{"id":602169,"date":"2022-05-13T10:00:48","date_gmt":"2022-05-13T10:00:48","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=602169"},"modified":"2022-05-14T05:37:15","modified_gmt":"2022-05-14T05:37:15","slug":"ml-aggarwal-solutions-for-class-8-maths-chapter-12-linear-equations-and-inequalities-in-one-variable","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ml-aggarwal-solutions-for-class-8-maths-chapter-12-linear-equations-and-inequalities-in-one-variable\/","title":{"rendered":"ML Aggarwal Solutions for Class 8 Maths Chapter 12- Linear Equations and Inequalities in One Variable"},"content":{"rendered":"\n

Class 8: Maths Chapter 12 solutions. Complete Class 8 Maths Chapter 12 Notes.<\/p>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 12- Linear Equations and Inequalities in One Variable<\/h2>\n\n\n\n

ML Aggarwal 8th Maths Chapter 12, Class 8 Maths Chapter 12 solutions<\/p>\n\n\n\n

Exercise 12.1<\/h4>\n\n\n\n

Solve the following equations:<\/strong><\/p>\n\n\n\n

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

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

(ii) 3x \u2013 7 = 3(5 \u2013 x)<\/strong><\/p>\n\n\n\n

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

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

5x \u2013 3x = \u2013 5 + 3<\/p>\n\n\n\n

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

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

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

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

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

3x \u2013 7 = 15 \u2013 3x<\/p>\n\n\n\n

3x + 3x = 15 + 7<\/p>\n\n\n\n

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

6x = 22<\/p>\n\n\n\n

x = 22 \/ 6<\/p>\n\n\n\n

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

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

(i) 4(2x + 1) = 3(x \u2013 1) + 7<\/strong><\/p>\n\n\n\n

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

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

(i) 4(2x + 1) = 3(x \u2013 1) + 7<\/p>\n\n\n\n

8x + 4 = 3x \u2013 3 + 7<\/p>\n\n\n\n

8x + 4 = 3x + 4<\/p>\n\n\n\n

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

8x \u2013 3x = 4 \u2013 4<\/p>\n\n\n\n

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

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

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

6p \u2013 3 = 5 \u2013 3p + 2<\/p>\n\n\n\n

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

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

9p = 10<\/p>\n\n\n\n

p = 10 \/ 9<\/p>\n\n\n\n

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

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

(i) 5y \u2013 2{y \u2013 3(y \u2013 5)} = 6<\/strong><\/p>\n\n\n\n

(ii) 0.3(6 \u2013 x) = 0.4(x + 8)<\/strong><\/p>\n\n\n\n

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

(i) 5y \u2013 2 {y \u2013 3(y \u2013 5)} = 6<\/p>\n\n\n\n

5y \u2013 2 (y \u2013 3y + 15) = 6<\/p>\n\n\n\n

5y \u2013 2 (- 2y + 15) = 6<\/p>\n\n\n\n

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

5y + 4y \u2013 30 = 6<\/p>\n\n\n\n

5y + 4y = 6 + 30<\/p>\n\n\n\n

9y = 36<\/p>\n\n\n\n

y = 36 \/ 9<\/p>\n\n\n\n

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

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

(ii) 0.3 (6 \u2013 x) = 0.4 (x + 8)<\/p>\n\n\n\n

1.8 \u2013 0.3x = 0.4x + 3.2<\/p>\n\n\n\n

\u2013 0.3x \u2013 0.4x = 3.2 \u2013 1.8<\/p>\n\n\n\n

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

\u2013 0.7x = 1.4<\/p>\n\n\n\n

x = \u2013 (1.4 \/ 0.7)<\/p>\n\n\n\n

x = \u2013 (14 \/ 7)<\/p>\n\n\n\n

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

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

(i) (x \u2013 1) \/ 3 = {(x + 2) \/ 6} + 3<\/strong><\/p>\n\n\n\n

(ii) (x + 7) \/ 3 = 1 + {(3x \u2013 2) \/ 5}<\/strong><\/p>\n\n\n\n

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

(i) {(x \u2013 1) \/ 3} = {(x + 2) \/ 6} + 3<\/p>\n\n\n\n

{(x \u2013 1) \/ 3} \u2013 {(x + 2) \/ 6} = 3<\/p>\n\n\n\n

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

{2 (x \u2013 1) \u2013 1(x + 2)} \/ 6 = 3<\/p>\n\n\n\n

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

(x \u2013 4) \/ 6 = 3<\/p>\n\n\n\n

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

x \u2013 4 = 6 \u00d7 3<\/p>\n\n\n\n

x \u2013 4 = 18<\/p>\n\n\n\n

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

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

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

(ii) (x + 7) \/ 3 = 1 + {(3x \u2013 2) \/ 5}<\/p>\n\n\n\n

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

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

{5 (x + 7) \u2013 3 (3x \u2013 2)} \/ 15 = 1<\/p>\n\n\n\n

{(5x + 35) \u2013 (9x \u2013 6)} \/ 15 = 1<\/p>\n\n\n\n

(5x \u2013 9x + 35 + 6) \/ 15 = 1<\/p>\n\n\n\n

(- 4x + 41) \/ 15 = 1<\/p>\n\n\n\n

\u2013 4x + 41 = 15<\/p>\n\n\n\n

\u2013 4x = 15 \u2013 41<\/p>\n\n\n\n

\u2013 4x = \u2013 26<\/p>\n\n\n\n

x = \u2013 26 \/ \u2013 4<\/p>\n\n\n\n

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

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

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

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

(i) {(y + 1) \/ 3} \u2013 {(y \u2013 1) \/ 2} = (1 + 2y) \/ 3<\/strong><\/p>\n\n\n\n

(ii) (p \/ 3) + (p \/ 4) = 55 \u2013 {(p + 40) \/ 5}<\/strong><\/p>\n\n\n\n

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

(i) {(y + 1) \/ 3} \u2013 {(y \u2013 1) \/ 2} = (1 + 2y) \/ 3<\/p>\n\n\n\n

{2 (y + 1) \u2013 3 (y \u2013 1)} \/ 6 = (1 + 2y) \/ 3<\/p>\n\n\n\n

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

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

(- y + 5) \/ 6 = (1 + 2y) \/ 3<\/p>\n\n\n\n

3 (- y + 5) = 6 (1 + 2y)<\/p>\n\n\n\n

\u2013 3y + 15 = 6 + 12y<\/p>\n\n\n\n

\u2013 3y \u2013 12y = 6 \u2013 15<\/p>\n\n\n\n

\u2013 15y = \u2013 9<\/p>\n\n\n\n

y = \u2013 9 \/ \u2013 15<\/p>\n\n\n\n

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

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

(ii) (p \/ 3) + (p \/ 4) = 55 \u2013 {(p + 40) \/ 5}<\/p>\n\n\n\n

(p \/ 3) + (p \/ 4) + {(p + 40)} \/ 5 = 55<\/p>\n\n\n\n

Here, L.C.M. of 3, 4, 5 is 60<\/p>\n\n\n\n

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

{20p +15p + 12(p + 40)} \/ 60 = 55<\/p>\n\n\n\n

(20p + 15p + 12p + 480) \/ 60 = 55<\/p>\n\n\n\n

(47p + 480) \/ 6 = 55<\/p>\n\n\n\n

47p + 480 = 55 \u00d7 60<\/p>\n\n\n\n

47p + 480 = 3300<\/p>\n\n\n\n

47p = 3300 \u2013 480<\/p>\n\n\n\n

47p = 2820<\/p>\n\n\n\n

p = 2820 \/ 47<\/p>\n\n\n\n

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

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

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

(i) n \u2013 {(n \u2013 1) \/ 2} = 1 \u2013 {(n \u2013 2) \/ 3}<\/strong><\/p>\n\n\n\n

(ii) {(3t \u2013 2) \/ 3} + {(2t + 3) \/ 2} = t + (7 \/ 6)<\/strong><\/p>\n\n\n\n

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

(i) n \u2013 {(n \u2013 1) \/ 2} = 1 \u2013 {(n \u2013 2) \/ 3}<\/p>\n\n\n\n

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

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

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

3n + 3 = 10 \u2013 2n<\/p>\n\n\n\n

3n + 2n = 10 \u2013 3<\/p>\n\n\n\n

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

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

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

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

(ii) {(3t \u2013 2) \/ 3} + {(2t + 3) \/ 2} = t + (7 \/ 6)<\/p>\n\n\n\n

{2 (3t \u2013 2) + 3 (2t + 3)} \/ 6 = (6t + 7) \/ 6<\/p>\n\n\n\n

(6t \u2013 4) + (6t + 9) = 6t + 7<\/p>\n\n\n\n

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

6t + 6t + 9 \u2013 4 = 6t + 7<\/p>\n\n\n\n

12t + 5 = 6t + 7<\/p>\n\n\n\n

12t \u2013 6t = 7 \u2013 5<\/p>\n\n\n\n

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

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

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

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

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

(i) 4 (3x + 2) \u2013 5 (6x \u2013 1) = 2 (x \u2013 8) \u2013 6 (7x \u2013 4)<\/strong><\/p>\n\n\n\n

(ii) 3 (5x + 7) + 5 (2x \u2013 11) = 3 (8x \u2013 5) \u2013 15<\/strong><\/p>\n\n\n\n

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

(i) 4 (3x + 2) \u2013 5 (6x \u2013 1) = 2 (x \u2013 8) \u2013 6 (7x \u2013 4)<\/p>\n\n\n\n

12x + 8 \u2013 30x + 5 = 2x \u2013 16 \u2013 42x + 24<\/p>\n\n\n\n

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

\u2013 18x + 13 = \u2013 40x + 8<\/p>\n\n\n\n

\u2013 18x + 40x = 8 \u2013 13<\/p>\n\n\n\n

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

22x = \u2013 5<\/p>\n\n\n\n

x = \u2013 5 \/ 22<\/p>\n\n\n\n

(ii) 3 (5x + 7) + 5 (2x \u2013 11) = 3 (8x \u2013 5) \u2013 15<\/p>\n\n\n\n

15x + 21 + 10x \u2013 55 = 24x \u2013 15 \u2013 15<\/p>\n\n\n\n

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

25x \u2013 34 = 24x \u2013 30<\/p>\n\n\n\n

25x \u2013 24x = \u2013 30 + 34<\/p>\n\n\n\n

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

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

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

(i) (3 \u2013 2x) \/ (2x + 5) = \u2013 (3 \/ 11)<\/strong><\/p>\n\n\n\n

(ii) (5p + 2) \/ (8 \u2013 2p) = 7 \/ 6<\/strong><\/p>\n\n\n\n

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

(i) (3 \u2013 2x) \/ (2x + 5) = \u2013 (3 \/ 11)<\/p>\n\n\n\n

11 (3 \u2013 2x) = \u2013 3 (2x + 5)<\/p>\n\n\n\n

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

33 \u2013 22x = \u2013 6x \u2013 15<\/p>\n\n\n\n

\u2013 22x + 6x = \u2013 15 \u2013 33<\/p>\n\n\n\n

\u2013 16x = \u2013 48<\/p>\n\n\n\n

x = 48 \/ 16<\/p>\n\n\n\n

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

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

(ii) (5p + 2) \/ (8 \u2013 2p) = 7 \/ 6<\/p>\n\n\n\n

6 (5p + 2) = 7 (8 \u2013 2p)<\/p>\n\n\n\n

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

30p + 12 = 56 \u2013 14p<\/p>\n\n\n\n

30p + 14p = 56 \u2013 12<\/p>\n\n\n\n

44p = 44<\/p>\n\n\n\n

p = 44 \/ 44<\/p>\n\n\n\n

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

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

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

(i) 5 \/ x = 7 \/ (x \u2013 4)<\/strong><\/p>\n\n\n\n

(ii) 4 \/ (2x + 3) = 5 \/ (x + 4)<\/strong><\/p>\n\n\n\n

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

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

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

5x \u2013 20 = 7x<\/p>\n\n\n\n

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

5x \u2013 7x = 20<\/p>\n\n\n\n

\u2013 2x = 20<\/p>\n\n\n\n

x = 20 \/ \u2013 2<\/p>\n\n\n\n

x = (- 20 \/ 2)<\/p>\n\n\n\n

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

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

(ii) 4 \/ (2x + 3) = 5 \/ (x + 4)<\/p>\n\n\n\n

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

4x + 16 = 10x + 15<\/p>\n\n\n\n

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

4x \u2013 10x = 15 \u2013 16<\/p>\n\n\n\n

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

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

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

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

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

(i) {(2x + 5) \/ 2} \u2013 {5x \/ (x \u2013 1)} = x<\/strong><\/p>\n\n\n\n

(ii) 1 \/ 5 {(1 \/ 3x) \u2013 5} = 1 \/ 3 {3 \u2013 (1 \/ x)}<\/strong><\/p>\n\n\n\n

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

(i) {(2x + 5) \/ 2} \u2013 {5x \/ (x \u2013 1)} = x<\/p>\n\n\n\n

{(2x + 5) (x \u2013 1) \u2013 (5x) (2)} \/ {2 (x \u2013 1)} = x<\/p>\n\n\n\n

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

{2x (x \u2013 1) + 5 (x \u2013 1) \u2013 10x} \/ (2x \u2013 2) = x<\/p>\n\n\n\n

(2x2<\/sup> \u2013 2x + 5x \u2013 5 \u2013 10x) \/ (2x \u2013 2) = x<\/p>\n\n\n\n

(2x2<\/sup> \u2013 7x \u2013 5) \/ (2x \u2013 2) = x<\/p>\n\n\n\n

2x2<\/sup> \u2013 7x \u2013 5 = x (2x \u2013 2)<\/p>\n\n\n\n

2x2<\/sup> \u2013 7x \u2013 5 = 2x2<\/sup> \u2013 2x<\/p>\n\n\n\n

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

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

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

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

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

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

(ii) 1 \/ 5 {(1 \/ 3x) \u2013 5} = 1 \/ 3 {3 \u2013 (1 \/ x)}<\/p>\n\n\n\n

1 \/ 5 [{1 \u2013 5 (3x)} \/ 3x] = 1 \/ 3 [{(3x \u2013 1)} \/ x]<\/p>\n\n\n\n

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

1 \/ 5 {(1 \u2013 15x) \/ 3x} = 1 \/ 3 {(3x \u2013 1) \/ x}<\/p>\n\n\n\n

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

(1 \u2013 15x) \/ 15x = (3x \u2013 1) \/ 3x<\/p>\n\n\n\n

3x (1 \u2013 15x) = 15x (3x \u2013 1)<\/p>\n\n\n\n

3 (1 \u2013 15x) = 15 (3x \u2013 1)<\/p>\n\n\n\n

3 \u2013 45x = 45x \u2013 15<\/p>\n\n\n\n

\u2013 45x \u2013 45x = \u2013 15 \u2013 3<\/p>\n\n\n\n

\u2013 90x = \u2013 18<\/p>\n\n\n\n

x = 18 \/ 90<\/p>\n\n\n\n

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

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

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

(i) {(2x \u2013 3) \/ (2x \u2013 1)} = {(3x \u2013 1) \/ (3x + 1)}<\/strong><\/p>\n\n\n\n

(ii) {(2y + 3) \/ (3y + 2)} = {(4y + 5) \/ (6y + 7)}<\/strong><\/p>\n\n\n\n

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

(i) {(2x \u2013 3) \/ (2x \u2013 1)} = {(3x \u2013 1) \/ (3x + 1)}<\/p>\n\n\n\n

(2x \u2013 3) (3x + 1) = (3x \u2013 1) (2x \u2013 1)<\/p>\n\n\n\n

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

6x2<\/sup> + 2x \u2013 9x \u2013 3 = 6x2<\/sup> \u2013 3x \u2013 2x + 1<\/p>\n\n\n\n

6x2<\/sup> \u2013 7x \u2013 3 = 6x2<\/sup> \u2013 5x + 1<\/p>\n\n\n\n

6x2<\/sup> \u2013 7x \u2013 6x2<\/sup> + 5x = 1 + 3<\/p>\n\n\n\n

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

\u2013 2x = 4<\/p>\n\n\n\n

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

x = 4 \/ \u2013 2<\/p>\n\n\n\n

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

(ii) {(2y + 3) \/ (3y + 2)} = {(4y + 5) \/ (6y + 7)}<\/p>\n\n\n\n

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

12y2<\/sup> + 14y + 18y + 21 = 12y2<\/sup> + 8y + 15y + 10<\/p>\n\n\n\n

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

32y + 21 = 23y + 10<\/p>\n\n\n\n

32y \u2013 23y = 10 \u2013 21<\/p>\n\n\n\n

9y = \u2013 11<\/p>\n\n\n\n

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

y = \u2013 11 \/ 9<\/p>\n\n\n\n

12. If x = p + 1, find the value of p from the equation (1 \/ 2) (5x \u2013 30) \u2013 (1\/ 3) (1 + 7p) = 1 \/ 4<\/strong><\/p>\n\n\n\n

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

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

x = p + 1 \u2026.. (1)<\/p>\n\n\n\n

(1 \/ 2) (5x \u2013 30) \u2013 (1 \/ 3) (1 + 7p) = 1 \/ 4 \u2026\u2026\u2026. (2)<\/p>\n\n\n\n

Substituting the value of x from (1) in (2), we get,<\/p>\n\n\n\n

(1 \/ 2) {5 (p + 1) \u2013 30} \u2013 (1 \/ 3) (1 + 7p) = 1 \/ 4<\/p>\n\n\n\n

1 \/ 2 (5p + 5 \u2013 30) \u2013 1 \/ 3 (1 + 7p) = 1 \/ 4<\/p>\n\n\n\n

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

1 \/ 2 (5p \u2013 25) \u2013 1 \/ 3 (1 + 7p) = 1 \/ 4<\/p>\n\n\n\n

(5p \u2013 25) \/ 2 \u2013 (1 + 7p) \/ 3 = 1 \/ 4<\/p>\n\n\n\n

{3 (5p \u2013 25) \u2013 2 (1 + 7p)} \/ 6 = 1 \/ 4<\/p>\n\n\n\n

(15p \u2013 75 \u2013 2 \u2013 14p) \/ 6 = 1 \/ 4<\/p>\n\n\n\n

(p \u2013 77) \/ 6 = 1 \/ 4<\/p>\n\n\n\n

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

4 (p \u2013 77) = 6 (1)<\/p>\n\n\n\n

4p \u2013 308 = 6<\/p>\n\n\n\n

4p = 6 + 308<\/p>\n\n\n\n

4p = 314<\/p>\n\n\n\n

p = 314 \/ 4<\/p>\n\n\n\n

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

p = 157 \/ 2<\/p>\n\n\n\n

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

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

Solve {(x + 3) \/ 3} \u2013 {(x \u2013 2) \/ 2} = 1, Hence find p if (1 \/ x) + P = 1<\/strong><\/p>\n\n\n\n

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

{(x + 3) \/ 3} \u2013 {(x \u2013 2) \/ 2} = 1<\/p>\n\n\n\n

{2 (x + 3) \u2013 3(x \u2013 2)} \/ 6 = 1<\/p>\n\n\n\n

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

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

(- x + 12) \/ 6 = 1<\/p>\n\n\n\n

\u2013 x + 12 = 1 \u00d7 6<\/p>\n\n\n\n

\u2013 x + 12 = 6<\/p>\n\n\n\n

\u2013 x = 6 \u2013 12<\/p>\n\n\n\n

\u2013 x = \u2013 6<\/p>\n\n\n\n

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

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

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

(1 \/ x) + P = 1<\/p>\n\n\n\n

Substituting x = 6, we get,<\/p>\n\n\n\n

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

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

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

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

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

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

Exercise 12.2<\/h4>\n\n\n\n

1. Three more than twice a number is equal to four less than the number. Find the number.<\/strong><\/p>\n\n\n\n

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

Let the number be x<\/p>\n\n\n\n

Twice the number = 2x<\/p>\n\n\n\n

As per the given statement,<\/p>\n\n\n\n

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

3 + 2x + 4 = x<\/p>\n\n\n\n

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

7 = x \u2013 2x<\/p>\n\n\n\n

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

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

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

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

Therefore, the required number is -7<\/p>\n\n\n\n

2. When four consecutive integers are added, the sum is 46. Find the integers.<\/strong><\/p>\n\n\n\n

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

Let the first integer be x, then the next three consecutive integers will be,<\/p>\n\n\n\n

(x + 1), (x + 2) and (x + 3)<\/p>\n\n\n\n

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

x + (x + 1) + (x + 2) + (x + 3) = 46<\/p>\n\n\n\n

x + x + 1 + x + 2 + x + 3 = 46<\/p>\n\n\n\n

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

4x + 6 = 46<\/p>\n\n\n\n

4x = 46 \u2013 6<\/p>\n\n\n\n

4x = 40<\/p>\n\n\n\n

x = 40 \/ 4<\/p>\n\n\n\n

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

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

Therefore, four consecutive integers are 10, (10 + 1), (10 + 2) and (10 + 3)<\/p>\n\n\n\n

i.e. 10, 11, 12 and 13<\/p>\n\n\n\n

3. Manjula thinks a number and subtracts 7 \/ 3 from it. She multiplies the result by 6. The result now obtained is 2 less than twice the same number she thought of. What is the number?<\/strong><\/p>\n\n\n\n

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

Let a number thought by Manjula be x<\/p>\n\n\n\n

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

(x \u2013 7 \/ 3) \u00d7 6 = 2x \u2013 2<\/p>\n\n\n\n

6x \u2013 14 = 2x \u2013 2<\/p>\n\n\n\n

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

6x \u2013 2x = \u2013 2 + 14<\/p>\n\n\n\n

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

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

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

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

Therefore, the required number is 3<\/p>\n\n\n\n

4. A positive number is 7 times another number. If 15 is added to both the numbers, then one of the new numbers becomes (5 \/ 2) times the other new number. What are the numbers?<\/strong><\/p>\n\n\n\n

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

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

Then the other number = x \/ 7<\/p>\n\n\n\n

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

x + 15 = 5 \/ 2 {(x \/ 7) + 15}<\/p>\n\n\n\n

2 (x + 15) = (5x \/ 7) + (5 \u00d7 15)<\/p>\n\n\n\n

2x + 30 = (5x \/ 7) + 75<\/p>\n\n\n\n

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

2x \u2013 (5 \/ 7) x = 75 \u2013 30<\/p>\n\n\n\n

{(14 \u2013 5) \/ 7}x = 45<\/p>\n\n\n\n

9x \/ 7 = 45<\/p>\n\n\n\n

x = 45 \u00d7 7 \/ 9<\/p>\n\n\n\n

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

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

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

One number = 35<\/p>\n\n\n\n

Other number = 35 \/ 7 = 5<\/p>\n\n\n\n

Therefore, the numbers are 35 and 5<\/p>\n\n\n\n

5. When three consecutive even integers are added, the sum is zero. Find the integers.<\/strong><\/p>\n\n\n\n

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

Let the first even integer = x<\/p>\n\n\n\n

Then next two consecutive even integers = (x + 2) and (x + 4)<\/p>\n\n\n\n

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

x + (x + 2) + (x + 4) = 0<\/p>\n\n\n\n

x + x + 2 + x + 4 = 0<\/p>\n\n\n\n

3x + 6 = 0<\/p>\n\n\n\n

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

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

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

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

Therefore, three consecutive integers are -2, (- 2 + 2) and (- 2 + 4) i.e. -2, 0 and 2<\/p>\n\n\n\n

6. Find two consecutive odd integers such that two-fifth of the smaller exceeds two-ninth of the greater by 4.<\/strong><\/p>\n\n\n\n

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

Let the first odd integer = x<\/p>\n\n\n\n

Then next consecutive odd integers = (x + 2)<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

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

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

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

(2x \/ 5) \u2013 {2 (x + 2)} \/ 9 = 4<\/p>\n\n\n\n

{18x \u2013 10 (x + 2)} \/ 45 = 4<\/p>\n\n\n\n

(18x \u2013 10x \u2013 20) \/ 45 = 4<\/p>\n\n\n\n

(8x \u2013 20) \/ 45 = 4<\/p>\n\n\n\n

8x \u2013 20 = 4 \u00d7 45<\/p>\n\n\n\n

8x \u2013 20 = 180<\/p>\n\n\n\n

8x = 180 + 20<\/p>\n\n\n\n

8x = 200<\/p>\n\n\n\n

x = 200 \/ 8<\/p>\n\n\n\n

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

x = 25<\/p>\n\n\n\n

So, two consecutive odd integers are x = 25 and<\/p>\n\n\n\n

(x + 2) = (25 + 2) = 27<\/p>\n\n\n\n

Therefore, two consecutive odd integers are 25 and 27<\/p>\n\n\n\n

7. The denominator of a fraction is 1 more than twice its numerator. If the numerator and denominator are both increased by 5, it becomes (3 \/ 5). Find the original fraction.<\/strong><\/p>\n\n\n\n

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

Let the numerator of the original fraction = x<\/p>\n\n\n\n

Then, its denominator = 2x + 1<\/p>\n\n\n\n

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

The fraction = x \/ (2x + 1)<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

(x + 5) \/ {(2x + 1) + 5} = 3 \/ 5<\/p>\n\n\n\n

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

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

5 (x + 5) = 3 (2x + 6)<\/p>\n\n\n\n

5x + 25 = 6x + 18<\/p>\n\n\n\n

5x \u2013 6x = 18 \u2013 25<\/p>\n\n\n\n

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

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

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

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

Original fraction = x \/ (2x + 1)<\/p>\n\n\n\n

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

= 7 \/ (14 + 1)<\/p>\n\n\n\n

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

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

Therefore, the original fraction is 7 \/ 15<\/p>\n\n\n\n

8. Find two positive numbers in the ratio 2: 5 such that their difference is 15.<\/strong><\/p>\n\n\n\n

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

Let the two numbers be 2x and 5x<\/p>\n\n\n\n

Because the ratio of these two numbers = 2x \/ 5x<\/p>\n\n\n\n

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

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

According to the given problem,<\/p>\n\n\n\n

5x \u2013 2x = 15<\/p>\n\n\n\n

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

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

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

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

So, the numbers are 2x = 2 \u00d7 5 = 10 and 5x = 5 \u00d7 5 = 25<\/p>\n\n\n\n

Therefore the required numbers are 10 and 25<\/p>\n\n\n\n

9. What number should be added to each of the numbers 12, 22, 42 and 72 so that the resulting numbers may be in proportion?<\/strong><\/p>\n\n\n\n

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

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

According to the given problem,<\/p>\n\n\n\n

(12 + x), (22 + x), (42 + x) and (72 + x) are in proportion<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

(12 + x) \/ (22 + x) = (42 + x) \/ (72 + x)<\/p>\n\n\n\n

On cross multiplication, we get,<\/p>\n\n\n\n

(12 + x) (72 + x) = (42 + x) (22 + x)<\/p>\n\n\n\n

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

12 (72 + x) + x (72 + x) = 42 (22 + x) + x (22 + x)<\/p>\n\n\n\n

864 + 12x + 72x + x2<\/sup> = 924 + 42x + 22x + x2<\/sup><\/p>\n\n\n\n

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

864 + 84x + x2<\/sup> = 924 + 64x + x2<\/sup><\/p>\n\n\n\n

864 + 84x + x2<\/sup> \u2013 924 \u2013 64x \u2013 x2<\/sup> = 0<\/p>\n\n\n\n

864 + 84x \u2013 64x \u2013 924 = 0<\/p>\n\n\n\n

84x \u2013 64x = 924 \u2013 864<\/p>\n\n\n\n

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

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

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

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

Therefore, the required number is 3<\/p>\n\n\n\n

10. The digits of a two-digit number differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What can be the original number?<\/strong><\/p>\n\n\n\n

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

Let one\u2019s digit of a two-digit number be x<\/p>\n\n\n\n

Given that the difference between both the digits is 3,<\/p>\n\n\n\n

Then ten\u2019s digit = x + 3<\/p>\n\n\n\n

Hence, the number = x + 10 (x + 3)<\/p>\n\n\n\n

= x + 10x + 30<\/p>\n\n\n\n

= 11x + 30<\/p>\n\n\n\n

By interchanging the digits, we get,<\/p>\n\n\n\n

One\u2019s digit of a new number = x + 3 and<\/p>\n\n\n\n

Ten\u2019s digit of a new number = x<\/p>\n\n\n\n

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

Number = x + 3 + 10x = 11x + 3<\/p>\n\n\n\n

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

11x + 30 + 11x + 3 = 143<\/p>\n\n\n\n

22x + 33 = 143<\/p>\n\n\n\n

22x = 143 \u2013 33<\/p>\n\n\n\n

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

22x = 110<\/p>\n\n\n\n

x = 110 \/ 22<\/p>\n\n\n\n

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

Therefore, original number = 11x + 30<\/p>\n\n\n\n

= 11 \u00d7 5 + 30<\/p>\n\n\n\n

= 55 + 30<\/p>\n\n\n\n

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

= 85<\/p>\n\n\n\n

Hence, the original number is 85<\/p>\n\n\n\n

11. Sum of the digits of a two-digit number is 11. When we interchange the digits, it is found that the resulting new number is greater than the original number by 63. Find the two- digit number.<\/strong><\/p>\n\n\n\n

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

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

Sum of the digits of a two-digit numbers = 11<\/p>\n\n\n\n

Let unit\u2019s digit of a 2-digit number be x<\/p>\n\n\n\n

Then ten\u2019s digit will be 11 \u2013 x<\/p>\n\n\n\n

So, number = x + 10 (11 \u2013 x)<\/p>\n\n\n\n

= x + 110 \u2013 10x<\/p>\n\n\n\n

= 110 \u2013 9x<\/p>\n\n\n\n

Now, by interchanging the digits, we get,<\/p>\n\n\n\n

One\u2019s digit of a new number = 11 \u2013 x<\/p>\n\n\n\n

And ten\u2019s digit will be = x<\/p>\n\n\n\n

Hence, number = 11 \u2013 x + 10x<\/p>\n\n\n\n

= 11 + 9x<\/p>\n\n\n\n

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

11 + 9x \u2013 (110 \u2013 9x) = 63<\/p>\n\n\n\n

11 + 9x \u2013 110 + 9x = 63<\/p>\n\n\n\n

18x = 63 \u2013 11 + 110<\/p>\n\n\n\n

18x = 162<\/p>\n\n\n\n

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

x = 162 \/ 18<\/p>\n\n\n\n

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

So, original number = 110 \u2013 9x<\/p>\n\n\n\n

= 110 \u2013 9 \u00d7 9<\/p>\n\n\n\n

= 110 \u2013 81<\/p>\n\n\n\n

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

Therefore, the original number is 29<\/p>\n\n\n\n

12. Ritu is now four times as old as his brother Raju. In 4 years time, her age will be twice of Raju\u2019s age. What are their present ages?<\/strong><\/p>\n\n\n\n

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

Let the age of Raju be x years<\/p>\n\n\n\n

Then the age of Ritu will be = 4 \u00d7 x years<\/p>\n\n\n\n

In 4 years time,<\/p>\n\n\n\n

Age of Raju = (x + 4) years<\/p>\n\n\n\n

Age of Ritu = (4x + 4) years<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

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

4x + 4 = 2x + 8<\/p>\n\n\n\n

4x \u2013 2x = 8 \u2013 4<\/p>\n\n\n\n

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

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

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

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

Therefore, Raju\u2019s age = 2 years and<\/p>\n\n\n\n

Ritu\u2019s age = 4 \u00d7 2 = 8 years<\/p>\n\n\n\n

13. A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. How old are they now?<\/strong><\/p>\n\n\n\n

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

Let the present age of son be x years<\/p>\n\n\n\n

Then, age of his father will be 7x years<\/p>\n\n\n\n

Two years ago, age of son = (x \u2013 2) years<\/p>\n\n\n\n

Two years ago, age of his father = (7x \u2013 2) years<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

7x \u2013 2 = 13 (x \u2013 2)<\/p>\n\n\n\n

7x \u2013 2 = 13x \u2013 26<\/p>\n\n\n\n

7x \u2013 13x = \u2013 26 + 2<\/p>\n\n\n\n

\u2013 6x = \u2013 24<\/p>\n\n\n\n

x = \u2013 24 \/- 6<\/p>\n\n\n\n

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

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

Therefore, age of son = 4 years and<\/p>\n\n\n\n

Age of his father = 7x = 7 \u00d7 4 = 28 years<\/p>\n\n\n\n

14. The ages of Sona and Sonali are in the ratio 5: 3. Five years hence, the ration of their ages will be 10: 7. Find their present ages.<\/strong><\/p>\n\n\n\n

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

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

Ratio of ages of Sona and Sonali = 5: 3<\/p>\n\n\n\n

Let us consider the present age of Sona and Sonali be 5x and 3x years respectively<\/p>\n\n\n\n

Five years hence,<\/p>\n\n\n\n

The age of Sona = 5x + 5 and<\/p>\n\n\n\n

The age of Sonali = 3x + 5<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

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

On cross multiplication, we get,<\/p>\n\n\n\n

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

35x + 35 = 30x + 50<\/p>\n\n\n\n

35x \u2013 30x = 50 \u2013 35<\/p>\n\n\n\n

5x = 15<\/p>\n\n\n\n

x = 15 \/ 5<\/p>\n\n\n\n

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

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

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

Present age of Sona = 5x = 5 \u00d7 3 = 15 years and<\/p>\n\n\n\n

Present age of Sonali = 3x = 3 \u00d7 3 = 9 years<\/p>\n\n\n\n

Therefore, the present age of Sona and Sonali is 15years and 9 years<\/p>\n\n\n\n

15. An employee works in a company on a contract of 30 days on the condition that he will receive Rs 200 for each day he works and he will be fined Rs 20 for each day if he is absent. If he receives Rs 3800 in all, for how many days did he remain absent?<\/strong><\/p>\n\n\n\n

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

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

Period of contract = 30 days<\/p>\n\n\n\n

If an employee works a day, he will get = Rs 200<\/p>\n\n\n\n

If he is absent, he will be fined = Rs 20 per day<\/p>\n\n\n\n

At the end of contract period, he gets = Rs 3800<\/p>\n\n\n\n

Let an employee remain absent for x days<\/p>\n\n\n\n

Then, number of days, he worked = (30 \u2013 x) days<\/p>\n\n\n\n

According to the given condition,<\/p>\n\n\n\n

(30 \u2013 x) \u00d7 200 \u2013 x \u00d7 20 = 3800<\/p>\n\n\n\n

6000 \u2013 200x \u2013 20x = 3800<\/p>\n\n\n\n

6000 \u2013 220x = 3800<\/p>\n\n\n\n

220x = 6000 \u2013 3800<\/p>\n\n\n\n

220x = 2200<\/p>\n\n\n\n

x = 2200 \/ 220<\/p>\n\n\n\n

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

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

Therefore, an employee remained absent for 10 days.<\/p>\n\n\n\n

16. I have a total of Rs 300 in coins of denomination Rs 1, Rs 2 and Rs 5. The number of coins is 3 times the number of Rs 5 coins. The total number of coins is 160. How many coins of each denomination are with me?<\/strong><\/p>\n\n\n\n

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

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

Amount of coins = Rs 300<\/p>\n\n\n\n

Total number of coins = 160<\/p>\n\n\n\n

Le t the number of coins of Rs 5 = x<\/p>\n\n\n\n

Then number of coins of Rs 2 = 3x<\/p>\n\n\n\n

And number of coins of Rs 1 = 160 \u2013 (x + 3x)<\/p>\n\n\n\n

= 160 \u2013 4x<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

(160 \u2013 4x) \u00d7 1 + 3x \u00d7 2 + x \u00d7 5 = 300<\/p>\n\n\n\n

160 \u2013 4x + 6x + 5x = 300<\/p>\n\n\n\n

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

160 + 7x = 300<\/p>\n\n\n\n

7x = 300 \u2013 160<\/p>\n\n\n\n

7x = 140<\/p>\n\n\n\n

x = 140 \/ 7<\/p>\n\n\n\n

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

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

Hence, 5 Rs coins = 20<\/p>\n\n\n\n

2 Rs coins = 3x = 3 \u00d7 20 = 60 and<\/p>\n\n\n\n

1 Rs coins = 160 \u2013 60 \u2013 20 = 80<\/p>\n\n\n\n

17. A local bus is carrying 40 passengers, some with Rs 5 tickets and the remaining with Rs 7.50 tickets. If the total receipts from these passengers are Rs 230, find the number of passengers with Rs 5 tickets.<\/strong><\/p>\n\n\n\n

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

Let the number of passengers with Rs 5 tickets = x<\/p>\n\n\n\n

Then, the number of passengers with Rs 7.50 tickets = (40 \u2013 x)<\/p>\n\n\n\n

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

5 \u00d7 x + (40 \u2013 x) \u00d7 7.50 = 230<\/p>\n\n\n\n

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

5x + 300 \u2013 7.5x = 230<\/p>\n\n\n\n

5x \u2013 7.5x = 230 \u2013 300<\/p>\n\n\n\n

-2.5x = \u2013 70<\/p>\n\n\n\n

x = -70 \/ \u2013 2.5<\/p>\n\n\n\n

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

x = 28<\/p>\n\n\n\n

Therefore, the number of passengers with Rs 5 tickets = 28<\/p>\n\n\n\n

18. On a school picnic, a group of students agree to pay equally for the use of a full boat and pay Rs 10 each. If there had been 3 more students in the group, each would have paid Rs 2 less. How many students were there in the group?<\/strong><\/p>\n\n\n\n

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

Let the number of students in a group be x<\/p>\n\n\n\n

If there are 3 more students in a group, then,<\/p>\n\n\n\n

The total number of students in a group = x + 3<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

10 \u00d7 x = (x + 3) \u00d7 (10 \u2013 2)<\/p>\n\n\n\n

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

10x = (x + 3) \u00d7 8<\/p>\n\n\n\n

10x = 8 (x + 3)<\/p>\n\n\n\n

10x = 8x + 24<\/p>\n\n\n\n

10x \u2013 8x = 24<\/p>\n\n\n\n

2x = 24<\/p>\n\n\n\n

x = 24 \/ 2<\/p>\n\n\n\n

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

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

Therefore, the total number of students in the group = 12<\/p>\n\n\n\n

19. Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.<\/strong><\/p>\n\n\n\n

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

Let the number of deer in the herd = x<\/p>\n\n\n\n

Number of deer grazing in the field = x \/ 2<\/p>\n\n\n\n

Remaining = x \u2013 (x \/ 2)<\/p>\n\n\n\n

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

Given that the (3 \/ 4) of the remaining deer are playing<\/p>\n\n\n\n

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

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

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

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

Rest of deer = (x \/ 2) \u2013 (3 \/ 8) x<\/p>\n\n\n\n

= (1 \/ 8) x<\/p>\n\n\n\n

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

(1 \/ 8) x = 9<\/p>\n\n\n\n

x = 9 \u00d7 8<\/p>\n\n\n\n

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

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

Hence, total number of deer in the herd = 72<\/p>\n\n\n\n

20. Sakshi takes some flowers in a basket and visits three temples one by one. At each temple, she offers one-half of the flowers from the basket. If she is left with 6 flowers at the end, find the number of flowers she had in the beginning.<\/strong><\/p>\n\n\n\n

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

Let the total number of flowers in the basket = x<\/p>\n\n\n\n

Flowers offered in first temple = x \/ 2<\/p>\n\n\n\n

Remaining flowers = x \u2013 (x \/ 2)<\/p>\n\n\n\n

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

Flowers offered in the second temple<\/p>\n\n\n\n

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

Remaining flowers = (x \/ 2) \u2013 (x \/ 4)<\/p>\n\n\n\n

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

Flowers offered in the third temple = (x \/ 4) \u00d7 (1 \/ 2)<\/p>\n\n\n\n

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

Remaining flowers = (x \/ 4) \u2013 (x \/ 8)<\/p>\n\n\n\n

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

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

(x \/ 8) = 6<\/p>\n\n\n\n

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

x = 6 \u00d7 8<\/p>\n\n\n\n

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

Therefore, number of flowers she had in the beginning = 48<\/p>\n\n\n\n

21. Two supplementary angles differ by 500<\/sup>. Find the measure of each angle.<\/strong><\/p>\n\n\n\n

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

Let the measure of angle be x<\/p>\n\n\n\n

Then, its supplementary angle = 1800<\/sup> \u2013 x<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

x \u2013 (1800<\/sup> \u2013x) = 500<\/sup><\/p>\n\n\n\n

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

x \u2013 1800<\/sup> + x = 500<\/sup><\/p>\n\n\n\n

2x \u2013 1800<\/sup> = 500<\/sup><\/p>\n\n\n\n

2x = 1800<\/sup> + 500<\/sup><\/p>\n\n\n\n

2x = 2300<\/sup><\/p>\n\n\n\n

x = 2300<\/sup> \/ 2<\/p>\n\n\n\n

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

x = 1150<\/sup><\/p>\n\n\n\n

Measurement of each angle = x = 1150<\/sup> and<\/p>\n\n\n\n

(1800<\/sup> \u2013 x) = 1800<\/sup> \u2013 1150<\/sup><\/p>\n\n\n\n

= 650<\/sup><\/p>\n\n\n\n

Therefore, the measurement of each angel is 1150<\/sup> and 650<\/sup><\/p>\n\n\n\n

22. If the angles of a triangle are in the ratio 5: 6: 7, find the angles.<\/strong><\/p>\n\n\n\n

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

Let the angles of a triangle are 5x, 6x and 7x<\/p>\n\n\n\n

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

5x + 6x + 7x = 1800<\/sup><\/p>\n\n\n\n

18x = 1800<\/sup><\/p>\n\n\n\n

x = 1800<\/sup> \/ 18<\/p>\n\n\n\n

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

x = 100<\/sup><\/p>\n\n\n\n

Now, the angles of a triangle are,<\/p>\n\n\n\n

5x = 5 \u00d7 100<\/sup> = 500<\/sup><\/p>\n\n\n\n

6x = 6 \u00d7 100<\/sup> = 600<\/sup><\/p>\n\n\n\n

7x = 7 \u00d7 100<\/sup> = 700<\/sup><\/p>\n\n\n\n

Therefore, the angles of a triangle are 500<\/sup>, 600<\/sup> and 700<\/sup><\/p>\n\n\n\n

23. Two equal sides of an isosceles triangle are 3x \u2013 1 and 2x + 2 units. The third side is 2x units. Find x and the perimeter of the triangle.<\/strong><\/p>\n\n\n\n

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

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

Two equal sides of an isosceles triangle are 3x \u2013 1 and 2x + 2<\/p>\n\n\n\n

That is, 3x \u2013 1 = 2x + 2<\/p>\n\n\n\n

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

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

Given that,<\/p>\n\n\n\n

Third side of triangle = 2x<\/p>\n\n\n\n

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

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

Equal sides of a triangle = 3x \u2013 1<\/p>\n\n\n\n

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

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

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

Perimeter of the triangle = 8 + 8 + 6<\/p>\n\n\n\n

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

Therefore, the perimeter of the triangle = 22 units<\/p>\n\n\n\n

24. If each side of a triangle is increased by 4 cm, the ratio of the perimeters of the new triangle and the given triangle is 7: 5. Find the perimeter of the given triangle.<\/strong><\/p>\n\n\n\n

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

Let the perimeter of original triangle be x cm<\/p>\n\n\n\n

If each side of a triangle is increased by 4, then,<\/p>\n\n\n\n

The perimeter will be = x + 4 \u00d7 3<\/p>\n\n\n\n

= (x + 12) cm<\/p>\n\n\n\n

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

Ratio of perimeter of new triangle and given triangle = 7: 5<\/p>\n\n\n\n

(x + 12) \/ x = 7 \/ 5<\/p>\n\n\n\n

On cross multiplication, we get,<\/p>\n\n\n\n

5 (x + 12) = 7x<\/p>\n\n\n\n

5x + 60 = 7x<\/p>\n\n\n\n

7x \u2013 5x = 60<\/p>\n\n\n\n

2x = 60<\/p>\n\n\n\n

x = 60 \/ 2<\/p>\n\n\n\n

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

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

Therefore, the perimeter of the given triangle is 30 cm<\/p>\n\n\n\n

25. The length of a rectangle is 5 cm less than twice its breadth. If the length is decreased by 3 cm and breadth increased by 2 cm, the perimeter of the resulting rectangle is 72 cm. Find the area of the original rectangle.<\/strong><\/p>\n\n\n\n

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

Let the breadth of the original rectangle be x cm<\/p>\n\n\n\n

Then, length of the original rectangle will be (2x \u2013 5) cm<\/p>\n\n\n\n

If the length is decreased by 3 cm, then,<\/p>\n\n\n\n

New length = {(2x \u2013 5) \u2013 3}<\/p>\n\n\n\n

= (2x \u2013 8) cm<\/p>\n\n\n\n

If breadth is increased by 2 cm, then,<\/p>\n\n\n\n

New breadth = (x + 2) cm<\/p>\n\n\n\n

New perimeter = 2 (new length + new breadth)<\/p>\n\n\n\n

= 2 {(2x \u2013 8) + (x + 2)}<\/p>\n\n\n\n

= 2 (2x \u2013 8 + x + 2)<\/p>\n\n\n\n

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

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

= 6x \u2013 12<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

6x \u2013 12 = 72<\/p>\n\n\n\n

6x = 72 + 12<\/p>\n\n\n\n

6x = 84<\/p>\n\n\n\n

x = 84 \/ 6<\/p>\n\n\n\n

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

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

Breadth of the original rectangle = 14 cm and<\/p>\n\n\n\n

Length of the original rectangle = (2x \u2013 5)<\/p>\n\n\n\n

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

= 28 \u2013 5<\/p>\n\n\n\n

= 23 cm<\/p>\n\n\n\n

Area of original rectangle = Length \u00d7 Breadth<\/p>\n\n\n\n

= (23 \u00d7 14) cm2<\/sup><\/p>\n\n\n\n

= 322 cm2<\/sup><\/p>\n\n\n\n

Therefore, area of the original rectangle is 322 cm2<\/sup><\/p>\n\n\n\n

26. A rectangle is 10 cm long and 8 cm wide. When each side of the rectangle is increased by x cm, its perimeter is doubled. Find the equation in x and hence find the area of the new rectangle.<\/strong><\/p>\n\n\n\n

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

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

Length of rectangle (l) = 10 cm and<\/p>\n\n\n\n

Breadth of the rectangle = 8 cm<\/p>\n\n\n\n

Perimeter = 2 (Length + Breadth)<\/p>\n\n\n\n

= 2 (10 + 8) cm<\/p>\n\n\n\n

= 2 \u00d7 18<\/p>\n\n\n\n

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

If each side of the rectangle is increased by x cm, then,<\/p>\n\n\n\n

Perimeter = 2 (10 + x + 8 + x)<\/p>\n\n\n\n

= 2 (18 + 2x)<\/p>\n\n\n\n

= (36 + 4x) cm<\/p>\n\n\n\n

According to the given condition,<\/p>\n\n\n\n

36 + 4x = 2 (36)<\/p>\n\n\n\n

36 + 4x = 72<\/p>\n\n\n\n

4x = 72 \u2013 36<\/p>\n\n\n\n

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

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

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

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

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

Length of new rectangle = 1 + x = 10 + 9 = 19 cm and<\/p>\n\n\n\n

Breadth of new rectangle = b + x = 8 + 9 = 17 cm<\/p>\n\n\n\n

Area = Length \u00d7 Breadth = 19 \u00d7 17 cm2<\/sup> = 323 cm2<\/sup><\/p>\n\n\n\n

27. A streamer travels 90 km downstream in the same time as it takes to travel 60 km upstream. If the speed of the steamer is 5 km\/ hr, find the speed of the steamer in still water.<\/strong><\/p>\n\n\n\n

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

Let the speed of the steamer = x km\/ h<\/p>\n\n\n\n

The speed downstream = (x + 5) km\/h and<\/p>\n\n\n\n

The speed upstream = (x \u2013 5) km\/ h<\/p>\n\n\n\n

According to the given problem,<\/p>\n\n\n\n

90 \/ (x + 5) = 60 \/ (x \u2013 5)<\/p>\n\n\n\n

On cross multiplication, we get,<\/p>\n\n\n\n

90 (x \u2013 5) = 60 (x + 5)<\/p>\n\n\n\n

90x \u2013 450 = 60x + 300<\/p>\n\n\n\n

90x \u2013 60x = 300 + 450<\/p>\n\n\n\n

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

30x = 750<\/p>\n\n\n\n

x = 750 \/ 30<\/p>\n\n\n\n

x = 25<\/p>\n\n\n\n

Therefore, the speed of the streamer in still water is 25 km\/ h<\/p>\n\n\n\n

28. A steamer goes downstream and covers the distance between two ports in 5 hours while it covers the same distance upstream in 6 hours. If the speed of the stream is 1 km\/h, find the speed of the streamer in still water and the distance between two ports.<\/strong><\/p>\n\n\n\n

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

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

Speed of the stream in still water = 1 km\/h<\/p>\n\n\n\n

Let speed of the streamer = x km\/h<\/p>\n\n\n\n

Speed downstream = (x + 1) km\/h<\/p>\n\n\n\n

Speed upstream = (x \u2013 1) km\/h<\/p>\n\n\n\n

According to the given condition,<\/p>\n\n\n\n

(x + 1) \u00d7 5 = (x \u2013 1) \u00d7 6<\/p>\n\n\n\n

5x + 5 = 6x \u2013 6<\/p>\n\n\n\n

Calculating further, we get,<\/p>\n\n\n\n

6x \u2013 5x = 5 + 6<\/p>\n\n\n\n

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

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

Speed of streamer in still water is 11 km\/h and<\/p>\n\n\n\n

Distance between two ports = (11 + 1) \u00d7 5 = 60 km\/h<\/p>\n\n\n\n

29. Distance between two places A and B is 350 km. Two cars start simultaneously from A and B towards each other and the distance between them after 4 hours is 62 km. If the speed of one car is 8 km\/h less than the speed of other cars, find the speed of each car.<\/strong><\/p>\n\n\n\n

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

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

Distance between two places A and B = 350 km<\/p>\n\n\n\n

Let the speed of car C1<\/sub> = x km\/h and<\/p>\n\n\n\n

Speed of car C2<\/sub> = (x \u2013 8) km\/h<\/p>\n\n\n\n

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

After 4 hours, the distance between two cars is 62 km<\/p>\n\n\n\n

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

x \u00d7 4 + (x \u2013 8) \u00d7 4 = 350 \u2013 62<\/p>\n\n\n\n

4x + 4x \u2013 32 = 288<\/p>\n\n\n\n

8x = 288 + 32<\/p>\n\n\n\n

8x = 320<\/p>\n\n\n\n

x = 320 \/ 8<\/p>\n\n\n\n

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

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

Therefore, speed of car C1<\/sub> = 40 km\/h<\/p>\n\n\n\n

Speed of car C2<\/sub> = (x \u2013 8) = (40 \u2013 8) = 32 km\/h<\/p>\n\n\n\n

Exercise 12.3<\/strong><\/h4>\n\n\n\n

1. If the replacement set = {-7, -5, -3, -1, 1, 3}, find the solution set of:<\/strong><\/p>\n\n\n\n

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

(ii) x < \u2013 2<\/strong><\/p>\n\n\n\n

(iii) x > 2<\/strong><\/p>\n\n\n\n

(iv) -5 < x \u2264 5<\/strong><\/p>\n\n\n\n

(v) -8 < x < 1<\/strong><\/p>\n\n\n\n

(vi) 0 \u2264 x \u2264 4<\/strong><\/p>\n\n\n\n

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

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

Replacement set = {-7, -5, -3, -1, 1, 3}<\/p>\n\n\n\n

The solution set for the given replacement set is as follows:<\/p>\n\n\n\n

(i) Solution set of x > \u2013 2 is {-1, 0, 1, 3}<\/p>\n\n\n\n

(ii) Solution set of x < \u2013 2 is {-7, -5, -3}<\/p>\n\n\n\n

(iii) Solution set of x > 2 is {3}<\/p>\n\n\n\n

(iv) Solution set of -5 < x \u2264 5 is {-3, -1, 0, 1, 3}<\/p>\n\n\n\n

(v) Solution set of -8 < x < 1 is {-7, -5, -3, -1, 0}<\/p>\n\n\n\n

(vi) Solution set of 0 \u2264 x \u2264 4 is {0, 1, 3}<\/p>\n\n\n\n

2. Represent the solution of the following inequalities graphically:<\/strong><\/p>\n\n\n\n

(i) x \u2264 4, x \u03b5 N<\/strong><\/p>\n\n\n\n

(ii) x < 5, x \u03b5 W<\/strong><\/p>\n\n\n\n

(iii) -3 \u2264 x < 3, x \u03b5 l<\/strong><\/p>\n\n\n\n

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

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

x \u2264 4, x \u03b5 N<\/p>\n\n\n\n

The solution set = {1, 2, 3,4}<\/p>\n\n\n\n

These four numbers are shown indicating with thick dots on the number line given below<\/p>\n\n\n\n

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

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

x < 5, x \u03b5 W<\/p>\n\n\n\n

The solution set = {0, 1, 2, 3, 4}<\/p>\n\n\n\n

These five numbers are shown indicating with thick dots on the number line given below<\/p>\n\n\n\n

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

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

-3 \u2264 x < 3, x \u03b5 l<\/p>\n\n\n\n

The solution set = {-3, -2, -1, 0, 1, 2}<\/p>\n\n\n\n

These six numbers are shown indicating with thick dots on the number line given below<\/p>\n\n\n\n

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

3. If the replacement set is {-6, -4, -2, 0, 2, 4, 6}; then represent the solution set of the inequality -4 \u2264 x < 4 graphically.<\/strong><\/p>\n\n\n\n

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

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

Replacement set = {-6, -4, -2, 0, 2, 4, 6} and<\/p>\n\n\n\n

Inequality = -4 \u2264 x < 4<\/p>\n\n\n\n

Solution set = {-4, -2, 0, 2}<\/p>\n\n\n\n

Graphically representation of solution set is as follows:<\/p>\n\n\n\n

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

4. Find the solution set of the inequality x < 4 if the replacement set is<\/strong><\/p>\n\n\n\n

(i) {1, 2, 3, \u2026.. , 10}<\/strong><\/p>\n\n\n\n

(ii) {-1, 0, 1, 2, 5, 8}<\/strong><\/p>\n\n\n\n

(iii) {-5, 10}<\/strong><\/p>\n\n\n\n

(iv) {5, 6, 7, 8, 9, 10}<\/strong><\/p>\n\n\n\n

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

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

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

Replacement set = {1, 2, 3, \u2026\u2026., 10}<\/p>\n\n\n\n

Therefore, solution set = {1, 2, 3}<\/p>\n\n\n\n

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

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

Replacement set = {-1, 0, 1, 2, 5, 8}<\/p>\n\n\n\n

Therefore, solution set = {-1, 0, 1, 2}<\/p>\n\n\n\n

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

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

Replacement set = {-5, 10}<\/p>\n\n\n\n

Therefore, solution set = {-5}<\/p>\n\n\n\n

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

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

Replacement set = {5, 6, 7, 8, 9, 10}<\/p>\n\n\n\n

Therefore, solution set = \u03d5 <\/p>\n\n\n\n

5. If the replacement set = {-6, -3, 0, 3, 6, 9, 12}, find the truth set of the following:<\/strong><\/p>\n\n\n\n

(i) 2x \u2013 3 > 7<\/strong><\/p>\n\n\n\n

(ii) 3x + 8 \u2264 2<\/strong><\/p>\n\n\n\n

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

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

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

Replacement set = {-6, -3, 0, 3, 6, 9, 12}<\/p>\n\n\n\n

(i) 2x \u2013 3 > 7<\/p>\n\n\n\n

2x > 7 + 3<\/p>\n\n\n\n

2x > 10<\/p>\n\n\n\n

x > 10 \/ 2<\/p>\n\n\n\n

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

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

Therefore, solution set = {6, 9, 12}<\/p>\n\n\n\n

(ii) 3x + 8 \u2264 2<\/p>\n\n\n\n

3x \u2264 2 \u2013 8<\/p>\n\n\n\n

3x \u2264 \u2013 6<\/p>\n\n\n\n

x \u2264 \u2013 6 \/ 3<\/p>\n\n\n\n

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

x \u2264 \u2013 2<\/p>\n\n\n\n

Therefore, solution set = {-6, -3}<\/p>\n\n\n\n

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

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

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

2x < 4<\/p>\n\n\n\n

x < 4 \/ 2<\/p>\n\n\n\n

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

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

Therefore, solution set = {-6, -3, 0}<\/p>\n\n\n\n

6. Solve the following inequations:<\/strong><\/p>\n\n\n\n

(i) 4x + 1 < 17, x \u03b5 N<\/strong><\/p>\n\n\n\n

(ii) 4x + 1 \u2264 17, x \u03b5 W<\/strong><\/p>\n\n\n\n

(iii) 4 > 3x \u2013 11, x \u03b5 N<\/strong><\/p>\n\n\n\n

(iv) -17 \u2264 9x \u2013 8, x \u03b5 Z<\/strong><\/p>\n\n\n\n

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

(i) 4x + 1 < 17<\/p>\n\n\n\n

4x < 17 \u2013 1<\/p>\n\n\n\n

4x < 16<\/p>\n\n\n\n

x < 16 \/ 4<\/p>\n\n\n\n

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

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

As x \u03b5 N<\/p>\n\n\n\n

Hence, solution set = {1, 2, 3}<\/p>\n\n\n\n

(ii) 4x + 1 \u2264 17<\/p>\n\n\n\n

4x \u2264 17 \u2013 1<\/p>\n\n\n\n

4x \u2264 16<\/p>\n\n\n\n

x \u2264 16 \/ 4<\/p>\n\n\n\n

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

x \u2264 4<\/p>\n\n\n\n

As x \u03b5 W<\/p>\n\n\n\n

Hence, solution set = {0, 1, 2, 3, 4}<\/p>\n\n\n\n

(iii) 4 > 3x \u2013 11<\/p>\n\n\n\n

4 + 11 > 3x<\/p>\n\n\n\n

15 > 3x<\/p>\n\n\n\n

15 \/ 3 > x<\/p>\n\n\n\n

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

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

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

As x \u03b5 N<\/p>\n\n\n\n

Hence, solution set = {1, 2, 3, 4}<\/p>\n\n\n\n

(iv) -17 \u2264 9x \u2013 8<\/p>\n\n\n\n

-17 + 8 \u2264 9x<\/p>\n\n\n\n

\u2013 9 \u2264 9x<\/p>\n\n\n\n

\u2013 9 \/ 9 \u2264 x<\/p>\n\n\n\n

-1 \u2264 x<\/p>\n\n\n\n

x \u2265 -1<\/p>\n\n\n\n

As x \u03b5 Z<\/p>\n\n\n\n

Hence, solution set = {-1,0, 1, 2,\u2026}<\/p>\n\n\n\n

7. Solve the following inequations:<\/strong><\/p>\n\n\n\n

(i) {(2y \u2013 1) \/ 5} \u2264 2, y \u03b5 N<\/strong><\/p>\n\n\n\n

(ii) {(2y + 1) \/ 3} + 1 \u2264 3, y \u03b5 W<\/strong><\/p>\n\n\n\n

(iii) (2 \/ 3)p + 5 < 9, p \u03b5 W<\/strong><\/p>\n\n\n\n

(iv) \u2013 2 (p + 3) > 5, p \u03b5 l<\/strong><\/p>\n\n\n\n

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

(i) {(2y \u2013 1) \/ 5} \u2264 2<\/p>\n\n\n\n

2y \u2013 1 \u2264 2 \u00d7 5<\/p>\n\n\n\n

2y \u2013 1 \u2264 10<\/p>\n\n\n\n

2y \u2264 10 + 1<\/p>\n\n\n\n

2y \u2264 11<\/p>\n\n\n\n

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

y \u2264 11 \/ 2<\/p>\n\n\n\n

As y \u03b5 N,<\/p>\n\n\n\n

Therefore, solution set = {1, 2, 3, 4, 5}<\/p>\n\n\n\n

(ii) {(2y + 1) \/ 3} + 1 \u2264 3<\/p>\n\n\n\n

{(2y + 1 + 3) \/ 3} \u2264 3<\/p>\n\n\n\n

{(2y + 4) \/ 3} \u2264 3<\/p>\n\n\n\n

(2y + 4) \u2264 3 \u00d7 3<\/p>\n\n\n\n

2y + 4 \u2264 9<\/p>\n\n\n\n

2y \u2264 9 \u2013 4<\/p>\n\n\n\n

2y \u2264 5<\/p>\n\n\n\n

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

y \u2264 5 \/ 2<\/p>\n\n\n\n

As y \u03b5 N,<\/p>\n\n\n\n

Therefore, solution set = {0, 1, 2}<\/p>\n\n\n\n

(iii) (2 \/ 3) p + 5 < 9<\/p>\n\n\n\n

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

(2 \/ 3) p < 4<\/p>\n\n\n\n

2p < 4 \u00d7 3<\/p>\n\n\n\n

2p < 12<\/p>\n\n\n\n

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

p < 12 \/ 2<\/p>\n\n\n\n

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

As p \u03b5 W,<\/p>\n\n\n\n

Therefore, solution set = {0, 1, 2, 3, 4, 5}<\/p>\n\n\n\n

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

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

\u2013 2p > 5 + 6<\/p>\n\n\n\n

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

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

p > (11 \/ \u2013 2)<\/p>\n\n\n\n

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

As p \u03b5 l,<\/p>\n\n\n\n

Therefore, solution set = {\u2026-8, \u2013 7, \u2013 6}<\/p>\n\n\n\n

8. Solve the following inequations:<\/strong><\/p>\n\n\n\n

(i) 2x \u2013 3 < x + 2, x \u03b5 N<\/strong><\/p>\n\n\n\n

(ii) 3 \u2013 x \u2264 5 \u2013 3x, x \u03b5 W<\/strong><\/p>\n\n\n\n

(iii) 3 (x \u2013 2) < 2 (x \u2013 1), x \u03b5 W<\/strong><\/p>\n\n\n\n

(iv) (3 \/ 2) \u2013 (x \/ 2) > \u2013 1, x \u03b5 N<\/strong><\/p>\n\n\n\n

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

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

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

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

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

As x \u03b5 N,<\/p>\n\n\n\n

Hence, solution set = {1, 2, 3, 4}<\/p>\n\n\n\n

(ii) 3 \u2013 x \u2264 5 \u2013 3x<\/p>\n\n\n\n

3x \u2013 x \u2264 5 \u2013 3<\/p>\n\n\n\n

2x \u2264 2<\/p>\n\n\n\n

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

x \u2264 1<\/p>\n\n\n\n

As x \u03b5 W,<\/p>\n\n\n\n

Hence, solution set = {0, 1}<\/p>\n\n\n\n

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

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

3x \u2013 2x < \u2013 2 + 6<\/p>\n\n\n\n

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

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

As x \u03b5 W,<\/p>\n\n\n\n

Hence, solution set = {0, 1, 2, 3}<\/p>\n\n\n\n

(iv) (3 \/ 2) \u2013 (x \/ 2) > \u2013 1<\/p>\n\n\n\n

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

{(3 + 2) \/ 2} > (x \/ 2)<\/p>\n\n\n\n

(5 \/ 2) > (x \/ 2)<\/p>\n\n\n\n

So,<\/p>\n\n\n\n

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

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

As x \u03b5 N,<\/p>\n\n\n\n

Hence, solution set = {1, 2, 3, 4}<\/p>\n\n\n\n

9. If the replacement set is {-3, -2, -1, 0, 1, 2, 3}, solve the inequation {(3x \u2013 1) \/ 2} < 2. Represent its solution on the number line.<\/strong><\/p>\n\n\n\n

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

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

Replacement set = {-3, -2, -1, 0, 1, 2, 3} and<\/p>\n\n\n\n

Inequation = {(3x \u2013 1) \/ 2} < 2<\/p>\n\n\n\n

3x \u2013 1 < 2 \u00d7 2<\/p>\n\n\n\n

3x \u2013 1 < 4<\/p>\n\n\n\n

3x < 4 + 1<\/p>\n\n\n\n

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

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

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

Therefore, solution set = {\u2026-3, -2, -1, 0, 1}<\/p>\n\n\n\n

Graphical representation of this solution set is as follows:<\/p>\n\n\n\n

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

10. Solve (x \/ 3) + (1 \/ 4) < (x \/ 6) + (1 \/ 2), x \u03b5 W. Also represent its solution on the number line.<\/strong><\/p>\n\n\n\n

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

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

(x \/ 3) + (1 \/ 4) < (x \/ 6) + (1 \/ 2)<\/p>\n\n\n\n

(x \/ 3) \u2013 (x \/ 6) < (1 \/ 2) \u2013 (1 \/ 4)<\/p>\n\n\n\n

(2x \u2013 x) \/ 6 < (2 \u2013 1) \/ 4<\/p>\n\n\n\n

x \/ 6 < 1 \/ 4<\/p>\n\n\n\n

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

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

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

As x \u03b5 W,<\/p>\n\n\n\n

Hence, solution set = {0, 1}<\/p>\n\n\n\n

Graphical representation of this solution set is as follows:<\/p>\n\n\n\n

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

11. Solve the following inequations and graph their solutions on a number line<\/strong><\/p>\n\n\n\n

(i) \u2013 4 \u2264 4x < 14, x \u03b5 N<\/strong><\/p>\n\n\n\n

(ii) \u2013 1 < (x \/ 2) + 1 \u2264 3, x \u03b5 l<\/strong><\/p>\n\n\n\n

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

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

\u2013 4 \u2264 4x < 14<\/p>\n\n\n\n

Dividing by 4, we get,<\/p>\n\n\n\n

(-4 \/ 4) \u2264 (4x \/ 4) < (14 \/ 4)<\/p>\n\n\n\n

-1 \u2264 x < 7 \/ 2<\/p>\n\n\n\n

As x \u03b5 N,<\/p>\n\n\n\n

Therefore, solution set = {1, 2, 3}<\/p>\n\n\n\n

The graphical representation for this solution set is as follows:<\/p>\n\n\n\n

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

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

\u2013 1 < (x \/ 2) + 1 \u2264 3<\/p>\n\n\n\n

By subtracting \u2013 1, we get,<\/p>\n\n\n\n

-1 \u2013 1 < {(x \/ 2) + 1} \u2013 1 \u2264 3 \u2013 1<\/p>\n\n\n\n

\u2013 2 < (x \/ 2) \u2264 \u2013 2<\/p>\n\n\n\n

Multiplying by 2, we get,<\/p>\n\n\n\n

\u2013 2 \u00d7 2 < (x \/ 2) \u00d7 2 \u2264 \u2013 2 \u00d7 2<\/p>\n\n\n\n

\u2013 4 < x \u2264 \u2013 4<\/p>\n\n\n\n

As x \u03b5 l,<\/p>\n\n\n\n

Therefore, solution set = {-3, -2, -1, 0, 1, 2, 3, 4}<\/p>\n\n\n\n

The graphical representation for this solution set is as follows:<\/p>\n\n\n\n

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

Download PDF<\/strong><\/h2>\n\n\n\n

ML Aggarwal Solutions for Class 8 Maths Chapter 12- Linear Equations and Inequalities in One Variable<\/p>\n\n\n\n

Download PDF<\/strong>: ML Aggarwal Solutions for Class 8 Maths Chapter 12- Linear Equations and Inequalities in One Variable PDF<\/a><\/p>\n\n\n\n

Chapterwise ML Aggarwal Solutions for Class 8 Maths :<\/strong><\/h2>\n\n\n\n