2)<\/p>\n\n\n\n

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

(v) ((-4\/3) \u00d7 (12\/-5)) + ((3\/7) \u00d7 (21\/15))<\/strong><\/p>\n\n\n\n

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

((-4\/3) \u00d7 (12\/-5)) + ((3\/7) \u00d7 (21\/15)) = ((-4\/1) \u00d7 (4\/-5)) + ((1\/1) \u00d7 (3\/5)) (divisible by 3, 7)<\/p>\n\n\n\n

= (-4\u00d74)\/(1\u00d7-5) + (1\u00d73)\/(1\u00d75)<\/p>\n\n\n\n

= -16\/-5 + 3\/5<\/p>\n\n\n\n

Since the denominators are same we can add directly<\/p>\n\n\n\n

= (16+3)\/5<\/p>\n\n\n\n

= 19\/5<\/p>\n\n\n\n

(vi) ((13\/5) \u00d7 (8\/3)) \u2013 ((-5\/2) \u00d7 (11\/3))<\/strong><\/p>\n\n\n\n

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

((13\/5) \u00d7 (8\/3)) \u2013 ((-5\/2) \u00d7 (11\/3)) = (13\u00d78)\/(5\u00d73) \u2013 (-5\u00d711)\/(2\u00d73)<\/p>\n\n\n\n

= 104\/15 + 55\/6<\/p>\n\n\n\n

By taking LCM for 15 and 6 which is 30<\/p>\n\n\n\n

= ((104\u00d72) + (55\u00d75))\/30<\/p>\n\n\n\n

= (208+275)\/30<\/p>\n\n\n\n

= 483\/30<\/p>\n\n\n\n

(vii) ((13\/7) \u00d7 (11\/26)) \u2013 ((-4\/3) \u00d7 (5\/6))<\/strong><\/p>\n\n\n\n

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

((13\/7) \u00d7 (11\/26)) \u2013 ((-4\/3) \u00d7 (5\/6)) = ((1\/7) \u00d7 (11\/2)) \u2013 ((-2\/3) \u00d7 (5\/3)) (divisible by 13, 2)<\/p>\n\n\n\n

= (1\u00d711)\/(7\u00d72) \u2013 (-2\u00d75)\/(3\u00d73)<\/p>\n\n\n\n

= 11\/14 + 10\/9<\/p>\n\n\n\n

By taking LCM for 14 and 9 which is 126<\/p>\n\n\n\n

= ((11\u00d79) + (10\u00d714))\/126<\/p>\n\n\n\n

= (99+140)\/126<\/p>\n\n\n\n

= 239\/126<\/p>\n\n\n\n

(viii) ((8\/5) \u00d7 (-3\/2)) + ((-3\/10) \u00d7 (11\/16))<\/strong><\/p>\n\n\n\n

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

((8\/5) \u00d7 (-3\/2)) + ((-3\/10) \u00d7 (11\/16)) = ((4\/5) \u00d7 (-3\/1)) + ((-3\/10) \u00d7 (11\/16)) (divisible by 2)<\/p>\n\n\n\n

= (4\u00d7-3)\/(5\u00d71) + (-3\u00d711)\/(10\u00d716)<\/p>\n\n\n\n

= -12\/5 \u2013 33\/160<\/p>\n\n\n\n

By taking LCM for 5 and 160 which is 160<\/p>\n\n\n\n

= ((-12\u00d732) \u2013 (33\u00d71))\/160<\/p>\n\n\n\n

= (-384 \u2013 33)\/160<\/p>\n\n\n\n

= -417\/160<\/p>\n\n\n\n

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

(i) ((3\/2) \u00d7 (1\/6)) + ((5\/3) \u00d7 (7\/2) \u2013 (13\/8) \u00d7 (4\/3))<\/strong><\/p>\n\n\n\n

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

((3\/2) \u00d7 (1\/6)) + ((5\/3) \u00d7 (7\/2) \u2013 (13\/8) \u00d7 (4\/3)) =<\/p>\n\n\n\n

((1\/2) \u00d7 (1\/2)) + ((5\/3) \u00d7 (7\/2) \u2013 (13\/2) \u00d7 (1\/3))<\/p>\n\n\n\n

(1\u00d71)\/(2\u00d72) + (5\u00d77)\/(3\u00d72) \u2013 (13\u00d71)\/(2\u00d73)<\/p>\n\n\n\n

1\/4 + 35\/6 \u2013 13\/6<\/p>\n\n\n\n

By taking LCM for 4 and 6 which is 24<\/p>\n\n\n\n

((1\u00d76) + (35\u00d74) \u2013 (13\u00d74))\/24<\/p>\n\n\n\n

(6 + 140 \u2013 52)\/24<\/p>\n\n\n\n

94\/24<\/p>\n\n\n\n

Further divide by 2 we get, 94\/24 = 47\/12<\/p>\n\n\n\n

(ii) ((1\/4) \u00d7 (2\/7)) \u2013 ((5\/14) \u00d7 (-2\/3) + (3\/7) \u00d7 (9\/2))<\/strong><\/p>\n\n\n\n

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

((1\/4) \u00d7 (2\/7)) \u2013 ((5\/14) \u00d7 (-2\/3) + (3\/7) \u00d7 (9\/2)) =<\/p>\n\n\n\n

((1\/2) \u00d7 (1\/7)) \u2013 ((5\/7) \u00d7 (-1\/3) + (3\/7) \u00d7 (9\/2))<\/p>\n\n\n\n

(1\u00d71)\/(2\u00d77) \u2013 (5\u00d7-1)\/(7\u00d73) + (3\u00d79)\/(7\u00d72)<\/p>\n\n\n\n

1\/14 + 5\/21 + 27\/14<\/p>\n\n\n\n

By taking LCM for 14 and 21 which is 42<\/p>\n\n\n\n

((1\u00d73) + (5\u00d72) + (27\u00d73))\/42<\/p>\n\n\n\n

(3 + 10 + 81)\/42<\/p>\n\n\n\n

94\/42<\/p>\n\n\n\n

Further divide by 2 we get, 94\/42 = 47\/21<\/p>\n\n\n\n

(iii) ((13\/9) \u00d7 (-15\/2)) + ((7\/3) \u00d7 (8\/5) + (3\/5) \u00d7 (1\/2))<\/strong><\/p>\n\n\n\n

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

((13\/3) \u00d7 (-5\/2)) + ((7\/3) \u00d7 (8\/5) + (3\/5) \u00d7 (1\/2)) =<\/p>\n\n\n\n

(13\u00d7-5)\/(3\u00d72) + (7\u00d78)\/(3\u00d75) + (3\u00d71)\/(5\u00d72)<\/p>\n\n\n\n

-65\/6 + 56\/15 + 3\/10<\/p>\n\n\n\n

By taking LCM for 6, 15 and 10 which is 30<\/p>\n\n\n\n

((-65\u00d75) + (56\u00d72) + (3\u00d73))\/30<\/p>\n\n\n\n

(-325 + 112 + 9)\/30<\/p>\n\n\n\n

-204\/30<\/p>\n\n\n\n

Further divide by 2 we get, -204\/30 = -102\/15<\/p>\n\n\n\n

(iv) ((3\/11) \u00d7 (5\/6)) \u2013 ((9\/12) \u00d7 (4\/3) + (5\/13) \u00d7 (6\/15))<\/strong><\/p>\n\n\n\n

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

((3\/11) \u00d7 (5\/6)) \u2013 ((9\/12) \u00d7 (4\/3) + (5\/13) \u00d7 (6\/15)) =<\/p>\n\n\n\n

((1\/11) \u00d7 (5\/2)) \u2013 ((1\/1) \u00d7 (1\/1) + (1\/13) \u00d7 (2\/1))<\/p>\n\n\n\n

(1\u00d75)\/(11\u00d72) \u2013 1\/1 + (1\u00d72)\/(13\u00d71)<\/p>\n\n\n\n

5\/22 \u2013 1\/1 + 2\/13<\/p>\n\n\n\n

By taking LCM for 22, 1 and 13 which is 286<\/p>\n\n\n\n

((5\u00d713) \u2013 (1\u00d7286) + (2\u00d722))\/286<\/p>\n\n\n\n

(65 \u2013 286 + 44)\/286<\/p>\n\n\n\n

-177\/286<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.6 PAGE NO: 1.31<\/h4>\n\n\n\n

1. Verify the property: x \u00d7 y = y \u00d7 x by taking:<\/strong><\/p>\n\n\n\n

(i) x = -1\/3, y = 2\/7<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 y = y \u00d7 x<\/p>\n\n\n\n

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

(-1\u00d72)\/(3\u00d77) = (2\u00d7-1)\/(7\u00d73)<\/p>\n\n\n\n

-2\/21 = -2\/21<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

(ii) x = -3\/5, y = -11\/13<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 y = y \u00d7 x<\/p>\n\n\n\n

-3\/5 \u00d7 -11\/13 = -11\/13 \u00d7 -3\/5<\/p>\n\n\n\n

(-3\u00d7-11)\/(5\u00d713) = (-11\u00d7-3)\/(13\u00d75)<\/p>\n\n\n\n

33\/65 = 33\/65<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

(iii) x = 2, y = 7\/-8<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 y = y \u00d7 x<\/p>\n\n\n\n

2 \u00d7 7\/-8 = 7\/-8 \u00d7 2<\/p>\n\n\n\n

(2\u00d77)\/-8 = (7\u00d72)\/-8<\/p>\n\n\n\n

14\/-8 = 14\/-8<\/p>\n\n\n\n

-14\/8 = -14\/8<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

(iv) x = 0, y = -15\/8<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 y = y \u00d7 x<\/p>\n\n\n\n

0 \u00d7 -15\/8 = -15\/8 \u00d7 0<\/p>\n\n\n\n

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

Hence, the property is satisfied.<\/p>\n\n\n\n

2. Verify the property: x \u00d7 (y \u00d7 z) = (x \u00d7 y) \u00d7 z by taking:<\/strong><\/p>\n\n\n\n

(i) x = -7\/3, y = 12\/5, z = 4\/9<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y \u00d7 z) = (x \u00d7 y) \u00d7 z<\/p>\n\n\n\n

-7\/3 \u00d7 (12\/5 \u00d7 4\/9) = (-7\/3 \u00d7 12\/5) \u00d7 4\/9<\/p>\n\n\n\n

(-7\u00d712\u00d74)\/(3\u00d75\u00d79) = (-7\u00d712\u00d74)\/(3\u00d75\u00d79)<\/p>\n\n\n\n

-336\/135 = -336\/135<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

(ii) x = 0, y = -3\/5, z = -9\/4<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y \u00d7 z) = (x \u00d7 y) \u00d7 z<\/p>\n\n\n\n

0 \u00d7 (-3\/5 \u00d7 -9\/4) = (0 \u00d7 -3\/5) \u00d7 -9\/4<\/p>\n\n\n\n

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

Hence, the property is satisfied.<\/p>\n\n\n\n

(iii) x = 1\/2, y = 5\/-4, z = -7\/5<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y \u00d7 z) = (x \u00d7 y) \u00d7 z<\/p>\n\n\n\n

1\/2 \u00d7 (5\/-4 \u00d7 -7\/5) = (1\/2 \u00d7 5\/-4) \u00d7 -7\/5<\/p>\n\n\n\n

(1\u00d75\u00d7-7)\/(2\u00d7-4\u00d75) = (1\u00d75\u00d7-7)\/(2\u00d7-4\u00d75)<\/p>\n\n\n\n

-35\/-40 = -35\/-40<\/p>\n\n\n\n

35\/40 = 35\/40<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

(iv) x = 5\/7, y = -12\/13, z = -7\/18<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y \u00d7 z) = (x \u00d7 y) \u00d7 z<\/p>\n\n\n\n

5\/7 \u00d7 (-12\/13 \u00d7 -7\/18) = (5\/7 \u00d7 -12\/13) \u00d7 -7\/18<\/p>\n\n\n\n

(5\u00d7-12\u00d7-7)\/(7\u00d713\u00d718) = (5\u00d7-12\u00d7-7)\/(7\u00d713\u00d718)<\/p>\n\n\n\n

420\/1638 = 420\/1638<\/p>\n\n\n\n

Hence, the property is satisfied.<\/p>\n\n\n\n

3. Verify the property: x \u00d7 (y + z) = x \u00d7 y + x \u00d7 z by taking:<\/strong><\/p>\n\n\n\n

(i) x = -3\/7, y = 12\/13, z = -5\/6<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y + z) = x \u00d7 y + x \u00d7 z<\/p>\n\n\n\n

-3\/7 \u00d7 (12\/13 + -5\/6) = -3\/7 \u00d7 12\/13 + -3\/7 \u00d7 -5\/6<\/p>\n\n\n\n

-3\/7 \u00d7 ((12\u00d76) + (-5\u00d713))\/78 = (-3\u00d712)\/(7\u00d713) + (-3\u00d7-5)\/(7\u00d76)<\/p>\n\n\n\n

-3\/7 \u00d7 (72-65)\/78 = -36\/91 + 15\/42<\/p>\n\n\n\n

-3\/7 \u00d7 7\/78 = (-36\u00d76 + 15\u00d713)\/546<\/p>\n\n\n\n

-1\/26 = (196-216)\/546<\/p>\n\n\n\n

= -21\/546<\/p>\n\n\n\n

= -1\/26<\/p>\n\n\n\n

Hence, the property is verified.<\/p>\n\n\n\n

(ii) x = -12\/5, y = -15\/4, z = 8\/3<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y + z) = x \u00d7 y + x \u00d7 z<\/p>\n\n\n\n

-12\/5 \u00d7 (-15\/4 + 8\/3) = -12\/5 \u00d7 -15\/4 + -12\/5 \u00d7 8\/3<\/p>\n\n\n\n

-12\/5 \u00d7 ((-15\u00d73) + (8\u00d74))\/12 = (-12\u00d7-15)\/(5\u00d74) + (-12\u00d78)\/(5\u00d73)<\/p>\n\n\n\n

-12\/5 \u00d7 (-45+32)\/12 = 180\/20 \u2013 96\/15<\/p>\n\n\n\n

-12\/5 \u00d7 -13\/12 = 9 \u2013 32\/5<\/p>\n\n\n\n

13\/5 = (9\u00d75 \u2013 32\u00d71)\/5<\/p>\n\n\n\n

= (45-32)\/5<\/p>\n\n\n\n

= 13\/5<\/p>\n\n\n\n

Hence, the property is verified.<\/p>\n\n\n\n

(iii) x = -8\/3, y = 5\/6, z = -13\/12<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y + z) = x \u00d7 y + x \u00d7 z<\/p>\n\n\n\n

-8\/3 \u00d7 (5\/6 + -13\/12) = -8\/3 \u00d7 5\/6 + -8\/3 \u00d7 -13\/12<\/p>\n\n\n\n

-8\/3 \u00d7 ((5\u00d72) \u2013 (13\u00d71))\/12 = (-8\u00d75)\/(3\u00d76) + (-8\u00d7-13)\/(3\u00d712)<\/p>\n\n\n\n

-8\/3 \u00d7 (10-13)\/12 = -40\/18 + 104\/36<\/p>\n\n\n\n

-8\/3 \u00d7 -3\/12 = (-40\u00d72 + 104\u00d71)\/36<\/p>\n\n\n\n

2\/3 = (-80+104)\/36<\/p>\n\n\n\n

= 24\/36<\/p>\n\n\n\n

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

Hence, the property is verified.<\/p>\n\n\n\n

(iv) x = -3\/4, y = -5\/2, z = 7\/6<\/strong><\/p>\n\n\n\n

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

By using the property<\/p>\n\n\n\n

x \u00d7 (y + z) = x \u00d7 y + x \u00d7 z<\/p>\n\n\n\n

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

-3\/4 \u00d7 ((-5\u00d73) + (7\u00d71))\/6 = (-3\u00d7-5)\/(4\u00d72) + (-3\u00d77)\/(4\u00d76)<\/p>\n\n\n\n

-3\/4 \u00d7 (-15+7)\/6 = 15\/8 \u2013 21\/24<\/p>\n\n\n\n

-3\/4 \u00d7 -8\/6 = (15\u00d73 \u2013 21\u00d71)\/24<\/p>\n\n\n\n

-3\/4 \u00d7 -4\/3 = (45-21)\/24<\/p>\n\n\n\n

1 = 24\/24<\/p>\n\n\n\n

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

Hence, the property is verified.<\/p>\n\n\n\n

4. Use the distributivity of multiplication of rational numbers over their addition to simplify:<\/strong><\/p>\n\n\n\n

(i) 3\/5 \u00d7 ((35\/24) + (10\/1))<\/strong><\/p>\n\n\n\n

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

3\/5 \u00d7 35\/24 + 3\/5 \u00d7 10<\/p>\n\n\n\n

1\/1 \u00d7 7\/8 + 6\/1<\/p>\n\n\n\n

By taking LCM for 8 and 1 which is 8<\/p>\n\n\n\n

7\/8 + 6 = (7\u00d71 + 6\u00d78)\/8<\/p>\n\n\n\n

= (7+48)\/8<\/p>\n\n\n\n

= 55\/8<\/p>\n\n\n\n

(ii) -5\/4 \u00d7 ((8\/5) + (16\/5))<\/strong><\/p>\n\n\n\n

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

-5\/4 \u00d7 8\/5 + -5\/4 \u00d7 16\/5<\/p>\n\n\n\n

-1\/1 \u00d7 2\/1 + -1\/1 \u00d7 4\/1<\/p>\n\n\n\n

-2 + -4<\/p>\n\n\n\n

-2 \u2013 4<\/p>\n\n\n\n

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

(iii) 2\/7 \u00d7 ((7\/16) \u2013 (21\/4))<\/strong><\/p>\n\n\n\n

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

2\/7 \u00d7 7\/16 \u2013 2\/7 \u00d7 21\/4<\/p>\n\n\n\n

1\/1 \u00d7 1\/8 \u2013 1\/1 \u00d7 3\/2<\/p>\n\n\n\n

1\/8 \u2013 3\/2<\/p>\n\n\n\n

By taking LCM for 8 and 2 which is 8<\/p>\n\n\n\n

1\/8 \u2013 3\/2 = (1\u00d71 \u2013 3\u00d74)\/8<\/p>\n\n\n\n

= (1 \u2013 12)\/8<\/p>\n\n\n\n

= -11\/8<\/p>\n\n\n\n

(iv) 3\/4 \u00d7 ((8\/9) \u2013 40)<\/strong><\/p>\n\n\n\n

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

3\/4 \u00d7 8\/9 \u2013 3\/4 \u00d7 40<\/p>\n\n\n\n

1\/1 \u00d7 2\/3 \u2013 3\/1 \u00d7 10<\/p>\n\n\n\n

2\/3 \u2013 30\/1<\/p>\n\n\n\n

By taking LCM for 3 and 1 which is 3<\/p>\n\n\n\n

2\/3 \u2013 30\/1 = (2\u00d71 \u2013 30\u00d73)\/3<\/p>\n\n\n\n

= (2 \u2013 90)\/3<\/p>\n\n\n\n

= -88\/3<\/p>\n\n\n\n

5. Find the multiplicative inverse (reciprocal) of each of the following rational numbers:<\/strong><\/p>\n\n\n\n

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

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

(iii) 12\/5<\/strong><\/p>\n\n\n\n

(iv) -7\/9<\/strong><\/p>\n\n\n\n

(v) -3\/-5<\/strong><\/p>\n\n\n\n

(vi) 2\/3 \u00d7 9\/4<\/strong><\/p>\n\n\n\n

(vii) -5\/8 \u00d7 16\/15<\/strong><\/p>\n\n\n\n

(viii) -2 \u00d7 -3\/5<\/strong><\/p>\n\n\n\n

(ix) -1<\/strong><\/p>\n\n\n\n

(x) 0\/3<\/strong><\/p>\n\n\n\n

(xi) 1<\/strong><\/p>\n\n\n\n

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

(i)<\/strong> The reciprocal of 9 is 1\/9<\/p>\n\n\n\n

(ii)<\/strong> The reciprocal of -7 is -1\/7<\/p>\n\n\n\n

(iii)<\/strong> The reciprocal of 12\/5 is 5\/12<\/p>\n\n\n\n

(iv)<\/strong> The reciprocal of -7\/9 is 9\/-7<\/p>\n\n\n\n

(v)<\/strong> The reciprocal of -3\/-5 is 5\/3<\/p>\n\n\n\n

(vi)<\/strong> The reciprocal of 2\/3 \u00d7 9\/4 is<\/p>\n\n\n\n

Firstly solve for 2\/3 \u00d7 9\/4 = 1\/1 \u00d7 3\/2 = 3\/2<\/p>\n\n\n\n

\u2234 The reciprocal of 3\/2 is 2\/3<\/p>\n\n\n\n

(vii)<\/strong> The reciprocal of -5\/8 \u00d7 16\/15<\/p>\n\n\n\n

Firstly solve for -5\/8 \u00d7 16\/15 = -1\/1 \u00d7 2\/3 = -2\/3<\/p>\n\n\n\n

\u2234 The reciprocal of -2\/3 is 3\/-2<\/p>\n\n\n\n

(viii)<\/strong> The reciprocal of -2 \u00d7 -3\/5<\/p>\n\n\n\n

Firstly solve for -2 \u00d7 -3\/5 = 6\/5<\/p>\n\n\n\n

\u2234 The reciprocal of 6\/5 is 5\/6<\/p>\n\n\n\n

(ix)<\/strong> The reciprocal of -1 is -1<\/p>\n\n\n\n

(x)<\/strong> The reciprocal of 0\/3 does not exist<\/p>\n\n\n\n

(xi)<\/strong> The reciprocal of 1 is 1<\/p>\n\n\n\n

6. Name the property of multiplication of rational numbers illustrated by the following statements:<\/strong><\/p>\n\n\n\n

(i) -5\/16 \u00d7 8\/15 = 8\/15 \u00d7 -5\/16<\/strong><\/p>\n\n\n\n

(ii) -17\/5 \u00d79 = 9 \u00d7 -17\/5<\/strong><\/p>\n\n\n\n

(iii) 7\/4 \u00d7 (-8\/3 + -13\/12) = 7\/4 \u00d7 -8\/3 + 7\/4 \u00d7 -13\/12<\/strong><\/p>\n\n\n\n

(iv) -5\/9 \u00d7 (4\/15 \u00d7 -9\/8) = (-5\/9 \u00d7 4\/15) \u00d7 -9\/8<\/strong><\/p>\n\n\n\n

(v) 13\/-17 \u00d7 1 = 13\/-17 = 1 \u00d7 13\/-17<\/strong><\/p>\n\n\n\n

(vi) -11\/16 \u00d7 16\/-11 = 1<\/strong><\/p>\n\n\n\n

(vii) 2\/13 \u00d7 0 = 0 = 0 \u00d7 2\/13<\/strong><\/p>\n\n\n\n

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

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

(i) -5\/16 \u00d7 8\/15 = 8\/15 \u00d7 -5\/16<\/p>\n\n\n\n

According to commutative law, a\/b \u00d7 c\/d = c\/d \u00d7 a\/b<\/p>\n\n\n\n

The above rational number satisfies commutative property.<\/p>\n\n\n\n

(ii) -17\/5 \u00d79 = 9 \u00d7 -17\/5<\/p>\n\n\n\n

According to commutative law, a\/b \u00d7 c\/d = c\/d \u00d7 a\/b<\/p>\n\n\n\n

The above rational number satisfies commutative property.<\/p>\n\n\n\n

(iii) 7\/4 \u00d7 (-8\/3 + -13\/12) = 7\/4 \u00d7 -8\/3 + 7\/4 \u00d7 -13\/12<\/p>\n\n\n\n

According to given rational number, a\/b \u00d7 (c\/d + e\/f) = (a\/b \u00d7 c\/d) + (a\/b \u00d7 e\/f)<\/p>\n\n\n\n

Distributivity of multiplication over addition satisfies.<\/p>\n\n\n\n

(iv) -5\/9 \u00d7 (4\/15 \u00d7 -9\/8) = (-5\/9 \u00d7 4\/15) \u00d7 -9\/8<\/p>\n\n\n\n

According to associative law, a\/b \u00d7 (c\/d \u00d7 e\/f ) = (a\/b \u00d7 c\/d) \u00d7 e\/f<\/p>\n\n\n\n

The above rational number satisfies associativity of multiplication.<\/p>\n\n\n\n

(v) 13\/-17 \u00d7 1 = 13\/-17 = 1 \u00d7 13\/-17<\/p>\n\n\n\n

Existence of identity for multiplication satisfies for the given rational number.<\/p>\n\n\n\n

(vi) -11\/16 \u00d7 16\/-11 = 1<\/p>\n\n\n\n

Existence of multiplication inverse satisfies for the given rational number.<\/p>\n\n\n\n

(vii) 2\/13 \u00d7 0 = 0 = 0 \u00d7 2\/13<\/p>\n\n\n\n

By using a\/b \u00d7 0 = 0 \u00d7 a\/b<\/p>\n\n\n\n

Multiplication of zero satisfies for the given rational number.<\/p>\n\n\n\n

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

According to distributive law, (a\/b \u00d7 c\/d) + (a\/b \u00d7 e\/f ) = a\/b \u00d7 (c\/d + e\/f)<\/p>\n\n\n\n

The above rational number satisfies distributive law.<\/p>\n\n\n\n

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

(i) The product of two positive rational numbers is always\u2026<\/strong><\/p>\n\n\n\n

(ii) The product of a positive rational number and a negative rational number is always\u2026.<\/strong><\/p>\n\n\n\n

(iii) The product of two negative rational numbers is always\u2026<\/strong><\/p>\n\n\n\n

(iv) The reciprocal of a positive rational numbers is\u2026<\/strong><\/p>\n\n\n\n

(v) The reciprocal of a negative rational numbers is\u2026<\/strong><\/p>\n\n\n\n

(vi) Zero has \u2026. Reciprocal.<\/strong><\/p>\n\n\n\n

(vii) The product of a rational number and its reciprocal is\u2026<\/strong><\/p>\n\n\n\n

(viii) The numbers \u2026 and \u2026 are their own reciprocals.<\/strong><\/p>\n\n\n\n

(ix) If a is reciprocal of b, then the reciprocal of b is.<\/strong><\/p>\n\n\n\n

(x) The number 0 is \u2026 the reciprocal of any number.<\/strong><\/p>\n\n\n\n

(xi) reciprocal of 1\/a, a \u2260 0 is \u2026<\/strong><\/p>\n\n\n\n

(xii) (17\u00d712)-1<\/sup> = 17-1<\/sup> \u00d7 \u2026<\/strong><\/p>\n\n\n\n

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

(i) The product of two positive rational numbers is always positive.<\/p>\n\n\n\n

(ii) The product of a positive rational number and a negative rational number is always negative.<\/p>\n\n\n\n

(iii) The product of two negative rational numbers is always positive.<\/p>\n\n\n\n

(iv) The reciprocal of a positive rational numbers is positive.<\/p>\n\n\n\n

(v) The reciprocal of a negative rational numbers is negative.<\/p>\n\n\n\n

(vi) Zero has no Reciprocal.<\/p>\n\n\n\n

(vii) The product of a rational number and its reciprocal is 1.<\/p>\n\n\n\n

(viii) The numbers 1 and -1 are their own reciprocals.<\/p>\n\n\n\n

(ix) If a is reciprocal of b, then the reciprocal of b is a.<\/p>\n\n\n\n

(x) The number 0 is not the reciprocal of any number.<\/p>\n\n\n\n

(xi) reciprocal of 1\/a, a \u2260 0 is a.<\/p>\n\n\n\n

(xii) (17\u00d712)-1<\/sup> = 17-1<\/sup> \u00d7 12-1<\/sup><\/p>\n\n\n\n

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

(i) -4 \u00d7 7\/9 = 79 \u00d7 \u2026<\/strong><\/p>\n\n\n\n

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

-4 \u00d7 7\/9 = 79 \u00d7 -4<\/p>\n\n\n\n

By using commutative property.<\/p>\n\n\n\n

(ii) 5\/11 \u00d7 -3\/8 = -3\/8 \u00d7 \u2026<\/strong><\/p>\n\n\n\n

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

5\/11 \u00d7 -3\/8 = -3\/8 \u00d7 5\/11<\/p>\n\n\n\n

By using commutative property.<\/p>\n\n\n\n

(iii) 1\/2 \u00d7 (3\/4 + -5\/12) = 1\/2 \u00d7 \u2026 + \u2026 \u00d7 -5\/12<\/strong><\/p>\n\n\n\n

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

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

By using distributive property.<\/p>\n\n\n\n

(iv) -4\/5 \u00d7 (5\/7 + -8\/9) = (-4\/5 \u00d7 \u2026) + -4\/5 \u00d7 -8\/9<\/strong><\/p>\n\n\n\n

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

-4\/5 \u00d7 (5\/7 + -8\/9) = (-4\/5 \u00d7 5\/7) + -4\/5 \u00d7 -8\/9<\/p>\n\n\n\n

By using distributive property.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.7 PAGE NO: 1.35<\/h4>\n\n\n\n

1. Divide:<\/strong><\/p>\n\n\n\n

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

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

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

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

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

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

(iii) -3\/4 by 9\/-16<\/strong><\/p>\n\n\n\n

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

(-3\/4) \/ (9\/-16)<\/p>\n\n\n\n

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

(iv) -7\/8 by -21\/16<\/strong><\/p>\n\n\n\n

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

(-7\/8) \/ (-21\/16)<\/p>\n\n\n\n

(-7\/8) \u00d7 16\/-21 = 2\/3<\/p>\n\n\n\n

(v) 7\/-4 by 63\/64<\/strong><\/p>\n\n\n\n

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

(7\/-4) \/ (63\/64)<\/p>\n\n\n\n

(7\/-4) \u00d7 64\/63 = -16\/9<\/p>\n\n\n\n

(vi) 0 by -7\/5<\/strong><\/p>\n\n\n\n

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

0 \/ (7\/5) = 0<\/p>\n\n\n\n

(vii) -3\/4 by -6<\/strong><\/p>\n\n\n\n

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

(-3\/4) \/ -6<\/p>\n\n\n\n

(-3\/4) \u00d7 1\/-6 = 1\/8<\/p>\n\n\n\n

(viii) 2\/3 by -7\/12<\/strong><\/p>\n\n\n\n

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

(2\/3) \/ (-7\/12)<\/p>\n\n\n\n

(2\/3) \u00d7 12\/-7 = -8\/7<\/p>\n\n\n\n

(ix) -4 by -3\/5<\/strong><\/p>\n\n\n\n

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

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

-4 \u00d7 5\/-3 = 20\/3<\/p>\n\n\n\n

(x) -3\/13 by -4\/65<\/strong><\/p>\n\n\n\n

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

(-3\/13) \/ (-4\/65)<\/p>\n\n\n\n

(-3\/13) \u00d7 (65\/-4) = 15\/4<\/p>\n\n\n\n

2. Find the value and express as a rational number in standard form:<\/strong><\/p>\n\n\n\n

(i) 2\/5 \u00f7 26\/15<\/strong><\/p>\n\n\n\n

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

(2\/5) \/ (26\/15)<\/p>\n\n\n\n

(2\/5) \u00d7 (15\/26)<\/p>\n\n\n\n

(2\/1) \u00d7 (3\/26) = (2\u00d73)\/ (1\u00d726) = 6\/26 = 3\/13<\/p>\n\n\n\n

(ii) 10\/3 \u00f7 -35\/12<\/strong><\/p>\n\n\n\n

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

(10\/3) \/ (-35\/12)<\/p>\n\n\n\n

(10\/3) \u00d7 (12\/-35)<\/p>\n\n\n\n

(10\/1) \u00d7 (4\/-35) = (10\u00d74)\/ (1\u00d7-35) = -40\/35 = -8\/7<\/p>\n\n\n\n

(iii) -6 \u00f7 -8\/17<\/strong><\/p>\n\n\n\n

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

-6 \/ (-8\/17)<\/p>\n\n\n\n

-6 \u00d7 (17\/-8)<\/p>\n\n\n\n

-3 \u00d7 (17\/-4) = (-3\u00d717)\/ (1\u00d7-4) = 51\/4<\/p>\n\n\n\n

(iv) -40\/99 \u00f7 -20<\/strong><\/p>\n\n\n\n

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

(-40\/99) \/ -20<\/p>\n\n\n\n

(-40\/99) \u00d7 (1\/-20)<\/p>\n\n\n\n

(-2\/99) \u00d7 (1\/-1) = (-2\u00d71)\/ (99\u00d7-1) = 2\/99<\/p>\n\n\n\n

(v) -22\/27 \u00f7 -110\/18<\/strong><\/p>\n\n\n\n

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

(-22\/27) \/ (-110\/18)<\/p>\n\n\n\n

(-22\/27) \u00d7 (18\/-110)<\/p>\n\n\n\n

(-1\/9) \u00d7 (6\/-5)<\/p>\n\n\n\n

(-1\/3) \u00d7 (2\/-5) = (-1\u00d72) \/ (3\u00d7-5) = 2\/15<\/p>\n\n\n\n

(vi) -36\/125 \u00f7 -3\/75<\/strong><\/p>\n\n\n\n

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

(-36\/125) \/ (-3\/75)<\/p>\n\n\n\n

(-36\/125) \u00d7 (75\/-3)<\/p>\n\n\n\n

(-12\/25) \u00d7 (15\/-1)<\/p>\n\n\n\n

(-12\/5) \u00d7 (3\/-1) = (-12\u00d73) \/ (5\u00d7-1) = 36\/5<\/p>\n\n\n\n

3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.<\/strong><\/p>\n\n\n\n

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

We know that the product of two rational numbers = 15<\/p>\n\n\n\n

One of the number = -10<\/p>\n\n\n\n

\u2234 other number can be obtained by dividing the product by the given number.<\/p>\n\n\n\n

Other number = 15\/-10<\/p>\n\n\n\n

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

4. The product of two rational numbers is -8\/9. If one of the numbers is -4\/15, find the other.<\/strong><\/p>\n\n\n\n

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

We know that the product of two rational numbers = -8\/9<\/p>\n\n\n\n

One of the number = -4\/15<\/p>\n\n\n\n

\u2234 other number is obtained by dividing the product by the given number.<\/p>\n\n\n\n

Other number = (-8\/9)\/(-4\/15)<\/p>\n\n\n\n

= (-8\/9) \u00d7 (15\/-4)<\/p>\n\n\n\n

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

= (-2\u00d75) \/(3\u00d7-1)<\/p>\n\n\n\n

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

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

5. By what number should we multiply -1\/6 so that the product may be -23\/9?<\/strong><\/p>\n\n\n\n

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

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 -1\/6 = -23\/9<\/p>\n\n\n\n

x = (-23\/9)\/(-1\/6)<\/p>\n\n\n\n

x = (-23\/9) \u00d7 (6\/-1)<\/p>\n\n\n\n

= (-23\/3) \u00d7 (2\u00d7-1)<\/p>\n\n\n\n

= (-23\u00d7-2)\/(3\u00d71)<\/p>\n\n\n\n

= 46\/3<\/p>\n\n\n\n

6. By what number should we multiply -15\/28 so that the product may be -5\/7?<\/strong><\/p>\n\n\n\n

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

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 -15\/28 = -5\/7<\/p>\n\n\n\n

x = (-5\/7)\/(-15\/28)<\/p>\n\n\n\n

x = (-5\/7) \u00d7 (28\/-15)<\/p>\n\n\n\n

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

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

7. By what number should we multiply -8\/13 so that the product may be 24?<\/strong><\/p>\n\n\n\n

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

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 -8\/13 = 24<\/p>\n\n\n\n

x = (24)\/(-8\/13)<\/p>\n\n\n\n

x = (24) \u00d7 (13\/-8)<\/p>\n\n\n\n

= (3) \u00d7 (13\u00d7-1)<\/p>\n\n\n\n

= -39<\/p>\n\n\n\n

8. By what number should -3\/4 be multiplied in order to produce 2\/3?<\/strong><\/p>\n\n\n\n

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

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 -3\/4 = 2\/3<\/p>\n\n\n\n

x = (2\/3)\/(-3\/4)<\/p>\n\n\n\n

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

= -8\/9<\/p>\n\n\n\n

9. Find (x+y) \u00f7 (x-y), if<\/strong><\/p>\n\n\n\n

(i) x= 2\/3, y= 3\/2<\/strong><\/p>\n\n\n\n

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

(x+y) \u00f7 (x-y)<\/p>\n\n\n\n

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

((2\u00d72 + 3\u00d73)\/6) \/ ((2\u00d72 \u2013 3\u00d73)\/6)<\/p>\n\n\n\n

((4+9)\/6) \/ ((4-9)\/6)<\/p>\n\n\n\n

(13\/6) \/ (-5\/6)<\/p>\n\n\n\n

(13\/6) \u00d7 (6\/-5)<\/p>\n\n\n\n

-13\/5<\/p>\n\n\n\n

(ii) x= 2\/5, y= 1\/2<\/strong><\/p>\n\n\n\n

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

(x+y) \u00f7 (x-y)<\/p>\n\n\n\n

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

((2\u00d72 + 1\u00d75)\/10) \/ ((2\u00d72 \u2013 1\u00d75)\/10)<\/p>\n\n\n\n

((4+5)\/10) \/ ((4-5)\/10)<\/p>\n\n\n\n

(9\/10) \/ (-1\/10)<\/p>\n\n\n\n

(9\/10) \u00d7 (10\/-1)<\/p>\n\n\n\n

-9<\/p>\n\n\n\n

(iii) x= 5\/4, y= -1\/3<\/strong><\/p>\n\n\n\n

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

(x+y) \u00f7 (x-y)<\/p>\n\n\n\n

(5\/4 \u2013 1\/3) \/ (5\/4 + 1\/3)<\/p>\n\n\n\n

((5\u00d73 \u2013 1\u00d74)\/12) \/ ((5\u00d73 + 1\u00d74)\/12)<\/p>\n\n\n\n

((15-4)\/12) \/ ((15+4)\/12)<\/p>\n\n\n\n

(11\/12) \/ (19\/12)<\/p>\n\n\n\n

(11\/12) \u00d7 (12\/19)<\/p>\n\n\n\n

11\/19<\/p>\n\n\n\n

(iv) x= 2\/7, y= 4\/3<\/strong><\/p>\n\n\n\n

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

(x+y) \u00f7 (x-y)<\/p>\n\n\n\n

(2\/7 + 4\/3) \/ (2\/7 \u2013 4\/3)<\/p>\n\n\n\n

((2\u00d73 + 4\u00d77)\/21) \/ ((2\u00d73 \u2013 4\u00d77)\/21)<\/p>\n\n\n\n

((6+28)\/21) \/ ((6-28)\/21)<\/p>\n\n\n\n

(34\/21) \/ (-22\/21)<\/p>\n\n\n\n

(34\/21) \u00d7 (21\/-22)<\/p>\n\n\n\n

-34\/22<\/p>\n\n\n\n

-17\/11<\/p>\n\n\n\n

(v) x= 1\/4, y= 3\/2<\/strong><\/p>\n\n\n\n

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

(x+y) \u00f7 (x-y)<\/p>\n\n\n\n

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

((1\u00d71 + 3\u00d72)\/4) \/ ((1\u00d71 \u2013 3\u00d72)\/4)<\/p>\n\n\n\n

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

(7\/4) \/ (-5\/4)<\/p>\n\n\n\n

(7\/4) \u00d7 (4\/-5) = -7\/5<\/p>\n\n\n\n

10. The cost of 723723 meters of rope is Rs 12 \u00be. Find the cost per meter.<\/strong><\/p>\n\n\n\n

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

We know that 23\/3 meters of rope = Rs 51\/4<\/p>\n\n\n\n

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 23\/3 = 51\/4<\/p>\n\n\n\n

x = (51\/4)\/(23\/3)<\/p>\n\n\n\n

x = (51\/4) \u00d7 (3\/23)<\/p>\n\n\n\n

= (51\u00d73) \/ (4\u00d723)<\/p>\n\n\n\n

= 153\/92<\/p>\n\n\n\n

= 1619216192<\/p>\n\n\n\n

\u2234 cost per meter is Rs 1619216192<\/p>\n\n\n\n

11. The cost of 213213 meters of cloth is Rs 75 \u00bc. Find the cost of cloth per meter.<\/strong><\/p>\n\n\n\n

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

We know that 7\/3 meters of cloth = Rs 301\/4<\/p>\n\n\n\n

Let us consider a number = x<\/p>\n\n\n\n

So, x \u00d7 7\/3 = 301\/4<\/p>\n\n\n\n

x = (301\/4)\/(7\/3)<\/p>\n\n\n\n

x = (301\/4) \u00d7 (3\/7)<\/p>\n\n\n\n

= (301\u00d73) \/ (4\u00d77)<\/p>\n\n\n\n

= (43\u00d73) \/ (4\u00d71)<\/p>\n\n\n\n

= 129\/4<\/p>\n\n\n\n

= 32.25<\/p>\n\n\n\n

\u2234 cost of cloth per meter is Rs 32.25<\/p>\n\n\n\n

12. By what number should -33\/16 be divided to get -11\/4?<\/strong><\/p>\n\n\n\n

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

Let us consider a number = x<\/p>\n\n\n\n

So, (-33\/16)\/x = -11\/4<\/p>\n\n\n\n

-33\/16 = x \u00d7 -11\/4<\/p>\n\n\n\n

x = (-33\/16) \/ (-11\/4)<\/p>\n\n\n\n

= (-33\/16) \u00d7 (4\/-11)<\/p>\n\n\n\n

= (-33\u00d74)\/(16\u00d7-11)<\/p>\n\n\n\n

= (-3\u00d71)\/(4\u00d7-1)<\/p>\n\n\n\n

= \u00be<\/p>\n\n\n\n

13. Divide the sum of -13\/5 and 12\/7 by the product of -31\/7 and -1\/2.<\/strong><\/p>\n\n\n\n

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

sum of -13\/5 and 12\/7<\/p>\n\n\n\n

-13\/5 + 12\/7<\/p>\n\n\n\n

((-13\u00d77) + (12\u00d75))\/35<\/p>\n\n\n\n

(-91+60)\/35<\/p>\n\n\n\n

-31\/35<\/p>\n\n\n\n

Product of -31\/7 and -1\/2<\/p>\n\n\n\n

-31\/7 \u00d7 -1\/2<\/p>\n\n\n\n

(-31\u00d7-1)\/(7\u00d72)<\/p>\n\n\n\n

31\/14<\/p>\n\n\n\n

\u2234 by dividing the sum and the product we get,<\/p>\n\n\n\n

(-31\/35) \/ (31\/14)<\/p>\n\n\n\n

(-31\/35) \u00d7 (14\/31)<\/p>\n\n\n\n

(-31\u00d714)\/(35\u00d731)<\/p>\n\n\n\n

-14\/35<\/p>\n\n\n\n

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

14. Divide the sum of 65\/12 and 12\/7 by their difference.<\/strong><\/p>\n\n\n\n

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

The sum is 65\/12 + 12\/7<\/p>\n\n\n\n

The difference is 65\/12 \u2013 12\/7<\/p>\n\n\n\n

When we divide, (65\/12 + 12\/7) \/ (65\/12 \u2013 12\/7)<\/p>\n\n\n\n

((65\u00d77 + 12\u00d712)\/84) \/ ((65\u00d77 \u2013 12\u00d712)\/84)<\/p>\n\n\n\n

((455+144)\/84) \/ ((455 \u2013 144)\/84)<\/p>\n\n\n\n

(599\/84) \/ (311\/84)<\/p>\n\n\n\n

599\/84 \u00d7 84\/311<\/p>\n\n\n\n

599\/311<\/p>\n\n\n\n

15. If 24 trousers of equal size can be prepared in 54 meters of cloth, what length of cloth is required for each trouser?<\/strong><\/p>\n\n\n\n

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

We know that total number trousers = 24<\/p>\n\n\n\n

Total length of the cloth = 54<\/p>\n\n\n\n

Length of the cloth required for each trouser = total length of the cloth\/number of trousers<\/p>\n\n\n\n

= 54\/24<\/p>\n\n\n\n

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

\u2234 9\/4 meters is required for each trouser.<\/p>\n\n\n\n

\n\n\n\n

EXERCISE 1.8 PAGE NO: 1.43<\/h4>\n\n\n\n

1. Find a rational number between -3 and 1.<\/strong><\/p>\n\n\n\n

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

Let us consider two rational numbers x and y<\/p>\n\n\n\n

We know that between two rational numbers x and y where x < y there is a rational number (x+y)\/2<\/p>\n\n\n\n

x < (x+y)\/2 < y<\/p>\n\n\n\n

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

So, the rational number between -3 and 1 is -1<\/p>\n\n\n\n

\u2234 -3 < -1 < 1<\/p>\n\n\n\n

2. Find any five rational numbers less than 2.<\/strong><\/p>\n\n\n\n

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

Five rational numbers less than 2 are 0, 1\/5, 2\/5, 3\/5, 4\/5<\/p>\n\n\n\n

3. Find two rational numbers between -2\/9 and 5\/9<\/strong><\/p>\n\n\n\n

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

The rational numbers between -2\/9 and 5\/9 is<\/p>\n\n\n\n

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

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

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

The rational numbers between -2\/9 and 1\/6 is<\/p>\n\n\n\n

(-2\/9 + 1\/6)\/2<\/p>\n\n\n\n

((-2\u00d72 + 1\u00d73)\/18)\/2<\/p>\n\n\n\n

(-4+3)\/36<\/p>\n\n\n\n

-1\/36<\/p>\n\n\n\n

\u2234 the rational numbers between -2\/9 and 5\/9 are -1\/36, 1\/6<\/p>\n\n\n\n

4. Find two rational numbers between 1\/5 and 1\/2<\/strong><\/p>\n\n\n\n

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

The rational numbers between 1\/5 and 1\/2 is<\/p>\n\n\n\n

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

((1\u00d72 + 1\u00d75)\/10)\/2<\/p>\n\n\n\n

(2+5)\/20 = 7\/20<\/p>\n\n\n\n

The rational numbers between 1\/5 and 7\/20 is<\/p>\n\n\n\n

(1\/5 + 7\/20)\/2<\/p>\n\n\n\n

((1\u00d74 + 7\u00d71)\/20)\/2<\/p>\n\n\n\n

(4+7)\/40<\/p>\n\n\n\n

11\/40<\/p>\n\n\n\n

\u2234 the rational numbers between 1\/5 and 1\/2 are 7\/20, 11\/40<\/p>\n\n\n\n

5. Find ten rational numbers between 1\/4 and 1\/2.<\/strong><\/p>\n\n\n\n

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

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.<\/p>\n\n\n\n

The LCM for 4 and 2 is 4.<\/p>\n\n\n\n

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

1\/2 = (1\u00d72)\/4 = 2\/4<\/p>\n\n\n\n

1\/4 = (1\u00d720 \/ 4\u00d720) = 20\/80<\/p>\n\n\n\n

1\/2 = (2\u00d720 \/ 4\u00d720) = 40\/80<\/p>\n\n\n\n

So, we now know that 21, 22, 23,\u202639 are integers between numerators 20 and 40.<\/p>\n\n\n\n

\u2234 the rational numbers between 1\/4 and 1\/2 are 21\/80, 22\/80, 23\/80, \u2026., 39\/80<\/p>\n\n\n\n

6. Find ten rational numbers between -2\/5 and 1\/2.<\/strong><\/p>\n\n\n\n

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

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.<\/p>\n\n\n\n

The LCM for 5 and 2 is 10.<\/p>\n\n\n\n

-2\/5 = (-2\u00d72)\/10 = -4\/10<\/p>\n\n\n\n

1\/2 = (1\u00d75)\/10 = 5\/10<\/p>\n\n\n\n

-2\/5 = (-4\u00d72 \/ 10\u00d72) = -8\/20<\/p>\n\n\n\n

1\/2 = (5\u00d72 \/ 10\u00d72) = 10\/20<\/p>\n\n\n\n

So, we now know that -7, -6, -5,\u202610 are integers between numerators -8 and 10.<\/p>\n\n\n\n

\u2234 the rational numbers between -2\/5 and 1\/2 are -7\/20, -6\/20, -5\/20, \u2026., 9\/20<\/p>\n\n\n\n

7. Find ten rational numbers between 3\/5 and 3\/4.<\/strong><\/p>\n\n\n\n

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

Firstly convert the given rational numbers into equivalent rational numbers with same denominators.<\/p>\n\n\n\n

The LCM for 5 and 4 is 20.<\/p>\n\n\n\n

3\/5 = 3\u00d7 20 \/ 5\u00d720 = 60\/100<\/p>\n\n\n\n

3\/4 = 3\u00d725 \/ 4\u00d725 = 75\/100<\/p>\n\n\n\n

So, we now know that 61, 62, 63,..74 are integers between numerators 60 and 75.<\/p>\n\n\n\n

\u2234 the rational numbers between 3\/5 and 3\/4 are 61\/100, 62\/100, 63\/100, \u2026., 74\/100<\/p>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 1: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 8 Maths Chapter 1\u2013Rational Numbers<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 8 Maths Chapter 1\u2013Rational Numbers PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n