Class X Chapter 4 – Linear Inequations(in one variable) Maths

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EXERCISE- 4 (A)

Question 1:
State, true or false:
(i) x < − y ⇒ − x > y
(ii) – 5x ≥ ⇒ x ≥ − 3
2𝑥 −7
(iii) 2x ≤ − 7 ⇒ −4 ≥ −4
1 1
(iv) 7 > 5 ⇒ 7 < 5
Solution 1:
(i) x < − y ⇒ − x > y
The given statement is true.
−5𝑥 15
(ii) – 5x ≥ 15 ⇒ 5 ≥ 5 x ≤ − 3
The given statement is false
2𝑥 −7
(iii) 2x ≤ − 7 ⇒ −4 ≥ −4
The given statement is true
1 1
(iv) 7 > 5 ⇒ 7 < 5
The given statement is true.

Question 2:
(i) a<b⇒a–c<b–c
(ii) If a > b ⇒ a + c > b + c
(iii) If a < b ⇒ ac < bc
𝑎 𝑏
(iv) If a > b ⇒ 𝑐 > 𝑐
(v) If a – c > b – d ⇒ a + d > b + c
(vi) If a < b ⇒ a – c < b – c (Since, c > O)
Where a, b, c and d are real numbers and c ≠ 0.
Solution 2:
(i) a < b ⇒ a – c < b – c
The given statement is true.
(ii) If a > b ⇒ a + c > b + c
The given statement is true.

(iii) If a < b ⇒ ac < bc

The given statement is false.
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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𝑎 𝑏
(iv) If a > b ⇒ 𝑐 > 𝑐
The given statement is false.

(v) If a – c > b – d ⇒ a + d > b + c

The given statement is true.

(vi) If a < b ⇒ a – c < b – c (Since, c > O)

The given statement is false.

Question 3:
If x ∈ N, find the solution set of in-equations.
(i) 5x + 3 ≤ 2x + 18
(ii) 3x – 2 < 19 – 4x
Solution 3:
(i) 5x + 3 ≤ 2x + 18
5x – 2x ≤ 18 – 3
3x ≤ 15
X≤5
Since, x ∈ N, therefore solution set is {1,2,3,4,5}

(ii) 3x – 2 < 19 – 4x
3x + 4x < 19 + 2
7x < 21
X<3
Since, x ∈ N, therefore solution set is {1,2}.

Question 4:
If the replacement set is the set of whole numbers, solve:
(i) x + 7 ≤ 11
(ii) 3x – 1 > 8
(iii) 8 – x > 5
1
(iv) 7 – 3x ≥ − 2
3 3
(v) x – 2 < 2 – 𝑥
(vi) 18 ≤ 3x – 2
Solution 4:
(i) x + 7 ≤ 11
X ≤ 11 – 7
X≤4
Since, the replacement set = W (set of whole numbers)
⟹ Solution set = {0,1,2,3,4}
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(ii) 3x – 1 > 8
3x > 8 + 1
X>3
Since, the replacement set = W (Set of whole numbers)
⟹ Solution set = {4, 5, 6……}

(iii) 8 – x > 5
−X > 5 – 8
−X >-3
X<3
Since, the replacement set = W (Set of whole numbers)
⟹ Solution set = {0, 1, 2 ……}
1
(iv) 7 – 3x ≥ − 2
1
-3x ≥ − 2 – 7
15
-3x ≥ − 2
5
X≤2
Since, the replacement set = W (set of whole numbers)
∴ Solution set = {0, 1, 2}
3 3
(v) x − 2 < −𝑥
2
3 3
x+x<2 + 2
2x < 3
3
X<2
Since, the replacement set = W (set of whole numbers)
∴ Solution set = {0, 1}

(vi) 18 ≤ 3x – 2
18 + 2 ≤ 3x
20 ≤ 3x
20
X≥ 3
Since, the replacement set = W (set of whole numbers)
∴ Solution set = {7, 8, 9….}

Question 5:
Solve the in-equation:
3 – 2x ≥ x – 12 given that x ∈ N
Solution 5:
3 – 2x ≥ x – 12
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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− 2x – x ≥ −12 – 3
− 3x ≥ − 15
X≤5
Since, x ∈ N, therefore,
Solution set = {1, 2, 3, 4, 5}

Question 6:
If 25 – 4x ≤ 16, find:
(i) the smallest value of x, when x is a real number,
(ii) the smallest value of x, when x is an integer.
Solution 6:
25 – 4x ≤ 16
− 4x ≤ 16 – 25
− 4x ≤ −9
9
X≥4
X ≥ 2.25
(i) The smallest value of x, when x is a real number, is 2.25.

(ii) The smallest value of x, when x is an integer, is 3.

Question 7:
If the replacement set is the set of real number, solve:
(i) 4x ≥ − 16
(ii) 8 – 3x ≤ 20
𝑥 𝑥
(iii) 5 + 4 > 5 + 9
𝑥+3 𝑥 −3
(iv) <
8 5
Solution 7:
(i) – 4x ≥ − 16
X≤4
Since, the replacement set of real numbers.
∴ solution set = { x:x ∈ R and x ≤ 4}

(ii) 8 – 3x ≤ 20
− 3x ≤ 20 – 8
− 3x ≤ 12
X≥−4
Since the replacement set of real numbers.
∴ solution set = { x:x ∈ R and x ≤ − 4}
𝑥 𝑥
(iii) 5 + 4 > +9
5
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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𝑥 𝑥
− > 9 –5
4 5
𝑥
>4
20
X > 80
Since the replacement set of real numbers.
∴ solution set = { x:x ∈ R and x > 80}
𝑥+3 𝑥 −3
(iv) < 5
8
5x + 15 < 8x – 24
5x – 8x < −24 – 15
− 3x < − 39
X > 13
Since the replacement set of real numbers.
∴ solution set = { x:x ∈ R and x > 13}

Question 8:
1 5
Find the smallest value of x for which 5 – 2x < 5 2 − 𝑥, where x is an integer.
3
Solution 8:
1 5
5 – 2x < 5 2 − 𝑥
3
5 11
−2x + 3 𝑥 < –5
2
−𝑥 1
<
3 2
3
−x < 2
−3
X> 2
X > −1.5
Thus, the required smallest value of x is -1.

Question 9:
Find the largest value of x for which
2(x −1) ≤ 9 – x and x ∈ W.
Solution 9:
2(x −1) ≤ 9 – x
2x – 2 ≤ 9 – x
2x + x ≤ 9 + 2
3x ≤11
11
x≤ 3
X ≤ 3.67
Since, x ∈ W, thus the required largest value of x is 3.
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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Question 10:
Solve the in-equation:
5
12 + 1 6 × ≤ 5 + 3x and x ∈ R.
Solution 10:
5
12 + 1 6 × ≤ 5 + 3x
11
X – 3X ≤ 5 – 12
6
−7
𝑋 ≤ −7
6
X≥6
∴ solution set = {x : x ∈ R and x ≥ 6}

Question 11:
Given x ∈ {Integers}, find the solution set of : − 5 ≤ 2x −3 < x + 2
Solution 11:
− 5 ≤ 2x −3 < x + 2
⇒ −5 ≤ 2x – 3 and 2x – 3 < x + 2
⇒ −5 + 3 ≤ 2x and 2x – x < 2 + 3
⇒ −2 ≤ 2x and x < 5
⇒ X ≥ −1 and x < 5

Since, x ∈ {integers}
∴ Solution set = { -1, 0, 1, 2, 3, 4}

Question 12:
Given x ∈ {whole numbers}, find the solution set of : −1 ≤ 3 + 4x < 23
Solution 12:
− 1 ≤ 3 + 4x < 23
⟹ − 1 ≤ 3 + 4x and 3 + 4x < 23
⟹ − 4 ≤ 4x and 4x < 20
⟹ x ≥ −1 and x < 5

Since, x 𝛜 { Whole numbers}

∴ solution set = { 0, 1, 2, 3, 4}
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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EXERCISE 4(B)

Question 1:
Represent the following in-equalities on real number lines:
(i) 2x – 1 < 5
(ii) 3x + 1 ≥ −5
(iii) 2(2x – 3) ≤ 6
(iv) – 4 < x < 4
(v) −2 ≤ x < 5
(vi) 8≥ x > −3
(vii) −5 < × ≤ −1
Solution 1:
(i) 2x – 1 < 5
2x < 6
X<3
Solution on number line is:

(ii) 3x + 1 ≥ −5
3x ≥ − 6
X≥−2
Solution on number line is:

(iii) 2(2x – 3) ≤ 6
2x – 3 ≤ 3
2x ≤ 6
X≤3
Solution on number line is:

(iv) – 4 < x < 4

Solution on number line is:

(v) −2 ≤ x < 5
Solution on number line is:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(vi) 8≥ x > −3
Solution on number line is:

(vii) −5 < × ≤ −1
Solution on number line is:

Question 2:
For each graph given alongside, write an in-equation taking x as the variable

Solution 2:
(i) x ≤ −1, × ∈ R
(ii) x ≥ 2, x ∈ R
(iii) −4 ≤ x < 3, x ∈ R
(iv) – 1 < x ≤ 5, x ∈ R

Question 3:
For the following in-equations, graph the solution set on the real number line:
(i) – 4 ≤ 3x – 1 < 8
(ii) x – 1 < 3 – x ≤ 5
Solution 3:
(i) – 4 ≤ 3x – 1 < 8
− 4 ≤ 3x – 1 and 3x – 1 < 8
− 1≤x and x < 3
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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The solution set on the real number line is:

(ii) x – 1 < 3 – x ≤ 5
X–1<3–x and 3 – x ≤ 5
2x < 4 and –x ≤ 2
X<2 and x ≥ − 2
The solution set on the real number line is:

Question 4:
Represent the solution of each of the following in-equalities on the real number line:
(i) 4x – 1 > x + 11
(ii) 7 – x ≤ 2 – 6x
(iii) x + 3 ≤ 2x + 9
(iv) 2 – 3x > 7 – 5x
(v) 1 + x ≥ 5x – 11
2𝑥+5
(vi) 3 > 3x – 3
Solution 4:
(i) 4x – 1 > x + 11
3x > 12
X>4
The solution on number line is:

(ii) 7 – x ≤ 2 – 6x
5x ≤ − 5
X≤−1
The solution on number line is:

(iii) x + 3 ≤ 2x + 9
− 6≤x
The solution on number line is:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(iv) 2 – 3x > 7 – 5x
2x > 5
5
X>2
X > 2.5
The solution on number line is:

(v) 1 + x ≥ 5x – 11
12 ≥ 4x
3≥x
The solution on number line is:

2𝑥+5
(vi) > 3x – 3
3
2x + 5 > 9x – 9
−7x > − 14
X<2
The solution on number line is:

Question 5:
X ∈ {real numbers} and − 1 < 3 – 2x ≤ 7 evaluate x and represent it on a number line.
Solution 5:
− 1 < 3 – 2x ≤ 7
− 1 < 3 – 2x and 3 – 2x ≤ 7
2x < 4 and −2x ≤ 4
X < 2 and x ≥ − 2
Solution set = { -2 ≤ x < 2, x ∈ R}
Thus, the solution can be represented on a number line as:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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Question 6:
List the elements of the solution set of the in-equation − 3 < x – 2 ≤ 9 – 2x; x ∈ N.
Solution 6:
− 3 < x – 2 ≤ 9 – 2x
− 3 < x – 2 and x – 2 ≤ 9 – 2x
− 1 < x and 3x ≤ 11
11
− 1<x≤ 3
Since, x ∈ N
∴ Solution set = {1, 2, 3}

Question 7:
Find the range of values of x which satisfies
2 1 1
− 2 3 ≤ 𝑥 + 3 < 3 3, x ∈ R.
Graph these values of x on the number line.
Solution 7:
2 1 1 1
− 23 ≤𝑥+ and x + 3 < 3 3
3
8 1 1 10
⟹−3 ≤𝑥+ and x + 3 <
3 3
8 1 10 1
⟹−3 − ≤ 𝑥 and x < −
3 3 3
9 9
⟹ − 3 ≤ 𝑥 and x < 3
⟹ − 3 ≤ x and x < 3
∴ − 3 ≤ x and x < 3
The required graph of the solution set is:

Question 8:
1 2𝑥 5
Find the values of x, which satisfy the in-equation: − 2 ≤ − <1 , x ∈ N.
2 3 6
Graph the solution on the number line.
Solution 8:
1 2𝑥 5
− 2≤2 − <1
3 6
1 2𝑥 1 2𝑥 5
− 2≤2 − and − <1
3 2 3 6
−5 2𝑥 −2𝑥 8
≤ − and <
2 3 3 6
15
≥ 𝑥 and x > − 2
4
3.75 ≥ 𝑥 and x > −2
Since, x ∈ N
∴ Solution set = {1, 2, 3}
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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The required graph of the solution set is:

Question 9:
Given x ∈ {real number}, find the range of values of x for which − 5 ≤ 2x – 3 < x + 2
And represent it on a real number line.
Solution 9:
− 5 ≤ 2x – 3 < x + 2
− 5 ≤ 2x – 3 and 2x – 3 < x + 2
− 2 ≤ 2x and x < 5
− 1 ≤ x and x < 5
∴ Required range is – 1 ≤ x < 5
The required graph is:

Question 10:
If 5x – 3 ≤ 5 + 3x ≤ 4x + 2, express it as a ≤ x ≤ b and then state the values of a and
b.
Solution 10:
5x – 3 ≤ 5 + 3x ≤ 4x + 2
5x – 3 ≤ 5 + 3x and 5 + 3x ≤ 4x + 2
2x ≤ 8 and – x ≤ −3
X ≤ 4 and x ≥ 3
Thus, 3 ≤ x ≤ 4
Hence, a = 3 and b = 4

Question 11:
Solve the following in-equation and graph the solution set on the number line:
2x – 3 < x + 2 ≤ 3x + 5; x ∈ R.
Solution 11:
2x – 3 < x + 2 ≤ 3x + 5
2x – 3 < x + 2 and x + 2 ≤ 3x + 5
X < 5 and – 3 ≤ 2x
X < 5 and – 1.5 ≤ x
Solution set = { −1.5 ≤ x < 5}
The solution set can be graphed on the number line as:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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Question 12:
Solve and graph the solution set of:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25; x ∈ R.
(ii) 2x – 9 ≤ 7 and 3x + 3x + 9 > 25; x ∈ I.
(iii) x + 5 ≥ 4(x – 1) and 3 – 2x < −7; x ∈ R.
Solution 12:
(i) 2x – 9 < 7 and 3x + 9 ≤ 25
2x < 16 and 3x ≤ 16
1
X < 8 and x ≤ 5
3
1
∴ Solution set = { x ≤ 5 3 , 𝑥 ∈ 𝑅}
The required graph on number line is:

(ii) 2x – 9 ≤ 7 and 3x + 3x + 9 > 25

2x ≤ 16 and 3x > 16
1
X ≤ 8 and x > 5 3
1
∴ Solution set = { 5 3 < 𝑥 ≤ 8, 𝑥 ∈ 𝐼} = {6, 7, 8}
The required graph on number line is:

(iii) x + 5 ≥ 4(x – 1) and 3 – 2x < −7

9 ≥ 3x and – 2x < − 10
3 ≥ x and x > 5
∴ solution set = Empty set

Question 13:
Solve and graph the solution set of:
(i) 3x – 2 > 19 or 3 – 2x ≥ −7; x ∈ R.
(ii) 5 > p - 1 > 2 or 7 ≤ 2p - 1 ≤ 17; p ∈ R.
Solution 13:
(i) 3x – 2 > 19 or 3 – 2x ≥ −7
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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3x > 21 or – 2x ≥ − 10
X > 7 or x ≤ 5
Graph of solution set of x > 7 or x ≤ 5 = Graph of points which belong to x > 7 or x
≤ 5 or both.
Thus, the graph of the solution set is:

(ii) 5 > p - 1 > 2 or 7 ≤ 2p - 1 ≤ 17

6 > p > 3 or 8 ≤ 2p ≤ 18
6 > p > 3 or 4 ≤ p ≤ 9
Graph of solution set of 6 > p > 3 or 4 ≤ p ≤ 9
= Graph of points which belong to 6 > p > 3 or 4 ≤ p ≤ 9 or both
= Graph of points which belong to 3 < p ≤ 9
Thus, the graph of the solution set is:

Question 14:
The diagram represents two in-equations A and B on real number lines:

A=

B=

(i) Write down A and B in set builder notation.

(ii) Represent A ∩ B and A ∩ B’ on two different number lines.
Solution 14:
(i) A = { x ∈ R: − 2 ≤ x < 5}
B = {x ∈ R: − 4 ≤ x < 3}

(ii) A ∩ B = { x ∈ R:- 2 ≤ x < 5}

It can be represented on number line as:

𝐵′ = { x ∈ R: 3 < x −4}
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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A ∩ B’ = { x ∈ R: 3 ≤ x < 5}
It can be represented on number line as:

Question 15:
Use real number line to find the range of values of x for which:
(i) x > 3 and 0 < x < 6.
(ii) x < 0 and – 3 ≤ x < 1
(iii) −1 < x ≤ 6 and −2 ≤ x ≤ 3
Solution 15:
(i) x > 3 and 0 < X < 6
Both the given in equations are true in the range where their graphs on the real
number lines overlap.
The graphs of the given in equations can be drawn as:
x>3

0<x<6

From both graphs, it is clear that their common range is

3<x<6

(ii) x < 0 and -3 ≤ x < 1

Both the given in equations are true in the range where their graphs on the real
number lines overlap.
The graphs of the given in equations can be drawn as:

X<0

− 3≤ x < 1
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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From both graphs, it is clear that their common range is

-3 ≤ x < 0

(iii) -1 < x ≤ 6 and -2 ≤ x ≤ 3

Both the given in equations are true in the range where their graphs on the real
number lines overlap.
The graphs of the given in equations can be drawn as:
− 1<x≤6

− 2≤x≤3

From both graphs, it is clear that their common range is

1<x≤3

Question 16:
Illustrate the set {x: −3 ≤ x < 0 or x > 2; x ∈ R} on a real number line.
Solution 16:
Graph of solution set of - 3 ≤ x < 0 or x > 2
= Graph of points which belong to -3 ≤ x < 0 or x > 2 or both
Thus, the required graph is:

Question 17:
Given A = {x: −1 < x ≤ 5, x ∈ R} and B = {x: − 4 ≤ x < 3, x ∈ R}
Represent on different number lines:
(i) A ∩ B (ii) A’ ∩ B (iii) A − B
Solution 17:
(i) A ∩ B = {x: -1 < x < 3, x ∈ R}
It can be represented on a number line as:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(ii) Numbers which belong to B but do not belong to A = B - A

A' ∩ B = {x: - 4 ≤ x ≤ -1, x ∈ R}
It can be represented on a number line as:

(iii) A - B = {x: 3 ≤ x ≤ 5, x ∈ R}
It can be represented on a number line as:

Question 18:
P is the solution set of 7X – 2 > 4X + 1 and Q is the solution set of 9x – 45 ≥ 5 (x – 5);
where x ∈ R, Represent:
(i) P ∩ Q
(ii) P – Q
(iii) P ∩ Q’ on different number lines.
Solution 18:
P = { X : 7X – 2 > 4X + 1, X ∈ R }
7x – 2 > 4x + 1
7x – 4x > 1 + 2
3x > 3
X>1
and
Q = {x: 9x – 45 ≥ 5 (x – 5), x ∈ R}
9x – 45 ≥ 5x – 25
9x – 5x ≥ − 25 + 45
4x ≥ 20
X≥5

(i) P ∩ Q = {x : x ≥ 5, x ∈ R}

(ii) P – Q = { X : 1 < X < 5, X ∈ R}

Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(iii) P ∩ Q’ = {x : 1 < x < 5, x ∈ R}

Question 19:
If P = {X: 7X – 4 > 5X + 2, X ∈ R} and Q = {X : X – 19 ≥ 1 – 3X, X ∈ R}; find the range
of set P ∩ Q and represent it on a number line.
Solution 19:
P = {X : 7X – 4 > 5X + 2, X ∈ R}
7X – 4 > 5X + 2
7X – 5X > 2 + 4
2X > 6
X>3

Q = { X:X – 19 ≥ 1 – 3X, X ∈ R}
X – 19 ≥ 1 – 3X
X + 3X ≥ 1 + 19
4X ≥ 20
X≥5
P ∩ Q = { X:X ≥ 5, X ∈ R}

Question 20:
Find the range of values of x, which satisfy:
𝟏 𝑿 𝟐 𝟏
− 𝟑 ≤ 𝟐 +𝟏 𝟑 <𝟓 𝟔
Graph, in each of the following cases, the values of x on the different real number lines:
(i) x ∈ w (ii) x ∈ z (iii) x ∈ R.
Solution 20:
𝟏 𝑿 𝟐 𝟏
− ≤ +𝟏 <𝟓
𝟑 𝟐 𝟑 𝟔
𝟏 𝟓 𝑿 𝟑𝟏 𝟓
− − ≤ < −
𝟑 𝟑 𝟐 𝟔 𝟑
𝟔 𝑿 𝟐𝟏
− 𝟑 ≤ 𝟐 < 𝟔
− 4≤X<7

(i) If x ∈ W, range of value of x is {0, 1, 2, 3, 4, 5, 6}

Class X Chapter 4 – Linear Inequations(in one variable) Maths
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(ii) If x ∈ Z, range of values of x is {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}.

(iii) If x ∈ R, range of values of x is – 4 ≤ x < 7.

Question 21:
Given: A = {x : −8 < 5x + 2 ≤ 17, x ∈ I}
B = {x: − 2 ≤ 7 + 3x < 17, x ∈ R}
Where R = {Real numbers} and I = {integers}. Represent A and B on two different
number lines. Write down the elements of A ∩ B.

Solution 21:
A = {x: -8 < 5x + 2 ≤ 17, x ∈ I}
= {x: − 10 < 5x ≤ 15, x ∈ I}
= {x: − 2 < x ≤ 3, x ∈ I}
It can be represented on number line as:

B = {x: − 2 ≤ 7 + 3x < 17, x ∈ R}

= {x: −9 ≤ 3x < 10, x ∈ R}
= {x: −3 ≤ x < 3.33, x ∈ R}
It can be represented on number line as:

A ∩ B = {-1, 0, 1, 2, 3}

Question 22:
Solve the following in-equation and represent the solution set on the number line.
2x − 5 ≤ 5x + 4 < 11, where x ∈ I.
Solution 22:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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2x − 5 ≤ 5x + 4 and 5x + 4 < 11
2x − 5x ≤ 4 − 5 and 5x < 11 − 4
3x ≤ - 1 and 5x < 7
7
x ≥ − 1 and x < 5
x ≥ − 1 and x < 1.4
Since x ∈ I, the solution set is {− 3, −2 −1, 0, 1}
And the number line representation is

Question 23:
Given that x ∈ I, solve the in-equation and graph the solution on the number line:
𝑥 −4 𝑥
3≥ 2 + 3 ≥2
Solution 23:
𝑥 −4 𝑥
3≥ + ≥2
2 3
3𝑥 −12 +2𝑥
3≥ ≥2
6
18 ≥ 5x – 12 ≥ 12
30 ≥ 5x ≥ 24
6 ≥ x ≥ 4.8
Solution set = {5,6}
It can be graphed on number line as:

Question 24:
Given:
A = {x: 11x - 5 > 7x + 3, x ∈ R} and
B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R}
Find the range of set A ∩ B and represent it on a number line.
Solution 24:
A = {x: 11x - 5 > 7x + 3, x ∈ R}
= {x : 4x > 8, x ∈ R}
= {x : x > 2, x ∈ R}

B = {x : 18x - 9 ≥ 15 + 12x, x ∈ R}
= {x : 6x ≥ 24, x ∈ R}
= {x : x ≥ 4, x ∈ R}
A ∩ B = {x : x ≥ 4, x ∈ R}
It can be represented on number line as:
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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Question 25:
Find the set of value of x, Satisfying:
𝑥 5
7X + 3 ≥ 3X – 5 and 4 − 5 ≤ 4 − x,
Where x ∈ N.
Solution 25:
7X + 3 ≥ 3X – 5
4X ≥ − 8
X≥-2

𝑿 𝟓
−𝟓 ≤ –𝑿
𝟒 𝟒
𝑋 5
−𝑋 ≤ +5
4 4
5𝑋 25
≤
4 4
X≤5

Since, x ∈ N
∴ Solution set = {1, 2, 3, 4, 5}

Question 26:
Solve:
𝒙 𝒙
(i) 𝟐 + 𝟓 ≤ + 𝟔, where x is a positive odd integer.
𝟑
2𝑥 +3 3𝑥 −1
(ii) ≥ , Where x is a positive even integer.
3 4
Solution 26:
𝒙 𝒙
(i) 𝟐 + 𝟓 ≤ 𝟑 + 𝟔
𝒙 𝒙
− ≤ 𝟔 –𝟓
𝟐
𝒙
𝟑
≤𝟏
𝟔
𝒙 ≤𝟔
Since, x is a positive odd integer
∴ Solution set = {1, 3, 5}
2𝑥 +3 3𝑥 −1
(ii) ≥ 4
3
8x + 12 ≥ 9x – 3
−X ≥ − 15
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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X ≤ 15
Since, x is a positive even integer
∴ Solution set = {2, 4, 6, 8, 10, 12, 14}

Question 27:
Solve the in-equation:
1 4𝑥 4
− 2 2 + 2𝑥 ≤ 5 ≤ 3 + 2𝑥, x ∈ W.
Graph the solution set on the number line.
Solution 27:
1 4𝑥 4
- 2 2 + 2𝑥 ≤ ≤ + 2𝑥
5 3
1 4𝑥 4
-22≤ − 2𝑥 ≤
5 3
5 6𝑥 4
−2 ≤ − ≤
5 3
25 10
≥𝑥 ≥ − 9
12
2.083 ≥ x ≥ − 1.111
Since, x ∈ W
∴ Solution set = {0, 1, 2}
The solution set can be represented on number line as:

Question 28:
Find three consecutive largest positive integers such that the sum of one-third of first,
one-fourth of second and one-fifth of third is atmost 20.
Solution 28:
Let the required integers be x, x + 1 and x + 2.
According to the given statement,
1 1 1
𝑥 + (𝑥 + 1) + (𝑥 + 2) ≤ 20
3 4 5
20𝑥 +15𝑥 +15 +12𝑥 +24
≤ 20
60
47x + 39 ≤ 1200
47x ≤ 1161
X ≤ 24. 702
Thus, the largest value of the positive integer x is 24.
Hence, the required integers are 24, 25 and 26.

Question 29:
Solve the given in-equation and graph the solution on the number line.
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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Solution 29:
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
⟹ 2y – 3 – y < y + 1 – y ≤ 4y + 7 – y
⟹ y – 3 < 1 ≤ 3y + 7
⟹ y – 3 < 1 and 1 ≤ 3y + 7
⟹ y < 4 and 3y ≥ −6 ⇒ y ≥ -2
⟹ -2 ≤ y < 4

The graph of the given equation can be represented on a number line as:

Question 30:
Solve the inequation:
3z – 5 ≤ z + 3 < 5z – 9; z ∈ R.
Graph the solution set on the number line.
Solution 30:
3z – 5 ≤ z + 3 < 5z – 9
3z – 5 ≤ z + 3 and z + 3 < 5z – 9
2z ≤ 8 and 12 < 4z
Z ≤ 4 and 3 < z
Since, z ∈ R
∴ Solution set = {3 < z ≤ 4, x ∈ R}
It can be represented on a number line as:

Question 31:
Solve the following in equation and represent the solution set on the number line.
1 2𝑋 5
− 3 < − 2 − 3 ≤ 6, x ∈ R
Solution 31:
𝟏 𝟐𝑿 𝟓
− 3<−𝟐 − 𝟑 ≤ 𝟔
Multiply by 6, we get
⇒ − 18 < − 3 – 4x ≤ 5
⇒ − 15 < − 4x ≤ 8
Dividing by – 4, We get

− 15 8
⇒ −4
>𝑥 ≥ −4
Class X Chapter 4 – Linear Inequations(in one variable) Maths
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15
⇒−2≤x< 4
15
⇒ x ∈ (−2, 4 )
The solution set can be represented on a number line as:

Question 32:
Solve the following in equation and represent the solution set on the number line:
3𝑥 −2
4x – 19 < 5 − 2 ≤ 5 + 𝑥, 𝑥 ∈ 𝑅
Solution 32:
3𝑥 −2
4x – 19 < −2 ≤ + 𝑥, 𝑥 ∈ 𝑅
5 5
3𝑋 −2
⇒ 4X – 19 + 2 < −2 +2 ≤ + 𝑋 + 2, 𝑋 ∈ 𝑅
5 5
3𝑋 8
⇒ 4X – 17 < ≤𝑋 + ,𝑋 ∈ 𝑅
5 5
3𝑋 −8 3𝑥
⇒ 4X – < 17 and ≤𝑥 − ,𝑥 ∈ 𝑅
5 5 5
20𝑋 −3𝑋 −8 5𝑋−3𝑋
⇒ < 17 and ≤ ,𝑋 ∈ 𝑅
5 5 5
17𝑥 −8 2𝑥
⇒ < 17 and ≤ ,𝑥 ∈ 𝑅
5 5 5
𝑥
⇒ 5 < 1 and – 4 ≤ x, x ∈ 𝑅
⇒ x < 5 and – 4 ≤ x, x ∈ R

The solution set can be represented on a number line as: