Class 11 RD Sharma Solutions - Chapter 15 Linear Inequations - Exercise 15.1 | Set 1

Question 1. Solve: 12x < 50, when

(i) x ∈ R

Solution:

Given: 12x < 50

Dividing both sides by 12, we get

12x/ 12 < 50/12

⇒ x < 25/6

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

(ii) x ∈ Z

Solution:

Since, 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

Solution:

Since 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.

Hence the solution of the given inequation is {1, 2, 3, 4}.

Question 2. Solve: − 4x > 30, when

(i) x ∈ R

Solution:

Given: − 4x > 30

So when we divide by 4, we get

⇒ − 4x/4 > 30/4

⇒ − x > 15/2

⇒ x < – 15/2

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

(ii) x ∈ Z

Solution:

Since, − 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

Solution:

As natural numbers start from 1 and can never be negative.

Hence when x is a natural number, the solution of the given inequation is ∅.

Question 3. Solve: 4x-2 < 8, when

(i) x ∈ R

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

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

(ii) x ∈ Z

Solution:

Since, 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

Solution:

Since, 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}.

Question 4. Solve: 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, ∞).

Question 5. Solve: 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).

Question 6. Solve: 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, ∞).

Question 7. Solve: 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].

Question 8. Solve:\frac{3x – 2}{5} ≤\frac{4x – 3}{2}

Solution:

Given:\frac{3x – 2}{5} ≤\frac{4x – 3}{2}

Multiplying both the sides by 5 we get,

\frac{3x – 2}{5} × 5 ≤\frac{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, ∞).

Question 9. Solve: –(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).

Question 10. Solve:\frac{x}{5} <\frac{3x-2}{4} –\frac{5x-3}{5}

Solution:

Given:\frac{x}{5} <\frac{3x-2}{4} –\frac{5x-3}{5}

⇒\frac{x}{5} <\frac{[5(3x-2) – 4(5x-3)]}{4(5)}

⇒\frac{x}{5} <\frac{[15x – 10 – 20x + 12]}{20}

⇒\frac{x}{5} <\frac{[2 – 5x]}{20}

Multiplying both the sides by 20 we get,

\frac{x}{5} × 20 <\frac{[2 – 5x]}{20} × 20

⇒ 4x < 2 – 5x

⇒ 4x + 5x < 2 – 5x + 5x

⇒ 9x < 2

Divide both sides by 9, we get

9x/9 < 2/9

⇒ x < 2/9

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

Question 11. Solve:\frac{[2(x-1)]}{5} ≤\frac{[3(2+x)]}{7}

Solution:

Given:\frac{[2(x-1)]}{5} ≤\frac{[3(2+x)]}{7}

⇒\frac{(2x – 2)}{5} ≤\frac{(6 + 3x)}{7}

Multiply both the sides by 5 we get,

\frac{(2x – 2)}{5} × 5 ≤\frac{(6 + 3x)}{7} × 5

⇒ 2x – 2 ≤\frac{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, ∞).

Question 12. Solve: 5x/2 + 3x/4 ≥ 39/4

Solution:

Given: 5x/2 + 3x/4 ≥ 39/4

By taking LCM, we get:

\frac{[2(5x)+3x]}{4} ≥ 39/4

⇒ 13x/4 ≥ 39/4

Multiplying both the 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, ∞).

Question 13. Solve:\frac{(x – 1)}{3} + 4 <\frac{(x – 5)}{5} – 2

Solution:

Given:\frac{(x – 1)}{3} + 4 <\frac{(x – 5)}{5} – 2

Subtracting both sides by 4 we get,

⇒\frac{(x – 1)}{3} + 4 – 4 <\frac{(x – 5)}{5} – 2 – 4

⇒\frac{(x – 1)}{3} <\frac{(x – 5)}{5} – 6

⇒\frac{(x – 1)}{3} <\frac{(x – 5 – 30)}{5}

⇒\frac{(x – 1)}{3} <\frac{(x – 35)}{5}

After Cross multiplying, 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).

Question 14. Solve:\frac{(2x + 3)}{4} – 3 <\frac{(x – 4)}{3} – 2

Solution:

Given:\frac{(2x + 3)}{4} – 3 <\frac{(x – 4)}{3} – 2

Adding 3 on both sides we get,

\frac{(2x + 3)}{4} – 3 + 3 <\frac{(x – 4)}{3} – 2 + 3

⇒\frac{(2x + 3)}{4} <\frac{(x – 4)}{3} + 1

⇒\frac{(2x + 3)}{4} <\frac{(x – 4 + 3)}{3}

⇒\frac{(2x + 3)}{4} <\frac{(x – 1)}{3}

After 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).