RD Sharma Class 11 Solutions Chapter 15 Exercise 15.1 (Updated for 2024)

1. Solve: 12x < 50, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

12x < 50

So when we divide by 12, we get

12x/ 12 < 50/12

x < 25/6

(i) x ∈ R

When x is a real number, the solution of the given inequation is (-∞, 25/6).

(ii) x ∈ Z

When, 4 < 25/6 < 5

So, when x is an integer, the maximum possible value of x is 4.

The solution of the given inequation is {…, –2, –1, 0, 1, 2, 3, 4}.

(iii) x ∈ N

When, 4 < 25/6 < 5

So, when x is a natural number, the maximum possible value of x is 4. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2, 3, 4}.

2. Solve: -4x > 30, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

-4x > 30

So when we divide by 4, we get

-4x/4 > 30/4

-x > 15/2

x < – 15/2

(i) x ∈ R

When x is a real number, the solution of the given inequation is (-∞, -15/2).

(ii) x ∈ Z

When, -8 < -15/2 < -7

So, when x is an integer, the maximum possible value of x is -8.

The solution of the given inequation is {…, –11, –10, -9, -8}.

(iii) x ∈ N

As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is ∅.

3. Solve: 4x-2 < 8, when

(i) x ∈ R

(ii) x ∈ Z

(iii) x ∈ N

Solution:

Given:

4x – 2 < 8

4x – 2 + 2 < 8 + 2

4x < 10

So dividing by 4 on both sides, we get,

4x/4 < 10/4

x < 5/2

(i) x ∈ R

When x is a real number, the solution of the given inequation is (-∞, 5/2).

(ii) x ∈ Z

When, 2 < 5/2 < 3

So, when x is an integer, the maximum possible value of x is 2.

The solution of the given inequation is {…, –2, –1, 0, 1, 2}.

(iii) x ∈ N

When, 2 < 5/2 < 3

So, when x is a natural number, the maximum possible value of x is 2. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2}.

4. 3x – 7 > x + 1

Solution:

Given:

3x – 7 > x + 1

3x – 7 + 7 > x + 1 + 7

3x > x + 8

3x – x > x + 8 – x

2x > 8

Dividing both sides by 2, we get

2x/2 > 8/2

x > 4

∴ The solution of the given inequation is (4, ∞).

5. x + 5 > 4x – 10

Solution:

Given: x + 5 > 4x – 10

x + 5 – 5 > 4x – 10 – 5

x > 4x – 15

4x – 15 < x

4x – 15 – x < x – x

3x – 15 < 0

3x – 15 + 15 < 0 + 15

3x < 15

Dividing both sides by 3, we get

3x/3 < 15/3

x < 5

∴ The solution of the given inequation is (-∞, 5).

6. 3x + 9 ≥ –x + 19

Solution:

Given: 3x + 9 ≥ –x + 19

3x + 9 – 9 ≥ –x + 19 – 9

3x ≥ –x + 10

3x + x ≥ –x + 10 + x

4x ≥ 10

Dividing both sides by 4, we get

4x/4 ≥ 10/4

x ≥ 5/2

∴ The solution of the given inequation is [5/2, ∞).

7. 2 (3 – x) ≥ x/5 + 4

Solution:

Given: 2 (3 – x) ≥ x/5 + 4

6 – 2x ≥ x/5 + 4

6 – 2x ≥ (x+20)/5

5(6 – 2x) ≥ (x + 20)

30 – 10x ≥ x + 20

30 – 20 ≥ x + 10x

10 ≥11x

11x ≤ 10

Dividing both sides by 11, we get

11x/11 ≤ 10/11

x ≤ 10/11

∴ The solution of the given inequation is (-∞, 10/11].

8. (3x – 2)/5 ≤ (4x – 3)/2

Solution:

Given:

(3x – 2)/5 ≤ (4x – 3)/2

Multiplying both the sides by 5 we get,

(3x – 2)/5 × 5 ≤ (4x – 3)/2 × 5

(3x – 2) ≤ 5(4x – 3)/2

3x – 2 ≤ (20x – 15)/2

Multiplying both the sides by 2 we get,

(3x – 2) × 2 ≤ (20x – 15)/2 × 2

6x – 4 ≤ 20x – 15

20x – 15 ≥ 6x – 4

20x – 15 + 15 ≥ 6x – 4 + 15

20x ≥ 6x + 11

20x – 6x ≥ 6x + 11 – 6x

14x ≥ 11

Dividing both sides by 14, we get

14x/14 ≥ 11/14

x ≥ 11/14

∴ The solution of the given inequation is [11/14, ∞).

9. –(x – 3) + 4 < 5 – 2x

Solution:

Given: –(x – 3) + 4 < 5 – 2x

–x + 3 + 4 < 5 – 2x

–x + 7 < 5 – 2x

–x + 7 – 7 < 5 – 2x – 7

–x < –2x – 2

–x + 2x < –2x – 2 + 2x

x < –2

∴ The solution of the given inequation is (–∞, –2).

10. x/5 < (3x-2)/4 – (5x-3)/5

Solution:

Given: x/5 < (3x-2)/4 – (5x-3)/5

x/5 < [5(3x-2) – 4(5x-3)]/4(5)

x/5 < [15x – 10 – 20x + 12]/20

x/5 < [2 – 5x]/20

Multiplying both the sides by 20 we get,

x/5 × 20 < [2 – 5x]/20 × 20

4x < 2 – 5x

4x + 5x < 2 – 5x + 5x

9x < 2

Dividing both sides by 9, we get

9x/9 < 2/9

x < 2/9

∴ The solution of the given inequation is (-∞, 2/9).

11. [2(x-1)]/5 ≤ [3(2+x)]/7

Solution:

Given:

(2x – 2)/5 ≤ (6 + 3x)/7

Multiplying both sides by 5 we get,

(2x – 2)/5 × 5 ≤ (6 + 3x)/7 × 5

2x – 2 ≤ 5(6 + 3x)/7

7 (2x – 2) ≤ 5 (6 + 3x)

14x – 14 ≤ 30 + 15x

14x – 14 + 14 ≤ 30 + 15x + 14

14x ≤ 44 + 15x

14x – 44 ≤ 44 + 15x – 44

14x – 44 ≤ 15x

15x ≥ 14x – 44

15x – 14x ≥ 14x – 44 – 14x

x ≥ –44

∴ The solution of the given inequation is [–44, ∞).

12. 5x/2 + 3x/4 ≥ 39/4

Solution:

Given:

5x/2 + 3x/4 ≥ 39/4

By taking LCM

13x/4 ≥ 39/4

Multiplying both sides by 4 we get,

13x/4 × 4 ≥ 39/4 × 4

13x ≥ 39

Divide both sides by 13, we get

13x/13 ≥ 39/13

x ≥ 39/13

x ≥ 3

∴ The solution of the given inequation is [3, ∞).

13. (x – 1)/3 + 4 < (x – 5)/5 – 2

Solution:

Given:

(x – 1)/3 + 4 < (x – 5)/5 – 2

Subtract both sides by 4 we get,

(x – 1)/3 + 4 – 4 < (x – 5)/5 – 2 – 4

(x – 1)/3 < (x – 5)/5 – 6

(x – 1)/3 < (x – 5 – 30)/5

(x – 1)/3 < (x – 35)/5

Cross multiply we get,

5 (x – 1) < 3 (x – 35)

5x – 5 < 3x – 105

5x – 5 + 5 < 3x – 105 + 5

5x < 3x – 100

5x – 3x < 3x – 100 – 3x

2x < –100

Divide both sides by 2, we get

2x/2 < -100/2

x < -50

∴ The solution of the given inequation is (-∞, -50).

14. (2x + 3)/4 – 3 < (x – 4)/3 – 2

Solution:

Given:

(2x + 3)/4 – 3 < (x – 4)/3 – 2

Add 3 on both sides we get,

(2x + 3)/4 – 3 + 3 < (x – 4)/3 – 2 + 3

(2x + 3)/4 < (x – 4)/3 + 1

(2x + 3)/4 < (x – 4 + 3)/3

(2x + 3)/4 < (x – 1)/3

Cross multiplying, we get,

3 (2x + 3) < 4 (x – 1)

6x + 9 < 4x – 4

6x + 9 – 9 < 4x – 4 – 9

6x < 4x – 13

6x – 4x < 4x – 13 – 4x

2x < –13

Dividing both sides by 2, we get

2x/2 < -13/2

x < -13/2

∴ The solution of the given inequation is (-∞, -13/2).