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# RD Chapter 15- Linear Inequations Ex-15.1 Interview Questions Answers

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Question 1 : Solve the following linear Inequations in R.
Solve: 12x < 50, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

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, 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, 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}.

Question 2 :
Solve: -4x > 30, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

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, 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 ∅.

Question 3 :
Solve: 4x-2 < 8, when
(i) x ∈ R
(ii) x ∈ Z
(iii) x ∈ N

Given:
4x – 2 < 8
4x – 2 + 2 < 8 + 2
4x < 10
So divide 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, 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, 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 : 3x – 7 > x + 1

Given:
3x – 7 > x + 1
3x – 7 + 7 > x + 1 + 7
3x > x + 8
3x – x > x + 8 – x
2x > 8
Divide both sides by 2, we get
2x/2 > 8/2
x > 4
∴ The solution of the given inequation is (4, ∞).

Question 5 : x + 5 > 4x – 10

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
Divide both sides by 3, we get
3x/3 < 15/3
x < 5
∴ The solution of the given inequation is (-∞, 5).

Question 6 : 3x + 9 ≥ –x + 19

Given: 3x + 9 ≥ –x + 19
3x + 9 – 9 ≥ –x + 19 – 9
3x ≥ –x + 10
3x + x ≥ –x + 10 + x
4x ≥ 10
Divide both sides by 4, we get
4x/4 ≥ 10/4
x ≥ 5/2
∴ The solution of the given inequation is [5/2, ∞).

Question 7 : 2 (3 – x) ≥ x/5 + 4

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
Divide both sides by 11, we get
11x/11 ≤ 10/11
x ≤ 10/11
∴ The solution of the given inequation is (-∞, 10/11].

Question 8 : (3x – 2)/5 ≤ (4x – 3)/2

Given:
(3x – 2)/5 ≤ (4x – 3)/2
Multiply 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
Multiply 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
Divide both sides by 14, we get
14x/14 ≥ 11/14
x ≥ 11/14
∴ The solution of the given inequation is [11/14, ∞).

Question 9 : –(x – 3) + 4 < 5 – 2x

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 : x/5 < (3x-2)/4 – (5x-3)/5

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
Multiply both the sides by 20 we get,
x/5 × 20 < [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).

krishan