{"id":601795,"date":"2022-05-13T06:54:15","date_gmt":"2022-05-13T06:54:15","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=601795"},"modified":"2022-05-14T04:20:20","modified_gmt":"2022-05-14T04:20:20","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-1-rational-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-1-rational-numbers\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers"},"content":{"rendered":"\n

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

ML Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 1, Class 8 Maths Chapter 1 solutions<\/p>\n\n\n\n

Exercise 1.1<\/h4>\n\n\n\n

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

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

(ii) 7 \/ \u2013 13 and 4 \/ \u2013 13<\/strong><\/p>\n\n\n\n

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

(i) Given<\/p>\n\n\n\n

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

Adding both the numbers<\/p>\n\n\n\n

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

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

= 9 \/ 7<\/p>\n\n\n\n

\u2234 The addition of 4 \/ 7 and 5 \/ 7 is 9 \/ 7<\/p>\n\n\n\n

(ii) Given<\/p>\n\n\n\n

7 \/ \u2013 13 and 4 \/ \u2013 13<\/p>\n\n\n\n

Consider<\/p>\n\n\n\n

7 \/ \u2013 13 = {7 \u00d7 (-1)} \/ {- 13 \u00d7 (-1)}<\/p>\n\n\n\n

= \u2013 7 \/ 13<\/p>\n\n\n\n

Also,<\/p>\n\n\n\n

4 \/ \u2013 13 = {4 \u00d7 (-1)} \/ {- 13 \u00d7 (-1)}<\/p>\n\n\n\n

= \u2013 4 \/ 13<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Adding both the numbers<\/p>\n\n\n\n

(7 \/ \u2013 13) + (4 \/ \u2013 13) = (- 7 \u2013 4) \/ 13<\/p>\n\n\n\n

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

= \u2013 11 \/ 13<\/p>\n\n\n\n

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

(i) 5 \/ 11 + <\/strong><\/p>\n\n\n\n

(ii) \u2013 4 \/ 9 + <\/strong><\/p>\n\n\n\n

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

(i) Given<\/p>\n\n\n\n

5 \/ 11 +
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

5 \/ 11 + 39 \/ 9<\/p>\n\n\n\n

Taking L.C.M we get,<\/p>\n\n\n\n

5 \/ 11 = (5 \u00d7 9) \/ (11 \u00d7 9)<\/p>\n\n\n\n

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

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

39 \/ 9 = (39 \u00d7 11) \/ (9 \u00d7 11)<\/p>\n\n\n\n

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

= 429 \/ 99<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Adding both the numbers,<\/p>\n\n\n\n

45 \/ 99 + 429 \/ 99 = (45 + 429) \/ 99<\/p>\n\n\n\n

= 474 \/ 99<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

Dividing numerator and denominator by 3,<\/p>\n\n\n\n

= 4 (78 \u00f7 3) \/ (99 \u00f7 3)<\/p>\n\n\n\n

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

=
\"ML<\/p>\n\n\n\n

(ii) Given<\/p>\n\n\n\n

\u2013 4 \/ 9 +
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

\u2013 4 \/ 9 + 38 \/ 13<\/p>\n\n\n\n

Taking L.C.M we get,<\/p>\n\n\n\n

\u2013 4 \/ 9 = (-4 \u00d7 13) \/ (9 \u00d7 13)<\/p>\n\n\n\n

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

= \u2013 52 \/ 117<\/p>\n\n\n\n

38 \/ 13 = (38 \u00d7 9) \/ (13 \u00d7 9)<\/p>\n\n\n\n

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

= 342 \/ 117<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Adding both the numbers,<\/p>\n\n\n\n

\u2013 52 \/ 117 + 342 \/ 117 = (- 52 + 342) \/ 117<\/p>\n\n\n\n

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

= 290 \/ 117<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

3. Verify commutative property of addition for the following pairs of rational numbers.<\/strong><\/p>\n\n\n\n

(i) \u2013 4 \/ 3 and 3 \/ 7<\/strong><\/p>\n\n\n\n

(ii) \u2013 2 \/ \u2013 5 and 1 \/ 3<\/strong><\/p>\n\n\n\n

(iii) 9 \/ 11 and 2 \/ 13<\/strong><\/p>\n\n\n\n

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

(i) \u2013 4 \/ 3 and 3 \/ 7<\/p>\n\n\n\n

Adding both the numbers,<\/p>\n\n\n\n

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

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 28 + 9) \/ 21<\/p>\n\n\n\n

= \u2013 19 \/ 21<\/p>\n\n\n\n

And<\/p>\n\n\n\n

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

Again taking L.C.M. we get,<\/p>\n\n\n\n

= (9 \u2013 28) \/ 21<\/p>\n\n\n\n

= \u2013 19 \/ 21<\/p>\n\n\n\n

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

\u2013 4 \/ 3 + 3 \/ 7 = 3 \/ 7 + (- 4 \/ 3)<\/p>\n\n\n\n

(ii) \u2013 2 \/ \u2013 5 and 1 \/ 3<\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

\u2013 2 \/ \u2013 5 = { \u2013 2 \u00d7 (- 1)} \/ {- 5 \u00d7 (- 1)}<\/p>\n\n\n\n

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

Now,<\/p>\n\n\n\n

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

Taking L.C.M. we get,<\/p>\n\n\n\n

= (6 + 5) \/ 15<\/p>\n\n\n\n

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

And 1 \/ 3 + 2 \/ 5<\/p>\n\n\n\n

Again taking L.C.M. we get,<\/p>\n\n\n\n

= (5 + 6) \/ 15<\/p>\n\n\n\n

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

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

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

(iii) 9 \/ 11 and 2 \/ 13<\/p>\n\n\n\n

Adding both the numbers,<\/p>\n\n\n\n

= 9 \/ 11 + 2 \/ 13<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (117 + 22) \/ 143<\/p>\n\n\n\n

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

= 139 \/ 143<\/p>\n\n\n\n

And 2 \/ 13 + 9 \/ 11<\/p>\n\n\n\n

Again taking L.C.M. we get,<\/p>\n\n\n\n

= (22 + 117) \/ 143<\/p>\n\n\n\n

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

= 139 \/ 143<\/p>\n\n\n\n

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

9 \/ 11 + 2 \/ 13 = 2 \/ 13 + 9 \/ 11<\/p>\n\n\n\n

4. Find the additive inverse of the following rational numbers:<\/strong><\/p>\n\n\n\n

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

(ii) \u2013 7 \/ \u2013 12<\/strong><\/p>\n\n\n\n

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

(i) Given<\/p>\n\n\n\n

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

Additive inverse of<\/p>\n\n\n\n

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

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

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

(ii) Given<\/p>\n\n\n\n

\u2013 7 \/ -12<\/p>\n\n\n\n

Additive inverse of<\/p>\n\n\n\n

\u2013 7 \/ \u2013 12 = \u2013 (- 7 \/ \u2013 12)<\/p>\n\n\n\n

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

= \u2013 7 \/ 12<\/p>\n\n\n\n

5. Verify that \u2013 (- x) = x for<\/strong><\/p>\n\n\n\n

(i) x = 10 \/ 13<\/strong><\/p>\n\n\n\n

(ii) x = \u2013 15 \/ 17<\/strong><\/p>\n\n\n\n

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

(i) x = 10 \/ 13<\/p>\n\n\n\n

\u2013 x = \u2013 10 \/ 13<\/p>\n\n\n\n

\u2013 (- x) = \u2013 (- 10 \/ 13)<\/p>\n\n\n\n

= 10 \/ 13<\/p>\n\n\n\n

Hence, \u2013 (- x) = x<\/p>\n\n\n\n

(ii) x = \u2013 15 \/ 17<\/p>\n\n\n\n

\u2013 x = 15 \/ 17<\/p>\n\n\n\n

\u2013 (- x) = \u2013 (15 \/ 17)<\/p>\n\n\n\n

= \u2013 15 \/ 17<\/p>\n\n\n\n

Hence, \u2013 (- x) = x<\/p>\n\n\n\n

6. Using appropriate properties of addition, find the following:<\/strong><\/p>\n\n\n\n

(i) 4 \/ 5 + 11 \/ 7 + (-7 \/ 5) + (- 2 \/ 7)<\/strong><\/p>\n\n\n\n

(ii) 3 \/ 7 + 4 \/ 9 + (- 5 \/ 21) + (2 \/ 3)<\/strong><\/p>\n\n\n\n

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

(i) 4 \/ 5 + 11 \/ 7 + (- 7 \/ 5) + (- 2 \/ 7)<\/p>\n\n\n\n

= 4 \/ 5 + (- 7 \/ 5) + 11 \/ 7 + (- 2 \/ 7)<\/p>\n\n\n\n

= {4 + (- 7)} \/ 5 + {11 + (- 2)}\/ 7<\/p>\n\n\n\n

= (4 \u2013 7) \/ 5 + (11 \u2013 2) \/ 7<\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

= \u2013 3 \/ 5 + 9 \/ 7<\/p>\n\n\n\n

Now, taking L.C.M. we get,<\/p>\n\n\n\n

= (- 21 + 45) \/ 35<\/p>\n\n\n\n

= 24 \/ 35<\/p>\n\n\n\n

(ii) 3 \/ 7 + 4 \/ 9 + (- 5 \/ 21) + 2 \/ 3<\/p>\n\n\n\n

= 3 \/ 7 + (- 5 \/ 21) + 4 \/ 9 + 2 \/ 3<\/p>\n\n\n\n

On simplifying, we get,<\/p>\n\n\n\n

= {9 + (-5)} \/ 21 + (4 + 6) \/ 9<\/p>\n\n\n\n

= 4 \/ 21 + 10 \/ 9<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (12 + 70) \/ 63<\/p>\n\n\n\n

= 82 \/ 63<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

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

(i) (- 4 \/ 9) + (2 \/ 7) is a \u2026\u2026\u2026.. number<\/strong><\/p>\n\n\n\n

(ii) (43 \/ 89) + (- 51 \/ 47) = \u2026\u2026.. + (43 \/ 89)<\/strong><\/p>\n\n\n\n

(iii) 2 \/ 7 + \u2026\u2026 = 2 \/ 7 = 0 + \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

(iv) 4 \/ 11 + {(- 7 \/ 12) + 9 \/ 10} = {(4 \/ 11) + (- 7 \/ 12)} + \u2026..<\/strong><\/p>\n\n\n\n

(v) 5 \/ 9 + \u2026\u2026 = 0 = (- 5 \/ 9) + \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

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

(i) (- 4 \/ 9) + (2 \/ 7) is a rational number<\/p>\n\n\n\n

(ii) (43 \/ 89) + (- 51 \/ 47) = (- 51 \/ 47) + (43 \/ 89) (Commutative property)<\/p>\n\n\n\n

(iii) 2 \/ 7 + 0 = 2 \/ 7 = 0 + 2 \/ 7 (Commutative property)<\/p>\n\n\n\n

(iv) 4 \/ 11 + {(- 7 \/ 12) + 9 \/ 10} = {(4 \/ 11) + (- 7 \/ 12)} + 9 \/ 10 (Associative property)<\/p>\n\n\n\n

(v) 5 \/ 9 + (- 5 \/ 9) = 0 = (- 5 \/ 9) + 5 \/ 9 (Existance of zero property)<\/p>\n\n\n\n

8. If a = \u2013 11 \/ 27, b = 4 \/ 9 and c = \u2013 5 \/ 18, then verify that a + (b + c) = (a + b) + c<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

a = \u2013 11 \/ 27, b = 4 \/ 9 and c = \u2013 5 \/ 18<\/p>\n\n\n\n

a + (b + c) = (a + b) + c<\/p>\n\n\n\n

Consider,<\/p>\n\n\n\n

L.H.S. = a + (b + c)<\/p>\n\n\n\n

= \u2013 11 \/ 27 + {4 \/ 9 + (- 5 \/ 18)}<\/p>\n\n\n\n

= \u2013 11 \/ 27 + (4 \/ 9 \u2013 5 \/ 18)<\/p>\n\n\n\n

On simplification, we get<\/p>\n\n\n\n

= \u2013 11 \/ 27 + (8 \u2013 5) \/ 18<\/p>\n\n\n\n

= \u2013 11 \/ 27 + 3 \/ 18<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 22 + 9) \/ 54<\/p>\n\n\n\n

= \u2013 13 \/ 54<\/p>\n\n\n\n

R.H.S. = (a + b) + c<\/p>\n\n\n\n

= (- 11 \/ 27 + 4 \/ 9) + (- 5 \/ 18)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= {(- 11 + 12) \/ 27} + (- 5 \/ 18)<\/p>\n\n\n\n

= (1 \/ 27) + (- 5 \/ 18)<\/p>\n\n\n\n

= (2 \u2013 15) \/ 54<\/p>\n\n\n\n

= \u2013 13 \/ 54<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

L.H.S. = R.H.S.<\/p>\n\n\n\n

Exercise 1.2<\/h4>\n\n\n\n

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

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

(ii) \u2013 4 \/ 9 from <\/strong><\/p>\n\n\n\n

(iii)  from <\/strong><\/p>\n\n\n\n

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

(i)
\"MLfrom \u2013 3 \/ 7<\/p>\n\n\n\n

= \u2013 3 \/ 7 \u2013 (13 \/ 5)<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 15 \u2013 91) \/ 35<\/p>\n\n\n\n

= \u2013 106 \/ 35<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

Hence, the subtraction of
\"MLfrom \u2013 3 \/ 7 is
\"ML<\/p>\n\n\n\n

(ii) \u2013 4 \/ 9 from
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

\u2013 4 \/ 9 from 29 \/ 8<\/p>\n\n\n\n

= 29 \/ 8 \u2013 (- 4 \/ 9)<\/p>\n\n\n\n

= 29 \/ 8 + 4 \/ 9<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (261 + 32) \/ 72<\/p>\n\n\n\n

= 293 \/ 72<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

(iii)
\"MLfrom
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

= \u2013 16 \/ 5 from \u2013 43 \/ 9<\/p>\n\n\n\n

= \u2013 43 \/ 9 \u2013 (- 16 \/ 5)<\/p>\n\n\n\n

= \u2013 43 \/ 9 + 16 \/ 5<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 215 + 144) \/ 45<\/p>\n\n\n\n

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

= \u2013 71 \/ 45<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

2. Sum of two rational numbers is 3 \/ 5. If one of them is \u2013 2 \/ 7, find the other.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Sum of two rational numbers is 3 \/ 5<\/p>\n\n\n\n

One of the number is \u2013 2 \/ 7<\/p>\n\n\n\n

Hence, the other number is calculated as follows:<\/p>\n\n\n\n

Other number = 3 \/ 5 \u2013 (- 2 \/ 7)<\/p>\n\n\n\n

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

Taking L.C.M. we get,<\/p>\n\n\n\n

= (21 + 10) \/ 35<\/p>\n\n\n\n

= 31 \/ 35<\/p>\n\n\n\n

Therefore, the other number is 31 \/ 35<\/p>\n\n\n\n

3. What rational number should be added to \u2013 5 \/ 11 to get \u2013 7 \/ 8?<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

According to the statement,<\/p>\n\n\n\n

Sum of two numbers = \u2013 7 \/ 8<\/p>\n\n\n\n

One number = \u2013 5 \/ 11<\/p>\n\n\n\n

Hence, the other number is calculated as below:<\/p>\n\n\n\n

Other number = \u2013 7 \/ 8 \u2013 (- 5 \/ 11)<\/p>\n\n\n\n

= \u2013 7 \/ 8 + 5 \/ 11<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 77 + 40) \/ 88<\/p>\n\n\n\n

= \u2013 37 \/ 88<\/p>\n\n\n\n

Therefore, the other number is \u2013 37 \/ 88<\/p>\n\n\n\n

4. What rational number should be subtracted from  to get ?<\/strong><\/p>\n\n\n\n

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

The required number can be calculated as follows:<\/p>\n\n\n\n

(
\"ML) \u2013 (
\"ML)<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

(- 23 \/ 5) + (7 \/ 2)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= (- 46 + 35) \/ 10<\/p>\n\n\n\n

= \u2013 11 \/ 10<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

Therefore, the required number is
\"ML<\/p>\n\n\n\n

5. Subtract the sum of \u2013 5 \/ 7 and \u2013 8 \/ 3 from the sum of 5 \/ 2 and \u2013 11 \/ 12.<\/strong><\/p>\n\n\n\n

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

Sum of \u2013 5 \/ 7 and \u2013 8 \/ 3 can be calculated as,<\/p>\n\n\n\n

\u2013 5 \/ 7 and \u2013 8 \/ 3 = (- 5 \/ 7) + (- 8 \/ 3)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= (- 15 \u2013 56) \/ 21<\/p>\n\n\n\n

= \u2013 71 \/ 21<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Sum of 5 \/ 2 and \u2013 11 \/ 12 can be calculated as,<\/p>\n\n\n\n

5 \/ 2 + (- 11 \/ 12) = 5 \/ 2 \u2013 11 \/ 12<\/p>\n\n\n\n

On simplification, we get,<\/p>\n\n\n\n

= (30 \u2013 11) \/ 12<\/p>\n\n\n\n

= 19 \/ 12<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

19 \/ 12 \u2013 (- 71 \/ 21)<\/p>\n\n\n\n

= 19 \/ 12 + 71 \/ 21<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (133 + 284) \/ 84<\/p>\n\n\n\n

= 417 \/ 84<\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

=<\/strong>
\"ML<\/p>\n\n\n\n

=<\/strong>
\"ML<\/p>\n\n\n\n

6. If x = \u2013 4 \/ 7 and y = 2 \/ 5, then verify that x \u2013 y \u2260 y \u2013 x<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

x = \u2013 4 \/ 7 and y = 2 \/ 5<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

x \u2013 y = \u2013 4 \/ 7 \u2013 (2 \/ 5)<\/p>\n\n\n\n

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

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 20 \u2013 14) \/ 35<\/p>\n\n\n\n

= \u2013 34 \/ 35<\/p>\n\n\n\n

And<\/p>\n\n\n\n

y \u2013 x = 2 \/ 5 \u2013 (- 4 \/ 7)<\/p>\n\n\n\n

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

Taking L.C.M. we get,<\/p>\n\n\n\n

= (14 + 20) \/ 35<\/p>\n\n\n\n

= 34 \/ 35<\/p>\n\n\n\n

Therefore, x \u2013 y \u2260 y \u2013 x<\/p>\n\n\n\n

7. If x = 4 \/ 9, y = \u2013 7 \/ 12 and z = \u2013 2 \/ 3, then verify that x \u2013 (y \u2013 z) \u2260 (x \u2013 y) \u2013 z<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

x = 4 \/ 9, y = \u2013 7 \/ 12, z = \u2013 2 \/ 3<\/p>\n\n\n\n

x \u2013 (y \u2013 z) \u2260 (x \u2013 y) \u2013 z<\/p>\n\n\n\n

L.H.S. = x \u2013 (y \u2013 z)<\/p>\n\n\n\n

= 4 \/ 9 \u2013 {- 7 \/ 12 \u2013 (- 2 \/ 3)}<\/p>\n\n\n\n

= 4 \/ 9 \u2013 (- 7 \/ 12 + 2 \/ 3)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= 4 \/ 9 \u2013 {(- 7 + 8) \/ 12}<\/p>\n\n\n\n

= 4 \/ 9 \u2013 (1 \/ 12)<\/p>\n\n\n\n

= 4 \/ 9 \u2013 1 \/ 12<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (16 \u2013 3) \/ 36<\/p>\n\n\n\n

= 13 \/ 36<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

R.H.S = (x \u2013 y) \u2013 z<\/p>\n\n\n\n

= {4 \/ 9 \u2013 (- 7 \/ 12)} \u2013 (- 7 \/ 12)<\/p>\n\n\n\n

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

On further calculation, we get<\/p>\n\n\n\n

= {(16 + 21) \/ 36} + 7 \/ 12<\/p>\n\n\n\n

= 37 \/ 36 + 7 \/ 12<\/p>\n\n\n\n

Again taking L.C.M. we get,<\/p>\n\n\n\n

= (37 + 21) \/ 36<\/p>\n\n\n\n

= 58 \/ 36<\/p>\n\n\n\n

Therefore, x \u2013 (y \u2013 z) \u2260 (x \u2013 y) \u2013 z<\/p>\n\n\n\n

8. Which of the following statement is true \/ false?<\/strong><\/p>\n\n\n\n

(i) 2 \/ 3 \u2013 4 \/ 5 is not a rational number.<\/strong><\/p>\n\n\n\n

(ii) \u2013 5 \/ 7 is the additive inverse of 5 \/ 7.<\/strong><\/p>\n\n\n\n

(iii) 0 is the additive inverse of its own.<\/strong><\/p>\n\n\n\n

(iv) Commutative property holds for subtraction of rational numbers.<\/strong><\/p>\n\n\n\n

(v) Associative property does not hold for subtraction of rational numbers.<\/strong><\/p>\n\n\n\n

(vi) 0 is the identity element for subtraction of rational numbers.<\/strong><\/p>\n\n\n\n

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

(i) 2 \/ 3 \u2013 4 \/ 5<\/p>\n\n\n\n

Taking L.C.M<\/p>\n\n\n\n

= (10 \u2013 12) \/ 15<\/p>\n\n\n\n

= \u2013 2 \/ 15<\/p>\n\n\n\n

Is a rational number<\/p>\n\n\n\n

Hence, the given statement is false<\/strong><\/p>\n\n\n\n

(ii) The given statement is true<\/strong><\/p>\n\n\n\n

(iii) The given statement is true<\/strong><\/p>\n\n\n\n

(iv) Let us take,<\/p>\n\n\n\n

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

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

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

2 \/ 4 \u2260 \u2013 2 \/ 4<\/p>\n\n\n\n

Therefore, the given statement is false<\/strong><\/p>\n\n\n\n

(v) The given statement is true<\/strong><\/p>\n\n\n\n

(vi) Let us take,<\/p>\n\n\n\n

7 \/ 8 \u2013 0 = 7 \/ 8<\/p>\n\n\n\n

But 0 \u2013 7 \/ 8 = \u2013 7 \/ 8<\/p>\n\n\n\n

7 \/ 8 \u2260 \u2013 7 \/ 8<\/p>\n\n\n\n

Therefore, the given statement is false<\/strong><\/p>\n\n\n\n

Exercise 1.3<\/strong><\/h4>\n\n\n\n

1. Multiply and express the result in the lowest form:<\/strong><\/p>\n\n\n\n

(i) 6 \/ \u2013 7 \u00d7 14 \/ 30<\/strong><\/p>\n\n\n\n

(ii)<\/strong>
\"ML \u00d7<\/strong>
\"ML<\/p>\n\n\n\n

(iii) 25 \/ \u2013 9 \u00d7 \u2013 3 \/ 10<\/strong><\/p>\n\n\n\n

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

(i) 6 \/ \u2013 7 \u00d7 14 \/ 30<\/p>\n\n\n\n

= (6 \u00d7 14) \/ (- 7 \u00d7 30)<\/p>\n\n\n\n

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

= 84 \/ \u2013 210<\/p>\n\n\n\n

= (84 \u00f7 42) \/ (- 210 \u00f7 42)<\/p>\n\n\n\n

\u2235 HCF of 84, 210 = 42<\/p>\n\n\n\n

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

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

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

(ii)
\"ML\u00d7
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

= 20 \/ 3 \u00d7 9 \/ 7<\/p>\n\n\n\n

= (20 \u00d7 9) \/ (3 \u00d7 7)<\/p>\n\n\n\n

= 180 \/ 21<\/p>\n\n\n\n

= (180 \u00f7 3) \/ (21 \u00f7 3)<\/p>\n\n\n\n

\u2235 HCF of 180, 21 = 3<\/p>\n\n\n\n

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

= 60 \/ 7<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

(iii) 25 \/ \u2013 9 \u00d7 \u2013 3 \/ 10<\/p>\n\n\n\n

= {25 \u00d7 (- 3)} \/ {(- 9) \u00d7 10}<\/p>\n\n\n\n

= \u2013 75 \/ \u2013 90<\/p>\n\n\n\n

= {- 75 \u00f7 (- 15)} \/ {- 90 \u00f7 (- 15)}<\/p>\n\n\n\n

\u2235 HCF of 75, 90 = 15<\/p>\n\n\n\n

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

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

2. Verify commutative property of multiplication for the following pairs of rational numbers:<\/strong><\/p>\n\n\n\n

(i) 4 \/ 5 and \u2013 7 \/ 8<\/strong><\/p>\n\n\n\n

(ii)  and <\/strong><\/p>\n\n\n\n

(iii) \u2013 7 \/ \u2013 20 and 5 \/ \u2013 14<\/strong><\/p>\n\n\n\n

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

(i) 4 \/ 5 and \u2013 7 \/ 8<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

4 \/ 5 \u00d7 \u2013 7 \/ 8<\/p>\n\n\n\n

= {4 \u00d7 (- 7)} \/ 5 \u00d7 8<\/p>\n\n\n\n

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

= \u2013 28 \/ 40<\/p>\n\n\n\n

and<\/p>\n\n\n\n

\u2013 7 \/ 8 \u00d7 4 \/ 5<\/p>\n\n\n\n

= (- 7 \u00d7 4) \/ (8 \u00d7 5)<\/p>\n\n\n\n

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

= \u2013 28 \/ 40<\/p>\n\n\n\n

Therefore, 4 \/ 5 \u00d7 (- 7 \/ 8) = \u2013 7 \/ 8 \u00d7 4 \/ 5<\/p>\n\n\n\n

(ii)
\"MLand
\"ML<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

40 \/ 3 and 9 \/ 8<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

40 \/ 3 \u00d7 9 \/ 8<\/p>\n\n\n\n

= (40 \u00d7 9) \/ (3 \u00d7 8)<\/p>\n\n\n\n

= 360 \/ 24<\/p>\n\n\n\n

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

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

and<\/p>\n\n\n\n

9 \/ 8 \u00d7 40 \/ 3<\/p>\n\n\n\n

= (9 \u00d7 40) \/ (8 \u00d7 3)<\/p>\n\n\n\n

= 360 \/ 24<\/p>\n\n\n\n

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

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

Therefore, 40 \/ 3 \u00d7 9 \/ 8 = 9 \/ 8 \u00d7 40 \/ 3<\/p>\n\n\n\n

(iii) \u2013 7 \/ \u2013 20 and 5 \/ \u2013 14<\/p>\n\n\n\n

\u2013 7 \/ \u2013 20 = {- 7 \u00d7 (- 1)} \/ {- 20 \u00d7 (- 1)}<\/p>\n\n\n\n

= 7 \/ 20<\/p>\n\n\n\n

Now, 7 \/ 20 and 5 \/ \u2013 14<\/p>\n\n\n\n

7 \/ 20 \u00d7 5 \/ \u2013 14<\/p>\n\n\n\n

= (7 \u00d7 5) \/ 20 \u00d7 (- 14)<\/p>\n\n\n\n

= 35 \/ \u2013 280<\/p>\n\n\n\n

and<\/p>\n\n\n\n

5 \/ \u2013 14 \u00d7 7 \/ 20<\/p>\n\n\n\n

= (5 \u00d7 7) \/ (- 14 \u00d7 20)<\/p>\n\n\n\n

= 35 \/ \u2013 280<\/p>\n\n\n\n

Therefore, 7 \/ 20 \u00d7 5 \/ \u2013 14 = 5 \/ \u2013 14 \u00d7 7 \/ 20<\/p>\n\n\n\n

3. Verify the following and name the property also:<\/strong><\/p>\n\n\n\n

(i) 3 \/ 5 \u00d7 (- 4 \/ 7 \u00d7 \u2013 8 \/ 9) = (3 \/ 5 \u00d7 \u2013 4 \/ 7) \u00d7 \u2013 8 \/ 9<\/strong><\/p>\n\n\n\n

(ii) 5 \/ 9 \u00d7 (- 3 \/ 2 + 7 \/ 5) = 5 \/ 9 \u00d7 \u2013 3 \/ 2 + 5 \/ 9 \u00d7 7 \/ 5<\/strong><\/p>\n\n\n\n

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

(i) 3 \/ 5 \u00d7 (- 4 \/ 7 \u00d7 \u2013 8 \/ 9) = (3 \/ 5 \u00d7 \u2013 4 \/ 7) \u00d7 \u2013 8 \/ 9<\/p>\n\n\n\n

L.H.S. = 3 \/ 5 \u00d7 (- 4 \/ 7 \u00d7 \u2013 8 \/ 9)<\/p>\n\n\n\n

= 3 \/ 5 \u00d7 (- 4 \u00d7 \u2013 8) \/ 7 \u00d7 9<\/p>\n\n\n\n

= 3 \/ 5 \u00d7 32 \/ 63<\/p>\n\n\n\n

= (3 \u00d7 32) \/ (5 \u00d7 63)<\/p>\n\n\n\n

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

= 96 \/ 315<\/p>\n\n\n\n

R.H.S. = (3 \/ 5 \u00d7 \u2013 4 \/ 7) \u00d7 \u2013 8 \/ 9<\/p>\n\n\n\n

= \u2013 12 \/ 35 \u00d7 \u2013 8 \/ 9<\/p>\n\n\n\n

= {- 12 \u00d7 (- 8)}\/ (35 \u00d7 9)<\/p>\n\n\n\n

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

= 96 \/ 315<\/p>\n\n\n\n

Hence, 3 \/ 5 \u00d7 (- 4 \/ 7 \u00d7 \u2013 8 \/ 9) = (3 \/ 5 \u00d7 \u2013 4 \/ 7) \u00d7 \u2013 8 \/ 9<\/p>\n\n\n\n

The name of the property is Associative property of multiplication<\/p>\n\n\n\n

(ii) 5 \/ 9 \u00d7 (- 3 \/ 2 + 7 \/ 5) = 5 \/ 9 \u00d7 \u2013 3 \/ 2 + 5 \/ 9 \u00d7 7 \/ 5<\/p>\n\n\n\n

L.H.S = 5 \/ 9 \u00d7 (- 3 \/ 2 + 7 \/ 5)<\/p>\n\n\n\n

= 5 \/ 9 \u00d7 {(- 15 + 14) \/ 10}<\/p>\n\n\n\n

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

= 5 \/ 9 \u00d7 (- 1 \/ 10)<\/p>\n\n\n\n

= \u2013 5 \/ 90<\/p>\n\n\n\n

= (- 5 \u00f7 5) \/ (90 \u00f7 5)<\/p>\n\n\n\n

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

= \u2013 1 \/ 18<\/p>\n\n\n\n

R.H.S. = 5 \/ 9 \u00d7 (- 3 \/ 2) + 5 \/ 9 \u00d7 7 \/ 5<\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

= \u2013 15 \/ 18 + 35 \/ 45<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 75 + 70) \/ 90<\/p>\n\n\n\n

= \u2013 5 \/ 90<\/p>\n\n\n\n

= (- 5 \u00f7 5) \/ (90 \u00f7 5)<\/p>\n\n\n\n

= \u2013 1 \/ 18<\/p>\n\n\n\n

Hence, L.H.S. = R.H.S.<\/p>\n\n\n\n

4. Find the multiplication inverse of the following:<\/strong><\/p>\n\n\n\n

(i) 12<\/strong><\/p>\n\n\n\n

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

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

(iv) \u2013 3 \/ 8 \u00d7 (- 7 \/ 13)<\/strong><\/p>\n\n\n\n

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

(i) The multiplication inverse of 12 is 1 \/ 12<\/p>\n\n\n\n

(ii) The multiplication inverse of 2 \/ 3 is 3 \/ 2<\/p>\n\n\n\n

(iii) The multiplication inverse of \u2013 4 \/ 7 is 7 \/ \u2013 4<\/p>\n\n\n\n

(iv) \u2013 3 \/ 8 \u00d7 (- 7 \/ 13) = 21 \/ 104<\/p>\n\n\n\n

The multiplication inverse of 21 \/ 104 is 104 \/ 21 =
\"ML<\/p>\n\n\n\n

5. Using the appropriate properties of operations of rational numbers, evaluate the following:<\/strong><\/p>\n\n\n\n

(i) 2 \/ 5 \u00d7 \u2013 3 \/ 7 \u2013 1 \/ 14 \u2013 3 \/ 7 \u00d7 3 \/ 5<\/strong><\/p>\n\n\n\n

(ii) 8 \/ 9 \u00d7 4 \/ 5 + 5 \/ 6 \u2013 9 \/ 5 \u00d7 8 \/ 9<\/strong><\/p>\n\n\n\n

(iii) \u2013 3 \/ 7 \u00d7 14 \/ 15 \u00d7 7 \/ 12 \u00d7 (- 30 \/ 35)<\/strong><\/p>\n\n\n\n

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

(i) 2 \/ 5 \u00d7 \u2013 3 \/ 7 \u2013 1 \/ 14 \u2013 3 \/ 7 \u00d7 3 \/ 5<\/p>\n\n\n\n

= 2 \/ 5 \u00d7 \u2013 3 \/ 7 \u2013 3 \/ 7 \u00d7 3 \/ 5 \u2013 1 \/ 14<\/p>\n\n\n\n

Taking common term, we get<\/p>\n\n\n\n

= \u2013 3 \/ 7 (2 \/ 5 + 3 \/ 5) \u2013 1 \/ 14<\/p>\n\n\n\n

= \u2013 3 \/ 7 \u00d7 (2 + 3) \/ 5 \u2013 1 \/ 14<\/p>\n\n\n\n

= \u2013 3 \/ 7 \u00d7 1 \u2013 1 \/ 14<\/p>\n\n\n\n

= \u2013 3 \/ 7 \u2013 1 \/ 14<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 6 \u2013 1) \/ 14<\/p>\n\n\n\n

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

= (- 7 \u00f7 7) \/ (14 \u00f7 7)<\/p>\n\n\n\n

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

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

(ii) 8 \/ 9 \u00d7 4 \/ 5 + 5 \/ 6 \u2013 9 \/ 5 \u00d7 8 \/ 9<\/p>\n\n\n\n

= 8 \/ 9 \u00d7 4 \/ 5 \u2013 9 \/ 5 \u00d7 8 \/ 9 + 5 \/ 6<\/p>\n\n\n\n

Taking common terms, we get,<\/p>\n\n\n\n

= 8 \/ 9 (4 \/ 5 \u2013 9 \/ 5) + 5 \/ 6<\/p>\n\n\n\n

= 8 \/ 9 {(4 \u2013 9) \/ 5} + 5 \/ 6<\/p>\n\n\n\n

= 8 \/ 9 \u00d7 \u2013 5 \/ 5 + 5 \/ 6<\/p>\n\n\n\n

= 8 \/ 9 \u00d7 (- 1) + 5 \/ 6<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= \u2013 8 \/ 9 + 5 \/ 6<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= (- 16 + 15) \/ 18<\/p>\n\n\n\n

= \u2013 1 \/ 18<\/p>\n\n\n\n

(iii) \u2013 3 \/ 7 \u00d7 14 \/ 15 \u00d7 7 \/ 12 \u00d7 (- 30 \/ 35)<\/p>\n\n\n\n

= (- 3 \/ 7 \u00d7 14 \/ 15) \u00d7 (7 \/ 12 \u00d7 \u2013 30 \/ 35)<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= \u2013 2 \/ 5 \u00d7 \u2013 1 \/ 2<\/p>\n\n\n\n

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

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

6. If p = \u2013 8 \/ 27, q = 3 \/ 4 and r = \u2013 12 \/ 15, then verify that<\/strong><\/p>\n\n\n\n

(i) p \u00d7 (q \u00d7 r) = (p \u00d7 q) \u00d7 r<\/strong><\/p>\n\n\n\n

(ii) p \u00d7 (q \u2013 r) = p \u00d7 q \u2013 p \u00d7 r<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

p = \u2013 8 \/ 27, q = 3 \/ 4 and r = \u2013 12 \/ 15<\/p>\n\n\n\n

(i) p \u00d7 (q \u00d7 r) = (p \u00d7 q) \u00d7 r<\/p>\n\n\n\n

L.H.S. = p \u00d7 (q \u00d7 r)<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 (3 \/ 4 \u00d7 \u2013 12 \/ 15)<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 \u2013 3 \/ 5<\/p>\n\n\n\n

On further calculation, we get,<\/p>\n\n\n\n

= {(- 8) \u00d7 (- 3)} \/ (27 \u00d7 5)<\/p>\n\n\n\n

= 24 \/ (27 \u00d7 5)<\/p>\n\n\n\n

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

= 8 \/ 45<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

R.H.S. = (p \u00d7 q) \u00d7 r<\/p>\n\n\n\n

= (- 8 \/ 27 \u00d7 3 \/ 4) \u00d7 \u2013 12 \/ 15<\/p>\n\n\n\n

= \u2013 2 \/ 9 \u00d7 \u2013 12 \/ 15<\/p>\n\n\n\n

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

= 8 \/ 45<\/p>\n\n\n\n

Therefore, L.H.S. = R.H.S.<\/p>\n\n\n\n

(ii) p \u00d7 (q \u2013 r) = p \u00d7 q \u2013 p \u00d7 r<\/p>\n\n\n\n

L.H.S. = p \u00d7 (q \u2013 r)<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 {(3 \/ 4) \u2013 (- 12 \/ 5)}<\/p>\n\n\n\n

Taking L.C.M. we get,<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 {(45 + 48) \/ 60}<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 93 \/ 60<\/p>\n\n\n\n

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

= \u2013 62 \/ 135<\/p>\n\n\n\n

R.H.S. = p \u00d7 q \u2013 p \u00d7 r<\/p>\n\n\n\n

= \u2013 8 \/ 27 \u00d7 3 \/ 4 \u2013 (8 \/ 27 \u00d7 \u2013 12 \/ 15)<\/p>\n\n\n\n

= \u2013 2 \/ 9 \u2013 32 \/ 135<\/p>\n\n\n\n

= (- 30 \u2013 32) \/ 135<\/p>\n\n\n\n

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

= \u2013 62 \/ 135<\/p>\n\n\n\n

Therefore, L.H.S. = R.H.S.<\/p>\n\n\n\n

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

(i) 2 \/ 3 \u00d7 \u2013 4 \/ 5 is a \u2026\u2026 number.<\/strong><\/p>\n\n\n\n

(ii) 54 \/ 81 \u00d7 \u2013 63 \/ 108 = \u2026\u2026\u2026. \u00d7 54 \/ 81<\/strong><\/p>\n\n\n\n

(iii) 4 \/ 5 \u00d7 1 = \u2026\u2026 = 1 \u00d7 \u2026\u2026<\/strong><\/p>\n\n\n\n

(iv) 5 \/ \u2013 12 \u00d7 \u2026\u2026 = 1 = \u2013 12 \/ 5 \u00d7 \u2026\u2026<\/strong><\/p>\n\n\n\n

(v) 3 \/ 7 \u00d7 (- 2 \/ 8 \u00d7 \u2026..) = (3 \/ 7 \u00d7 \u2013 2 \/ 8) \u00d7 5 \/ 9<\/strong><\/p>\n\n\n\n

(vi) \u2013 8 \/ 9 \u00d7 {4 \/ 13 + 5 \/ 17} = \u2013 8 \/ 9 \u00d7 4 \/ 13 + \u2026\u2026\u2026<\/strong><\/p>\n\n\n\n

(vii) \u2013 6 \/ 13 \u00d7 {8 \/ 9 \u2013 4 \/ 7} = \u2013 6 \/ 13 \u00d7 \u2026\u2026. \u2013 (- 6 \/ 13) \u00d7 (4 \/ 7)<\/strong><\/p>\n\n\n\n

(viii) 16 \/ 23 \u00d7 \u2026\u2026\u2026 = 0<\/strong><\/p>\n\n\n\n

(ix) The reciprocal of 0 is \u2026\u2026\u2026..<\/strong><\/p>\n\n\n\n

(x) The numbers \u2026\u2026 and \u2026\u2026 are their own reciprocals.<\/strong><\/p>\n\n\n\n

(xi) If y be the reciprocal of x, then the reciprocal of y2<\/sup> in terms of x will be \u2026\u2026<\/strong><\/p>\n\n\n\n

(xii) The product of a non-zero rational number and its reciprocal is \u2026\u2026\u2026.<\/strong><\/p>\n\n\n\n

(xiii) The reciprocal of a negative rational number is \u2026\u2026\u2026.<\/strong><\/p>\n\n\n\n

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

(i) 2 \/ 3 \u00d7 \u2013 4 \/ 5 is a rational number.<\/p>\n\n\n\n

(ii) 54 \/ 81 \u00d7 \u2013 63 \/ 108 = \u2026\u2026. \u00d7 54 \/ 81<\/p>\n\n\n\n

54 \/ 81 \u00d7 \u2013 63 \/ 108 = \u2013 63 \/ 108 \u00d7 54 \/ 81<\/p>\n\n\n\n

(iii) 4 \/ 5 \u00d7 1 = \u2026\u2026 = 1 \u00d7 \u2026\u2026<\/p>\n\n\n\n

4 \/ 5 \u00d7 1 = 4 \/ 5 = 1 \u00d7 4 \/ 5<\/p>\n\n\n\n

(iv) 5 \/ \u2013 12 \u00d7 \u2026\u2026 = 1 = \u2013 12 \/ 5 \u00d7 \u2026\u2026<\/p>\n\n\n\n

5 \/ \u2013 12 \u00d7 \u2013 12 \/ 5 = 1 = \u2013 12 \/ 5 \u00d7 5 \/ \u2013 12<\/p>\n\n\n\n

(v) 3 \/ 7 \u00d7 (- 2 \/ 8 \u00d7 \u2026.) = (3 \/ 7 \u00d7 \u2013 2 \/ 8) \u00d7 5 \/ 9<\/p>\n\n\n\n

3 \/ 7 \u00d7 (- 2 \/ 8 \u00d7 5 \/ 9) = (3 \/ 7 \u00d7 \u2013 2 \/ 8) \u00d7 5 \/ 9<\/p>\n\n\n\n

(vi) \u2013 8 \/ 9 \u00d7 (4 \/ 13 + 5 \/ 17) = \u2013 8 \/ 9 \u00d7 4 \/ 13 + \u2026\u2026\u2026<\/p>\n\n\n\n

\u2013 8 \/ 9 \u00d7 (4 \/ 13 + 5 \/ 17) = \u2013 8 \/ 9 \u00d7 4 \/ 13 + \u2013 8 \/ 9 \u00d7 5 \/ 17<\/p>\n\n\n\n

(vii) \u2013 6 \/ 13 \u00d7 (8 \/ 9 \u2013 4 \/ 7) = \u2013 6 \/ 13 \u00d7 \u2026.- (- 6 \/ 13) \u00d7 (4 \/ 7)<\/p>\n\n\n\n

\u2013 6 \/ 13 \u00d7 (8 \/ 9 \u2013 4 \/ 7) = \u2013 6 \/ 13 \u00d7 8 \/ 9 \u2013 (- 6 \/ 13) \u00d7 (4 \/ 7)<\/p>\n\n\n\n

(viii) 16 \/ 23 \u00d7 \u2026. = 0<\/p>\n\n\n\n

16 \/ 23 \u00d7 0 = 0<\/p>\n\n\n\n

(ix) The reciprocal of 0 is not defined<\/p>\n\n\n\n

(x) The numbers 1 and \u2013 1 are their own reciprocals<\/p>\n\n\n\n

(xi) If y be the reciprocal of x, then the reciprocal of y2<\/sup> in terms of x will be x2<\/sup><\/p>\n\n\n\n

(xii) The product of a non-zero rational number and its reciprocal is 1<\/p>\n\n\n\n

(xiii) The reciprocal of a negative rational number is a negative rational number<\/p>\n\n\n\n

8. If 4 \/ 5 the multiplicative inverse of ? Why or why not?<\/strong><\/p>\n\n\n\n

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

No, the multiplicative inverse of 4 \/ 5 is not \u2013 5 \/ 4<\/p>\n\n\n\n

The multiplicative inverse of 4 \/ 5 is 5 \/ 4<\/p>\n\n\n\n

9. Using distributivity, find<\/strong><\/p>\n\n\n\n

(i) {7 \/ 5 \u00d7 (- 3 \/ 12)} + {7 \/ 5 + 5 \/ 12}<\/strong><\/p>\n\n\n\n

(ii) {9 \/ 16 \u00d7 4 \/ 12} + {9 \/ 16 \u00d7 (- 3 \/ 9)}<\/strong><\/p>\n\n\n\n

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

(i) {7 \/ 5 \u00d7 (\u2013 <\/strong>3 \/ 12)} + {7 \/ 5 + 5 \/ 12}<\/p>\n\n\n\n

Taking common factor, we get<\/p>\n\n\n\n

= 7 \/ 5 <\/strong>\u00d7 (- 3 \/ 12 + 5 \/ 12)<\/p>\n\n\n\n

= 7 \/ 5 \u00d7 {(- 3 + 5) \/ 12}<\/p>\n\n\n\n

= 7 \/ 5 \u00d7 2 \/ 12<\/p>\n\n\n\n

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

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

(ii) {9 \/ 16 \u00d7 4 \/ 12} + {9 \/ 16 \u00d7 (- 3 \/ 9)}<\/p>\n\n\n\n

Taking common factor, we get<\/p>\n\n\n\n

= 9 \/ 16 \u00d7 {4 \/ 12 + (- 3 \/ 9)}<\/p>\n\n\n\n

= 9 \/ 16 \u00d7 (1 \/ 3 \u2013 1 \/ 3)<\/p>\n\n\n\n

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

= 9 \/ 16 \u00d7 0<\/p>\n\n\n\n

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

10. Find the sum of additive inverse and multiplication inverse of 9.<\/strong><\/p>\n\n\n\n

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

The additive inverse of 9 is \u2013 9<\/p>\n\n\n\n

The multiplicative inverse of 9 is 1 \/ 9<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2013 9 + 1 \/ 9 = (- 81 + 1) \/ 9<\/p>\n\n\n\n

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

= \u2013 80 \/ 9<\/p>\n\n\n\n

=<\/strong>
\"ML<\/p>\n\n\n\n

11. Find the product of additive inverse and multiplicative inverse of \u2013 3 \/ 7<\/strong><\/p>\n\n\n\n

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

The additive inverse of \u2013 3 \/ 7 is 3 \/ 7<\/p>\n\n\n\n

The multiplicative inverse of \u2013 3 \/ 7 is \u2013 7 \/ 3<\/p>\n\n\n\n

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

3 \/ 7 \u00d7 (- 7 \/ 3) = \u2013 1<\/p>\n\n\n\n

Exercise 1.4<\/h4>\n\n\n\n

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

(i) \u2013 3 \/ 7 \u00f7 4<\/strong><\/p>\n\n\n\n

(ii)  \u00f7 (- 4 \/ 9)<\/strong><\/p>\n\n\n\n

(iii) \u2013 8 \/ 9 \u00f7 \u2013 3 \/ 5<\/strong><\/p>\n\n\n\n

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

(i) \u2013 3 \/ 7 \u00f7 4<\/p>\n\n\n\n

= \u2013 3 \/ 7 \u00d7 1 \/ 4<\/p>\n\n\n\n

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

= \u2013 3 \/ 28<\/p>\n\n\n\n

Hence, the value of \u2013 3 \/ 7 \u00f7 4 = \u2013 3 \/ 28<\/p>\n\n\n\n

(ii)
\"ML\u00f7 (- 4 \/ 9)<\/p>\n\n\n\n

This can be written as,<\/p>\n\n\n\n

= 37 \/ 8 \u00f7 (- 4 \/ 9)<\/p>\n\n\n\n

= 37 \/ 8 \u00d7 9 \/ \u2013 4<\/p>\n\n\n\n

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

= 333 \/ \u2013 32<\/p>\n\n\n\n

= {333 \u00d7 (- 1)} \/ {- 32 \u00d7 (- 1)}<\/p>\n\n\n\n

= \u2013 333 \/ 32<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

(iii) \u2013 8 \/ 9 \u00f7 \u2013 3 \/ 5<\/p>\n\n\n\n

= \u2013 8 \/ 9 \u00d7 5 \/ \u2013 3<\/p>\n\n\n\n

= \u2013 40 \/ \u2013 27<\/p>\n\n\n\n

= {- 40 \u00d7 (- 1)} \/ {- 27 \u00d7 (- 1)}<\/p>\n\n\n\n

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

= 40 \/ 27<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

2. State whether the following statements are true or false:<\/strong><\/p>\n\n\n\n

(i) \u2013 9 \/ 13 \u00f7 2 \/ 7 is a rational number.<\/strong><\/p>\n\n\n\n

(ii) 4 \/ 13 \u00f7 11 \/ 12 = 11 \/ 12 \u00f7 4 \/ 13<\/strong><\/p>\n\n\n\n

(iii) \u2013 3 \/ 4 \u00f7 (5 \/ 9 \u00f7 \u2013 4 \/ 11) = (- 3 \/ 4 \u00f7 5 \/ 9) \u00f7 \u2013 4 \/ 11<\/strong><\/p>\n\n\n\n

(iv) 13 \/ 14 \u00f7 \u2013 5 \/ 7 \u2260 \u2013 5 \/ 7 \u00f7 13 \/ 14<\/strong><\/p>\n\n\n\n

(v) (- 7 \u00f7 4 \/ 5) \u00f7 \u2013 9 \/ 10 \u2260 \u2013 7 \u00f7 (4 \/ 5 \u00f7 \u2013 9 \/ 10)<\/strong><\/p>\n\n\n\n

(vi) \u2013 7 \/ 24 \u00f7 6 \/ 11 is not a rational number.<\/strong><\/p>\n\n\n\n

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

(i) The given statement is true<\/strong><\/p>\n\n\n\n

(ii) The given statement is false<\/strong><\/p>\n\n\n\n

Correct: Commutative property is not true for the division<\/p>\n\n\n\n

(iii) The given statement is false<\/strong><\/p>\n\n\n\n

Correct: Associative in division is not true<\/strong><\/p>\n\n\n\n

(iv) The given statement is true<\/strong><\/p>\n\n\n\n

(v) The given statement is true<\/strong><\/p>\n\n\n\n

(vi) The given statement is flase<\/strong><\/p>\n\n\n\n

Correct: It is a rational number<\/p>\n\n\n\n

3. The product of two rational numbers is \u2013 11 \/ 12. If one of them is , find the other.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Product of two rational numbers = \u2013 11 \/ 12<\/p>\n\n\n\n

One of the number =
\"ML= 22 \/ 9<\/p>\n\n\n\n

The other number is calculated as below<\/p>\n\n\n\n

\u2013 11 \/ 12 \u00f7 22 \/ 9<\/p>\n\n\n\n

= \u2013 11 \/ 12 \u00d7 9 \/ 22<\/p>\n\n\n\n

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

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

Therefore, the other number is \u2013 3 \/ 8<\/p>\n\n\n\n

4. By what rational number should \u2013 7 \/ 12 be multiplied to get the product as 5 \/ 14?<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Product = 5 \/ 14<\/p>\n\n\n\n

The required number can be calculated as below<\/p>\n\n\n\n

5 \/ 14 \u00f7 \u2013 7 \/ 12<\/p>\n\n\n\n

= 5 \/ 14 \u00d7 12 \/ \u2013 7<\/p>\n\n\n\n

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

= 30 \/ \u2013 49<\/p>\n\n\n\n

= {30 \u00d7 (- 1)} \/ {- 49 \u00d7 (- 1)}<\/p>\n\n\n\n

= \u2013 30 \/ 49<\/p>\n\n\n\n

Hence, the required number is \u2013 30 \/ 49<\/p>\n\n\n\n

5. By what rational number should \u2013 3 is divided to get \u2013 9 \/ 13?<\/strong><\/p>\n\n\n\n

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

The required number can be calculated as follows:<\/p>\n\n\n\n

\u2013 3 \u00f7 \u2013 9 \/ 13<\/p>\n\n\n\n

= \u2013 3 \u00d7 13 \/ \u2013 9<\/p>\n\n\n\n

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

= \u2013 13 \/ \u2013 3<\/p>\n\n\n\n

= {- 13 \u00d7 (- 1)} \/ {- 3 \u00d7 (- 1)}<\/p>\n\n\n\n

= 13 \/ 3<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

Therefore, the required number is
\"ML<\/p>\n\n\n\n

6. Divide the sum of \u2013 13 \/ 8 and 5 \/ 12 by their difference.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Sum of \u2013 13 \/ 8 and 5 \/ 12 is calculated as,<\/p>\n\n\n\n

= \u2013 13 \/ 8 + 5 \/ 12<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= (- 39 + 10) \/ 24<\/p>\n\n\n\n

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

= \u2013 29 \/ 24<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Difference of \u2013 13 \/ 8 and 5 \/ 12 is calculated as,<\/p>\n\n\n\n

= \u2013 13 \/ 8 \u2013 5 \/ 12<\/p>\n\n\n\n

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

= (- 39 \u2013 10) \/ 24<\/p>\n\n\n\n

= \u2013 49 \/ 24<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\u2013 29 \/ 24 \u00f7 \u2013 49 \/ 24<\/p>\n\n\n\n

= \u2013 29 \/ 24 \u00d7 24 \/ \u2013 49<\/p>\n\n\n\n

= \u2013 29 \/ \u2013 49<\/p>\n\n\n\n

= {- 29 \u00d7 (- 1)} \/ {- 49 \u00d7 (- 1)}<\/p>\n\n\n\n

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

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

7. Divide the sum of 8 \/ 3 and 4 \/ 7 by the product of \u2013 3 \/ 7 and 14 \/ 9.<\/strong><\/p>\n\n\n\n

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

Sum of 8 \/ 3 and 4 \/ 7 is calculated as below<\/p>\n\n\n\n

8 \/ 3 + 4 \/ 7 = (56 + 12) \/ 21<\/p>\n\n\n\n

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

= 68 \/ 21<\/p>\n\n\n\n

Product of \u2013 3 \/ 7 and 14 \/ 9 is calculated as follows:<\/p>\n\n\n\n

\u2013 3 \/ 7 \u00d7 14 \/ 9 = \u2013 2 \/ 3<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

68 \/ 21 \u00f7 \u2013 2 \/ 3 = 68 \/ 21 \u00d7 3 \/ \u2013 2<\/p>\n\n\n\n

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

= 34 \/ \u2013 7<\/p>\n\n\n\n

= {34 \u00d7 (- 1)} \/ {- 7 \u00d7 (- 1)}<\/p>\n\n\n\n

= \u2013 34 \/ 7<\/p>\n\n\n\n

=
\"ML<\/p>\n\n\n\n

8. If p = \u2013 3 \/ 2, q = 4 \/ 5 and r = \u2013 7 \/ 12, then verify that (p \u00f7 q) \u00f7 r \u2260 p \u00f7 (q \u00f7 r)<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

p = \u2013 3 \/ 2, q = 4 \/ 5 and r = \u2013 7 \/ 12<\/p>\n\n\n\n

(p \u00f7 q) \u00f7 r \u2260 p \u00f7 (q \u00f7 r)<\/p>\n\n\n\n

LHS = (p \u00f7 q) \u00f7 r<\/p>\n\n\n\n

= (- 3 \/ 2 \u00f7 4 \/ 5) \u00f7 (- 7 \/ 12)<\/p>\n\n\n\n

= (- 3 \/ 2 \u00d7 5 \/ 4) \u00f7 (- 7 \/ 12)<\/p>\n\n\n\n

= \u2013 15 \/ 8 \u00f7 \u2013 7 \/ 12<\/p>\n\n\n\n

= \u2013 15 \/ 8 \u00d7 12 \/ \u2013 7<\/p>\n\n\n\n

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

= \u2013 45 \/ \u2013 14<\/p>\n\n\n\n

= {- 45 \u00d7 (- 1)} \/ {- 14 \u00d7 (- 1)}<\/p>\n\n\n\n

= 45 \/ 14<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

RHS = p \u00f7 (q \u00f7 r)<\/p>\n\n\n\n

= \u2013 3 \/ 2 \u00f7 (4 \/ 5) \u00f7 (- 7 \/ 12)<\/p>\n\n\n\n

= \u2013 3 \/ 2 \u00f7 (4 \/ 5\u00d7 12 \/ \u2013 7)<\/p>\n\n\n\n

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

= \u2013 3 \/ 2 \u00f7 48 \/ \u2013 35<\/p>\n\n\n\n

= \u2013 3 \/ 2 \u00d7 \u2013 35 \/ 48<\/p>\n\n\n\n

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

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

Therefore, LHS \u2260 RHS<\/p>\n\n\n\n

Exercise 1. 5<\/h4>\n\n\n\n

1. Represent the following rational numbers on the number line.<\/strong><\/p>\n\n\n\n

(i) 11 \/4<\/strong><\/p>\n\n\n\n

(ii)<\/strong>
\"ML<\/p>\n\n\n\n

(iii) \u2013 9 \/ 7<\/strong><\/p>\n\n\n\n

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

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

(i) 11 \/ 4 =<\/strong>
\"ML<\/p>\n\n\n\n

The given rational number on the number line is shown as below:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(ii) <\/strong><\/p>\n\n\n\n

The given rational number on the number line is shown as below<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iii) \u2013 9 \/ 7 =<\/strong>
\"ML<\/p>\n\n\n\n

The given rational number on the number line is shown as below<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

(iv) \u2013 2 \/ \u2013 5 = \u2013 2 \u00d7 (- 1) \/ \u2013 5 \u00d7 (- 1)<\/strong><\/p>\n\n\n\n

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

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

The given rational number on the number line is shown as below<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n

2. Write the rational numbers for each point labeled with a letter:<\/strong><\/p>\n\n\n\n

\"ML<\/figure>\n\n\n\n
\"ML<\/figure>\n\n\n\n

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

(i) The rational numbers for each point labeled with a letter are as follows:<\/p>\n\n\n\n

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

B = 7 \/ 7 = 1<\/p>\n\n\n\n

C = 8 \/ 7 =
\"ML<\/p>\n\n\n\n

D = 12 \/ 7 =
\"ML<\/p>\n\n\n\n

E = 13 \/ 7 =
\"ML<\/p>\n\n\n\n

(ii) The rational numbers for each point labeled with a letter are as follows:<\/p>\n\n\n\n

P = \u2013 3 \/ 8<\/p>\n\n\n\n

Q = \u2013 4 \/ 8 or \u2013 1 \/ 2<\/p>\n\n\n\n

R = \u2013 7 \/ 8<\/p>\n\n\n\n

S = \u2013 11 \/ 8<\/p>\n\n\n\n

T = \u2013 12 \/ 8 or \u2013 3 \/ 2<\/p>\n\n\n\n

3. Find twenty rational numbers between \u2013 3 \/ 7 and 2 \/ 3<\/strong><\/p>\n\n\n\n

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

Twenty rational numbers between \u2013 3 \/ 7 and 2 \/ 3 can be calculated as follows:<\/p>\n\n\n\n

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

LCM of 7, 3 = 21<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

\u2013 3 \/ 7 = (- 3 \u00d7 3) \/ (7 \u00d7 3)<\/p>\n\n\n\n

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

= \u2013 9 \/ 21<\/p>\n\n\n\n

2 \/ 3 = (2 \u00d7 7) \/ (3 \u00d7 7)<\/p>\n\n\n\n

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

= 14 \/ 21<\/p>\n\n\n\n

Now, twenty rational numbers between \u2013 9 \/ 21 and 14 \/ 21 are,<\/p>\n\n\n\n

\u2013 8 \/ 21, \u2013 7 \/ 21, \u2013 6 \/ 21, \u2013 5 \/ 21, \u2013 4 \/ 21, \u2013 3 \/ 21, \u2013 2 \/ 21, \u2013 1 \/ 21, 0, 1 \/ 21, 2 \/ 21, 3 \/ 21, 4 \/ 21, 5 \/ 21, 6 \/ 21, 7 \/ 21, 8 \/ 21, 9 \/ 21, 10 \/ 21, 11 \/ 21, 12 \/ 21 and 13 \/ 21<\/p>\n\n\n\n

4. Find six rational numbers between \u2013 1 \/ 2 and 5 \/ 4<\/strong><\/p>\n\n\n\n

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

Six rational numbers between \u2013 1 \/ 2 and 5 \/ 4 can be calculated as below<\/p>\n\n\n\n

LCM of 2, 4 = 4<\/p>\n\n\n\n

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

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

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

Now, six rational numbers between \u2013 1 \/ 2 and 5 \/ 4 are as follows:<\/p>\n\n\n\n

\u2013 1 \/ 4, 0, 1 \/ 4, 2 \/ 4, 3 \/ 4 and 4 \/ 4<\/p>\n\n\n\n

5. Find three rational numbers between \u2013 2 and \u2013 1<\/strong><\/p>\n\n\n\n

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

Three rational numbers between \u2013 2 and \u2013 1 can be calculated as below:<\/p>\n\n\n\n

First rational number = 1 \/ 2 (- 1 \u2013 2)<\/p>\n\n\n\n

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

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

Second rational number \u2013 2 and \u2013 3 \/ 2<\/p>\n\n\n\n

= 1 \/ 2 {- 2 \u2013 (3 \/ 2)}<\/p>\n\n\n\n

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

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

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

Third rational number between \u2013 3 \/ 2 and \u2013 1<\/p>\n\n\n\n

= 1 \/ 2 {(- 3 \/ 2) \u2013 1}<\/p>\n\n\n\n

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

= 1 \/ 2 \u00d7 \u2013 5 \/ 2<\/p>\n\n\n\n

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

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

Therefore, three rational numbers are \u2013 7 \/ 4, \u2013 3 \/ 2, \u2013 5 \/ 4<\/p>\n\n\n\n

6. Write ten rational numbers which are greater than 0.<\/strong><\/p>\n\n\n\n

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

Ten rational numbers which are greater than 0<\/p>\n\n\n\n

There can be the finite number of a rational number greater than 1.<\/p>\n\n\n\n

Here, we shall take only 10 rational numbers.<\/p>\n\n\n\n

The numbers are as follows:<\/p>\n\n\n\n

(1 \/ 2), 1, (3 \/ 2), 2, (5 \/ 2), 3, (7 \/ 2), 4, (9 \/ 2), 5 etc.<\/p>\n\n\n\n

7. Write five rational numbers which are smaller than \u2013 4<\/strong><\/p>\n\n\n\n

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

Five rational numbers which are smaller than \u2013 4<\/p>\n\n\n\n

These can be finite number of rational numbers smaller than \u2013 4<\/p>\n\n\n\n

Here, we shall take only 5 rational numbers.<\/p>\n\n\n\n

The numbers are as follows:<\/p>\n\n\n\n

(- 9 \/ 2), \u2013 5, (- 11 \/ 2), \u2013 6, (- 13 \/ 2), etc.<\/p>\n\n\n\n

8. Identify the rational number which is different from the other three. Explain your reasoning<\/strong><\/p>\n\n\n\n

(- 5 \/ 11), (- 1 \/ 2), (- 4 \/ 9), (- 7 \/ 3)<\/strong><\/p>\n\n\n\n

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

Given four rational number are,<\/p>\n\n\n\n

(- 5 \/ 11), (- 1 \/ 2), (- 4 \/ 9), (- 7 \/ 3)<\/p>\n\n\n\n

Among the given numbers,<\/p>\n\n\n\n

\u2013 7 \/ 3 is different from the other three numbers.<\/p>\n\n\n\n

Because in \u2013 7 \/ 3 its denominator is less than its numerator<\/p>\n\n\n\n

In other numbers, denominators are greater than their numerators respectively.<\/p>\n\n\n\n

Exercise 1.6<\/h4>\n\n\n\n

1. In a bag, there are 20 kg of fruits. If  kg of these fruits be oranges and  kg of these are apples and rest are grapes. Find the mass of the grapes in the bag.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Total fruits in a bag = 20 kg<\/p>\n\n\n\n

Oranges =
\"MLkg i.e 43 \/ 6 kg<\/p>\n\n\n\n

Apples =
\"MLkg i.e 26 \/ 3 kg<\/p>\n\n\n\n

Remaining fruits in a bag = 20 \u2013 {(43 \/ 6) + (26 \/ 3)} kg<\/p>\n\n\n\n

= 20 \u2013 {(43 + 52) \/ 6}<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= 20 \u2013 (95 \/ 6)<\/p>\n\n\n\n

= (120 \u2013 95) \/ 6<\/p>\n\n\n\n

= 25 \/ 6<\/p>\n\n\n\n

=
\"MLkg<\/p>\n\n\n\n

Therefore, the mass of the grapes in the bag is
\"MLkg<\/p>\n\n\n\n

2. The population of a city is 6, 63,432. If 1 \/ 2 of the population are adult males and 1 \/ 3 of the population are adult females, then find the number of children in the city.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Population of a city = 6, 63,432<\/p>\n\n\n\n

Population of adult males = (1 \/ 2) of 6,63,432<\/p>\n\n\n\n

= 3,31,716<\/p>\n\n\n\n

Population of adult females = (1 \/ 3) of 6,63,432<\/p>\n\n\n\n

= 2,21,144<\/p>\n\n\n\n

Remaining population can be calculated as below<\/p>\n\n\n\n

Remaining population = 6,63,432 \u2013 (3,31,716 + 2,21,144)<\/p>\n\n\n\n

= 6,63,432 \u2013 5,52,860<\/p>\n\n\n\n

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

= 1,10,572<\/p>\n\n\n\n

Therefore, number of children in a city are 1,10,572<\/p>\n\n\n\n

3. In an election of housing society, there are 30 voters. Each of them gives the vote. Three persons X, Y and Z are standing for the post of Secretary. If Mr X got 2 \/ 5 of the total votes and Mr Z got 1 \/ 3 of the total votes, then find the number of votes which Mr Y got.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Number of votes = 30<\/p>\n\n\n\n

Number of person for election = X, Y, Z<\/p>\n\n\n\n

X got (2 \/ 5) of total votes = (2 \/ 5) of 30<\/p>\n\n\n\n

= (2 \/ 5) \u00d7 30<\/p>\n\n\n\n

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

Z got 1 \/ 3 of total votes = 1 \/ 3 of 30<\/p>\n\n\n\n

= (1 \/ 3) \u00d7 30<\/p>\n\n\n\n

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

Remaining votes can be calculated as below<\/p>\n\n\n\n

= 30 \u2013 (12 + 10)<\/p>\n\n\n\n

= 30 \u2013 22<\/p>\n\n\n\n

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

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

Therefore, Mr Y got 8 votes<\/p>\n\n\n\n

4. A person earns Rs 100 in a day. If he spent Rs  on food and Rs  on petrol. How much did he save on that day?<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

A person\u2019s earning in a day = Rs 100<\/p>\n\n\n\n

Money spent on food = Rs
\"ML= Rs 100 \/ 7<\/p>\n\n\n\n

Money spent on petrol = Rs
\"ML= Rs 92 \/ 3<\/p>\n\n\n\n

The savings of a person is calculated as follows:<\/p>\n\n\n\n

Savings = Rs 100 \u2013 {(100 \/ 7 + 92 \/ 3)}<\/p>\n\n\n\n

= Rs 100 \u2013 {(300 + 644) \/ 21}<\/p>\n\n\n\n

On further calculation, we get<\/p>\n\n\n\n

= Rs 100 \u2013 (944 \/ 21)<\/p>\n\n\n\n

= (2100 \u2013 944) \/ 21<\/p>\n\n\n\n

= Rs 1156 \/ 21<\/p>\n\n\n\n

= Rs
\"ML<\/p>\n\n\n\n

Hence, a person saved Rs
\"MLon that day.<\/p>\n\n\n\n

5. In an examination, 400 students appeared. If 2 \/ 3 of the boys and all 130 girls passed in the examination, then find how many boys failed in an examination?<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Number of students appeared exams = 400<\/p>\n\n\n\n

(2 \/ 3) of total boys and all 130 girls passed in the examination<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Number of total boys = 400 \u2013 130<\/p>\n\n\n\n

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

Number of boys passed = (2 \/ 3) of 270<\/p>\n\n\n\n

= (2 \/ 3) \u00d7 270<\/p>\n\n\n\n

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

So, number of boys failed = 270 \u2013 180<\/p>\n\n\n\n

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

Hence, 90 boys failed in an examination.<\/p>\n\n\n\n

6. A car is moving at the speed of  km \/ h. Find how much distance will it cover in 9 \/ 10 hrs?<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Speed of a car =
\"MLkm \/ h = 122 \/ 3 km \/ h<\/p>\n\n\n\n

Distance covered in 9 \/ 10 hour can be calculated as follows:<\/p>\n\n\n\n

Distance = (122 \/ 3) \u00d7 (9 \/ 10)<\/p>\n\n\n\n

= 366 \/ 10<\/p>\n\n\n\n

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

= 36.6 km<\/p>\n\n\n\n

=
\"MLkm<\/p>\n\n\n\n

Therefore, the distance covered by the car in 9 \/ 10 hours is
\"MLkm<\/p>\n\n\n\n

7. Find the area of a square lawn whose one side is<\/strong>
\"ML m long.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

One side of a square lawn =
\"MLm = 52 \/ 9 m<\/p>\n\n\n\n

The area of a square lawn can be calculated as follows:<\/p>\n\n\n\n

Area = (side)2<\/sup><\/p>\n\n\n\n

= (52 \/ 9)2<\/sup><\/p>\n\n\n\n

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

= 2704 \/ 81 sq. m<\/p>\n\n\n\n

=
\"MLsq. m<\/p>\n\n\n\n

Therefore, the area of a square lawn is
\"MLsq. m<\/p>\n\n\n\n

8. Perimeter of a rectangle is  m. If the length is  m, find its breadth.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Perimeter of a rectangle =
\"MLm<\/p>\n\n\n\n

= 108 \/ 7 m<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

Length + Breadth = (108 \/ 7) \u00f7 2<\/p>\n\n\n\n

= (108 \/ 7) \u00d7 (1 \/ 2)<\/p>\n\n\n\n

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

= 54 \/ 7 m<\/p>\n\n\n\n

Given length =
\"ML<\/p>\n\n\n\n

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

Hence, breadth of a rectangle can be calculated as,<\/p>\n\n\n\n

Breadth = (54 \/ 7) \u2013 (30 \/ 7)<\/p>\n\n\n\n

= 24 \/ 7<\/p>\n\n\n\n

=
\"MLm<\/p>\n\n\n\n

Therefore, the breadth of a rectangle is
\"MLm<\/p>\n\n\n\n

9. Rahul had a rope of  m long. He cut off a  m long piece, then he divided the rest of the rope into 3 parts of equal length. Find the length of each part.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Length of a rope =
\"MLm<\/p>\n\n\n\n

Length of one piece of rope after cut off =
\"MLm<\/p>\n\n\n\n

Remaining length of a rope can be calculated as below<\/p>\n\n\n\n

=
\"ML\u2013
\"ML<\/p>\n\n\n\n

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

=
\"MLm<\/p>\n\n\n\n

= 876 \/ 5 m<\/p>\n\n\n\n

This length divided into three equal parts<\/p>\n\n\n\n

So, length of each part can be calcualted as follows:<\/p>\n\n\n\n

Length of each part = (876 \/ 5) \u00f7 3<\/p>\n\n\n\n

= (876 \/ 5) \u00d7 (1 \/ 3)<\/p>\n\n\n\n

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

= 292 \/ 5 m<\/p>\n\n\n\n

=
\"MLm<\/p>\n\n\n\n

Therefore, the length of each part of a rope is
\"MLm<\/p>\n\n\n\n

10. If  litre of petrol costs Rs , then find the cost of 4 litre of petrol.<\/strong><\/p>\n\n\n\n

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

Given<\/p>\n\n\n\n

Cost of
\"MLlitre = 7 \/ 2 litre of petrol = Rs
\"ML<\/p>\n\n\n\n

= Rs 2163 \/ 8<\/p>\n\n\n\n

Hence, the cost of one litre can be calculated as below:<\/p>\n\n\n\n

Cost of one litre = Rs (2163 \u00d7 2) \/ (8 \u00d7 7)<\/p>\n\n\n\n

The cost of 4 litre of petrol can be calculated as below<\/p>\n\n\n\n

Cost of 4 litre = Rs (2163 \u00d7 2 \u00d74) \/ (8 \u00d7 7)<\/p>\n\n\n\n

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

= Rs 309<\/p>\n\n\n\n

Therefore, the cost of 4 litre of petrol is Rs 309<\/p>\n\n\n\n

11. Ramesh earns Rs 40,000 per month. He spends 3 \/ 8 of the income on food, 1 \/ 5 of the remaining on LIC premium and then 1 \/ 2 of the remaining on other expenses. Find how much money is left with him?<\/strong><\/p>\n\n\n\n

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

Ramesh earnings per month = Rs 40,000<\/p>\n\n\n\n

Expenditure on food = (3 \/ 8) of Rs 40, 000<\/p>\n\n\n\n

= Rs 15,000<\/p>\n\n\n\n

Remaining amount = 40,000 \u2013 15,000<\/p>\n\n\n\n

= Rs 25,000<\/p>\n\n\n\n

Expenditure on LIC premium = (1 \/ 5) of Rs 25,000<\/p>\n\n\n\n

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

Remaining amount = Rs 25000 \u2013 Rs 5000<\/p>\n\n\n\n

= Rs 20,000<\/p>\n\n\n\n

Expenditure on other expenses = (1 \/ 2) of Rs 20,000<\/p>\n\n\n\n

= Rs 10,000<\/p>\n\n\n\n

Remaining amount left = Rs 20,000 \u2013 Rs 10,000<\/p>\n\n\n\n

= Rs 10,000<\/p>\n\n\n\n

Therefore, the remaining amount left with Ramesh is Rs 10,000<\/p>\n\n\n\n

12. A, B, C, D and E went to a restaurant for dinner. A paid 1 \/ 2 of the bill, B paid 1 \/ 5 of the bill and rest of the bill was shared equally by C, D and E. What fractions of the bill was paid by each?<\/strong><\/p>\n\n\n\n

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

Let us consider the total bill of the restaurant = 1<\/p>\n\n\n\n

Bill paid by A = 1 \/ 2<\/p>\n\n\n\n

Bill paid by B = 1 \/ 5<\/p>\n\n\n\n

Remaining bill can be calculated as below:<\/p>\n\n\n\n

Remaining bill = 1 \u2013 {(1 \/ 2) + (1 \/ 5)}<\/p>\n\n\n\n

= 1 \u2013 {(5 + 2) \/ 10}<\/p>\n\n\n\n

= 1 \u2013 (7 \/ 10)<\/p>\n\n\n\n

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

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

Shares of the three persons = (3 \/ 10) \u00f7 3<\/p>\n\n\n\n

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

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

Therefore, each paid (1 \/ 10) of the bill.<\/p>\n\n\n\n

13. 2 \/ 5 of total number of students of a school come by car while 1 \/ 4 of students come by bus to school. All the other students walk to school of which 1 \/ 3 walk on their own and the rest are escorted by their parents. If 224 students come to school walking on their own, how many students study in the school?<\/strong><\/p>\n\n\n\n

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

Let total number of students be 1<\/p>\n\n\n\n

Students who come by car = 2 \/ 5<\/p>\n\n\n\n

Students who come by bus = 1 \/ 4<\/p>\n\n\n\n

Students who come by walking = 1 \/ 3 of remaining<\/p>\n\n\n\n

Rest students = 1 \u2013 (2 \/ 5 + 1 \/ 4)<\/p>\n\n\n\n

= 1 \u2013 (8 + 5) \/ 20<\/p>\n\n\n\n

= 1 \u2013 (13 \/ 20)<\/p>\n\n\n\n

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

= 7 \/ 20<\/p>\n\n\n\n

Number of students who come by walking can be calculated as below<\/p>\n\n\n\n

Number of students who come by walking = 1 \/ 3 of 7 \/ 20<\/p>\n\n\n\n

= 7 \/ 60<\/p>\n\n\n\n

Now, 7 \/ 60 of total students = 224<\/p>\n\n\n\n

Total students = (224 \u00d7 60) \/ 7<\/p>\n\n\n\n

= 32 \u00d7 60<\/p>\n\n\n\n

= 1920<\/p>\n\n\n\n

Hence, 1920 students study in the school<\/p>\n\n\n\n

14. A mother and her two sons got a room constructed for Rs 60,000. The elder son contributes 3 \/ 8 of his mother\u2019s contribution while the younger son contributes 1 \/ 2 of his mother\u2019s share. How much do the three contribute individually?<\/strong><\/p>\n\n\n\n

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

The cost of a room = Rs 60,000<\/p>\n\n\n\n

Elder son contribution = 3 \/ 8 of his mother\u2019s contribution<\/p>\n\n\n\n

Younger son contribution = 1 \/ 2 of his mother\u2019s share<\/p>\n\n\n\n

Let the mother contribution be 1<\/p>\n\n\n\n

Elder son\u2019s contribution = 3 \/ 8<\/p>\n\n\n\n

Younger son\u2019s contribution = 1 \/ 2<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

Ratios in their share = 1: (3 \/ 8): (1 \/ 2)<\/p>\n\n\n\n

= 8: 3: 4<\/p>\n\n\n\n

Sum of ratios = 8 + 3 + 4<\/p>\n\n\n\n

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

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

Mother\u2019s share = (60000 \u00d7 8) \/ 15<\/p>\n\n\n\n

= Rs 32000<\/p>\n\n\n\n

Elder son\u2019s share = (60000 \u00d7 3) \/ 15<\/p>\n\n\n\n

= Rs 12000<\/p>\n\n\n\n

Younger son\u2019s share = (60000 \u00d7 4) \/ 15<\/p>\n\n\n\n

= Rs 16000<\/p>\n\n\n\n

15. In a class of 56 students, the number of boys is 2 \/ 5 th of the number of girls. Find the number of boys and girls.<\/strong><\/p>\n\n\n\n

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

Total number of students in a class = 56<\/p>\n\n\n\n

Let the number of girls be 1<\/p>\n\n\n\n

Then number of boys will be = 2 \/ 5 of 1<\/p>\n\n\n\n

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

Ratios in girls and boys = 1: (2 \/ 5)<\/p>\n\n\n\n

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

Number of girls = {56 \/ (5 + 2)} \u00d7 5<\/p>\n\n\n\n

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

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

And number of boys = (56 \/ 7) \u00d7 2<\/p>\n\n\n\n

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

Therefore, number of boys = 16 and number of girls = 40<\/p>\n\n\n\n

16. A man donated 1 \/ 10 of his money to a school, 1 \/ 6 th of the remaining to a church and the remaining money he distributed equally among his three children. If each child gets Rs 50000, how much money did the man originally have?<\/strong><\/p>\n\n\n\n

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

Let the money of a man be 1<\/p>\n\n\n\n

Money donated to a school = 1 \/ 10<\/p>\n\n\n\n

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

= 9 \/ 10<\/p>\n\n\n\n

Money donated to a church = 1 \/ 6 of 9 \/ 10<\/p>\n\n\n\n

= 3 \/ 20<\/p>\n\n\n\n

Hence, remaining money = (9 \/ 10) \u2013 (3 \/ 20)<\/p>\n\n\n\n

= (18 \u2013 3) \/ 20<\/p>\n\n\n\n

= 15 \/ 20<\/p>\n\n\n\n

A man divides equally to his three children<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

Share of each child = (15 \/ 20) \u00f7 3<\/p>\n\n\n\n

= (15 \/ 20) \u00d7 (1 \/ 3)<\/p>\n\n\n\n

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

Here, each child gets Rs 50000<\/p>\n\n\n\n

Therefore, his total money = Rs 50000 \u00d7 (4 \/ 1)<\/p>\n\n\n\n

= Rs 200000<\/p>\n\n\n\n

17. If 1 \/ 4 of a number is added to 1 \/ 3 of that number, the result is 15 greater than half of that number. Find the number.<\/strong><\/p>\n\n\n\n

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

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

Then as per the condition,<\/p>\n\n\n\n

(1 \/ 4) x + (1 \/ 3) x \u2013 (1 \/ 2) x = 15<\/p>\n\n\n\n

(3x + 4x \u2013 6x) \/ 12 = 15<\/p>\n\n\n\n

(1 \/ 12) x of a number = 15<\/p>\n\n\n\n

x = 15 \u00d7 12 \/ 1<\/p>\n\n\n\n

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

Therefore, the required number is 180<\/p>\n\n\n\n

18. A student was asked to multiply a given number by 4 \/ 5. By mistake, he divided the given number by 4 \/ 5. His answer was 36 more than the correct answer. What was the given number?<\/strong><\/p>\n\n\n\n

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

Let the given number be x<\/p>\n\n\n\n

According to the condition,<\/p>\n\n\n\n

x \u00d7 4 \/ 5 = (4 \/ 5) x<\/p>\n\n\n\n

But by mistake a student divides the given number<\/p>\n\n\n\n

Then,<\/p>\n\n\n\n

x \u00f7 4 \/ 5 = x \u00d7 5 \/ 4<\/p>\n\n\n\n

= (5 \/ 4) x<\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

(5 \/ 4) x \u2013 (4 \/ 5) x = 36<\/p>\n\n\n\n

(25x \u2013 16x) \/ 20 = 36<\/p>\n\n\n\n

9x \/ 20 = 36<\/p>\n\n\n\n

9x = 36 \u00d7 20<\/p>\n\n\n\n

x =(36 \u00d7 20) \/ 9<\/p>\n\n\n\n

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

x = 80<\/p>\n\n\n\n

Therefore, the given number is 80<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 1- Rational Numbers PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n