Self Evaluation on Linear Inequations in One Variable ICSE Class 10 Maths OP Malhotra (2026-27). We Provide Step by Step Solutions of self evaluation on Linear Inequations in one variable. Visit official Website CISCE  for detail information about ICSE Board Class-10 Mathematics.

Self Evaluation on Linear Inequations in One Variable ICSE Class 10 Maths OP Malhotra (2026-27)

Board	ICSE
Publications	S Chand
Subject	Maths
Class	10th
Chapter-4	Linear Inequations in one variable
Writer	OP Malhotra
Self Evaluation	Extra Practice Questions
Edition	2026-2027

Self Evaluation on Linear Inequations in One Variable

Self Evaluation on Linear Inequations in One Variable ICSE Class 10 Maths OP Malhotra (2026-27)

Que-1: Solve the following inequation and graph the solution set on the number line 2x – 3 < x + 2 ≤ 3x + 5, x ∈ R.

Sol: 2x – 3 < x + 2 ≤ 3x + 5, x ∈ R 2x – 3 < x + 2
⇒ 2x – x < 2 + 3 ⇒ x < 5 … (i)
and x + 2 ≤ 3x + 5 ⇒ 2 – 5 ≤ 3x – x
⇒ – 2 ≤ 2x ⇒ – 3/2 ≤ x … (ii)
From (i) and (ii)
Solution: {−3/2≤x<5,x∈R}
On number line

 

 

Que-2: Solve the inequation : 12 + 1 5/6 x ≤ 5 + 3x, x ∈ R. Represent the solution on a number line.

Sol:

 

 

 

 

 

 

 

 

Que-3: Solve the inequation : – 3 ≤ 3 – 2x < 9, x ∈ R. Represent your solution on a number line.

Sol:  – 3 ≤ 3 – 2x < 9, x ∈ R
– 3 ≤ 3 – 2x
– 3 – 3 ≤ – 2x
⇒ – 6 ≤ – 2x
⇒ 2x ≤ 6 ⇒ x ≤ 6/2
⇒ x ≤ 3 … (i)
and 3 – 2x < 9
⇒ – 2x < 9 – 3 ⇒ – 2x < 6 ⇒ x > 6/–2 ⇒ x > – 3
⇒ – 3 < x … (ii)
From (i) and (ii)
– 3 < x ≤ 3
∴ Solution = {x : – 3 < x ≤ 3}
On the number line

 

 

Que-4: Find the value of x, which satisfies the inequation – 2 ≤ 1/2 – 2x/3 ≤ 1(5/6), x ∈ N.
Graph the solution on the number line.

Sol: 

 

 

 

 

 

 

 

 

 

 

 

Que-5: Solve the following inequation, and graph the solution on the number line
(2x – 5 ≤ 5x + 4 < 11, x ∈ R)

Sol: 2x – 5 ≤ 5x + 4 < 11, x ∈ R
Now 2x – 5 ≤ 5x + 4
⇒ – 5 – 4 ≤ 5x – 2x ⇒ – 9 ≤ 3x
⇒ –9/3 ≤ x ⇒ – 3 ≤ x … (i)
and 5x + 4 < 11 ⇒ 5x < 11 – 4
⇒ 5x < 7 ⇒ x 7/5 ⇒ x < 1.4 … (ii)
From (i) and (ii)
Solution = {- 3 < x < 1.4, x ∈ R}
On number line

 

 

Que-6: Solve 2 ≤ 2x – 3 ≤ 5, x ∈ R and mark it on a number line.

Sol: 2 ≤ 2x – 3 ≤ 5, x ∈ R
Now 2 ≤ 2x – 3 ⇒ 2 + 3 ≤ 2x
⇒ 5 ≤ 2x ⇒ 5/2 ≤ x ⇒ 2.5 ≤ x … (i)
and 2x – 3 ≤ 5 ⇒ 2x ≤ 5 + 3
⇒ 2x ≤ 8 ⇒ x ≤ 8/2 ⇒ x ≤ 4 … (ii)
From (i) and (ii)
Solution = {2.5 ≤ x ≤ 4, x ∈ R}
On number line

 

 

Que-7: Given that x ∈ I, solve the inequation and graph the solution on the number line. 3 ≥ (x−4)/2 + x/3 ≥ 2

Sol: 3 ≥ (x−4)/2 + x/3 ≥ 2, x ∈ 1
Now 3 ≥ (x−4)/2 + x/3 ⇒ 3 ≥ (3x−12+2x)/6
⇒ 18 ≥ 5x – 12 ⇒ 18 + 12 ≥ 5x
⇒ 30 ≥ 5x ⇒ 30/5 ≥ x
6 ≥ x ⇒ x ≤ 6 … (i)
and (x−4)/2 + x/3 ≥ 2
(3x−12+2x)/6 ≥ 2 ⇒ 5x – 12 ≥ 12
⇒ 5x ≥ 12 + 12 ⇒ 5x ≥ 24
⇒ x ≥ 24/5 ⇒ x ≥ 4.8
⇒ 4.8 ≤ x … (ii)
From (i) and (ii)
4.8 ≤ x ≤ 6
∵ x ∈ I
∴ Solution set = {5, 6}
On the number line

 

Que-8: Give that x ∈ R, solve the following inequality and graph the solution on the number line inequality : – 1 ≤ 3 + 4x < 23.

Sol: – 1 ≤ 3 + 4x < 23, x ∈ R
Now – 1 ≤ 3 + 4x
⇒ – 1 – 3 ≤ 4x
⇒ – 4 ≤ 4x
⇒ −4/4 ≤ x ⇒ – 1 ≤ x … (i)
and 3 + 4x < 23 ⇒ 4x < 23 – 3
⇒ 4x < 20 ⇒ x < 20/4 … (ii)
From (i) and (ii)
Solution : {-1 < x < 5, x ∈ R}
On the number line

 

Que-9: Solve the following inequation and graph the solution on the number line :
-2(2/3) ≤ x + 2/3 < 3(1/3); x ∈ R.

Sol: -2(2/3) ≤ x + 2/3 < 3(1/3); x ∈ R.
−8/3 ≤ x + 1/3 < 10/3
Now −8/3 ≤ x + 1/3 ⇒ −8/3 – 1/3 ≤ x

 

 

 

 

 

Que-10: Solve the given inequation and graph the solution on the number line :
2y – 3 < y + 1 ≤ 4y + 7, y ∈ R.

Sol: 2y – 3 < y + 1 ≤ 4y + 7, y ∈ R
Now 2y – 3 < y + 1
⇒ 2y – y < 1 + 3 ⇒ y < 4 … (i)
and y + 1 ≤ 4y + 7 ⇒ 1 – 7 ≤ 4y – y
⇒ – 6 ≤ 3 y ⇒ −6/3 ≤ y
⇒ – 2 ≤ y … (ii)
From (i) and (ii)
Solution = {- 2 ≤ y < 4, y ∈ R}
On the number line

 

Que-11: Solve the inequation and represent the solution set on the number line
– 3 + x ≤ 8x/3 + 2 ≤ 14/3 + 2x, where x ∈ I.

Sol: Given :
– 3 + x ≤ 8x/3 + 2 ≤ 14/3 + 2x, where x ∈ I
⇒ – 3 + x ≤ 8x/3 + 2 … (i)
and 8x/3 + 2 ≤ 14/3 + 2x … (ii)
⇒ – 5 ≤ 5x/3 and 2x/3 ≤ 8x/3
⇒ x ≥ – 3 and x ≤ 4
∴ – 3 ≤ x ≤ 4
Solution set = {-3, -2, -1, 0, 1,2, 3, 4}
Number line

 

Que-12: Solve the following inequation and represent the solution set on the number line.
– 3 < – 1/2 – 2x/3 ≤ 5/6, x ∈ R.

Sol: – 3 < – 1/2 – 2x/3 ≤ 5/6, x ∈ R.
⇒ – 3 < (−3−4x)/6 ≤ 5/6
⇒ – 18 < – 3 – 4x ≤ 5
⇒ – 18 + 3 < – 4x ≤ 5 + 3
⇒ – 15 < – 4x ≤ 8 ⇒ −15/−4 > x ≥ 8/−4
⇒ 15/4 > x ≥ – 2
⇒ 3.75 > x ≥ – 2

 

Que-13: Solve the following inequation and represent the solution set on the number line
2x – 5 ≤ 5x + 4 < 11, where x ∈ 1.

Sol: 

 

 

 

 

 

 

Que-14: Solve the following inequation and represent the solution set on the number line:
4x – 19 < 3x/5 – 2 ≤ – 2/5 + x, x ∈ R.

Sol: we have 4x – 19 < 3x/5 – 2 ≤ – 2/5 + x, x ∈ R.
⇒ 4x – 19 < 3x/5 – 2 and 3x/5 – 2 ≤ –2/5 + x, x ∈ R.
⇒ 4x – 3x/5 < 17 and – 2 + 2/5 ≤ x – 3x/5, x ∈ R.
⇒ (20x–3x)/5 < 17 and (−10+2)/5 ≤ (5x−3x)/5, x ∈ R.
⇒ 17x/5 < 17 and −8/5 ≤ 2x/5 + x, x ∈ R.
⇒ x < 5 and – 4 ≤ x, x ∈ R.
⇒ – 4 ≤ x < 5, x ∈ R
Hence, solution set is {x: -4 ≤ x ≤ 5, x ∈ R} The solution set is represented on the number line as below:

 

Que-15: Solve the following inequation, write the solution set and represent it on the number line:
−x/3 ≤ x/2−1(1/3) < 1/6,x∈R.

Sol:

 

 

 

 

 

 

 

 

 

 

 

Que-16: Find the values of x, which satisfy the inequation −2(5/6) < 1/2−2x/3 ≤2, x ∈ W. Graph the solution set on the number line.

Sol:

 

 

 

 

 

 

 

 

Que-17: Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R Represent the solution set on a real number line.

Sol: 13x – 5 < 15x + 4 < 7x + 12, x ∈ R
13x – 5 < 15x + 4 and 15x + 4 < 7x + 12
13x < 15x + 9 and 15x < 7x + 8
0 < 2x + 9 and 8x < 8
– 9 < 2x and x < 1
– 9/2 < x and x < 1 9
∴ – 9/2 < x < 1
i.e. – 4.5 < x < 1
Hence, the solution set is {x : – 4.5 < x < 1, x ∈ R}
The required line is

 

 

Que-18: Solve the following inequation, write the solution set and represent it on the number line. – 3(x – 7) ≥ 15 – 7x > (x+1)/3, x ∈ R.

Sol: – 3(x – 7) ≥ 15 – 7x > (x+1)/3
⇒ – 3(x – 7) ≥ 15 – 7x > and 15 – 7x > (x+1)/3
⇒ – 3x + 21 ≥ 15 – 21 and 45 – 21x > x + 1
⇒ – 3x + 7x ≥ 15 – 21 and 45 – 1 > x + 2x
⇒ 4x ≥ – 6 and 44 > 22x
⇒ x ≥ –3/2 and 2 > x
Hence, the solution set is

 

 

 

Very Short Answer Type Questions (VSA)

1. If −x − 4 > 7, then x … .

Sol: −x − 4 > 7
⇒ −x > 11
⇒ x < −11 (sign changes on multiplying by −1)

Answer: x < −11

2. If −2x/3 > −2, x ∈ N, then x = … .

Sol: −2x/3 > −2
⇒ −2x > −6
⇒ x < 3

Since x ∈ N, possible values are 1 and 2.

Answer: x = 1, 2

Solve the following inequations and represent the solution set on the number line:

3. x + 3 < 5, x ∈ W.

Sol: x + 3 < 5
⇒ x < 2

Since x ∈ W, x = 0, 1.

Solution Set = {0, 1}

 

 

 

4. x − 3 ≤ −1, x ∈ R.

Sol: x − 3 ≤ −1
⇒ x ≤ 2

Solution Set = {x ∈ R : x ≤ 2}

 

 

 

5. 2x + 5 ≤ 11, x ∈ W.

Sol: 2x + 5 ≤ 11
⇒ 2x ≤ 6
⇒ x ≤ 3

Since x ∈ W, x = 0, 1, 2, 3.

Solution Set = {0, 1, 2, 3}

 

 

6. 3x − 5/6 > 1/3 , x ∈ {0, 1, 2, 3, 4, 5}.

Sol: 3x − 5/6 > 1/3
⇒ 3x > 1/3 + 5/6
⇒ 3x > 7/6
⇒ x > 7/18

From the given set, x = 1, 2, 3, 4, 5.

Solution Set = {1, 2, 3, 4, 5}

 

 

 

7. −3 ≤ x − 4 < 1, x ∈ N.

Sol: −3 ≤ x − 4 < 1
⇒ 1 ≤ x < 5

Since x ∈ N, x = 1, 2, 3, 4.

Solution Set = {1, 2, 3, 4}

 

 

 

8. 5 − 3x ≥ x − 15, x ∈ N.

Sol: 5 − 3x ≥ x − 15
⇒ 5 ≥ 4x − 15
⇒ 20 ≥ 4x
⇒ x ≤ 5

Since x ∈ N, x = 1, 2, 3, 4, 5.

Solution Set = {1, 2, 3, 4, 5}

 

 

9. Solve y + 2 < 5 and graph the solution set. The replacement set is {−1, 0, 2, 4}.

Sol:  y + 2 < 5
⇒ y < 3

Checking the replacement set:
−1 ✓, 0 ✓, 2 ✓, 4 ✗

Solution Set = {−1, 0, 2}

 

 

 

10. Solve 3x + 7 > 4x, x ∈ R and graph the solution set.

Sol: 3x + 7 > 4x
⇒ 7 > x
⇒ x < 7

Solution Set = {x ∈ R : x < 7}

 

 

Multiple Choice Questions (MCQs)

1. If x < 7, then
    (a) −x < −7     (b) −x ≤ −7     (c) −x > −7     (d) −x ≥ −7

Sol: (c) −x > −7
x < 7
Multiplying both sides by −1, the inequality sign changes.
Therefore,
−x > −7

Answer: (c) −x > −7

2. If −3x + 12 < −15, then
    (a) x > 9     (b) x < 9     (c) x ≥ 9     (d) x ≤ 9

Sol: (a) x > 9
−3x + 12 < −15
⇒ −3x < −27
⇒ x > 9

Answer: (a) x > 9

3. If x satisfies the inequations 2x − 7 < 11 and 3x + 4 < −5, then
    (a) x < 3     (b) x > −3     (c) x < −3     (d) x > 3

Sol: (c) x < −3
2x − 7 < 11
⇒ 2x < 18
⇒ x < 9

3x + 4 < −5
⇒ 3x < −9
⇒ x < −3

Common solution: x < −3

Answer: (c) x < −3

4. Solve the inequality 2x − 5 ≤ (4x-7)/3
    (a) x < 4     (b) x ≤ 4     (c) x ≤ 8     (d) x < −4

Sol: (b) x ≤ 4
3(2x − 5) ≤ 4x − 7
⇒ 6x − 15 ≤ 4x − 7
⇒ 2x ≤ 8
⇒ x ≤ 4

Answer: (b) x ≤ 4

5. Solve: −4x < −5x − 8
    (a) x < 8     (b) x > −8     (c) x < −8     (d) x ≤ −8

Sol: (c) x < −8
−4x < −5x − 8
⇒ x < −8

Answer: (c) x < −8

6. Solve: 3(2x − 8) < 5x
    (a) x > 24     (b) x ≤ 24     (c) x > −24     (d) x < 24

Sol: (d) x < 24
6x − 24 < 5x
⇒ x < 24

Answer: (d) x < 24

7. Solve: 2(3y + 7) < 5y
    (a) y < −14     (b) y > −14     (c) y ≤ −14     (d) y < 14

Sol: (a) y < −14
6y + 14 < 5y
⇒ y < −14

Answer: (a) y < −14

8. Solve: 4(2t − 3) > 12t + 2
    (a) t < −3     (b) t < −3¹⁄₂     (c) t > −3¹⁄₂     (d) t ≤ −3¹⁄₂

Sol: (a) t < −7/2
8t − 12 > 12t + 2
⇒ −4t > 14
⇒ t < −14/4
⇒ t < −7/2

Answer: (a) t < −7/2

9. The value of x for which −11 ≤ 4x − 3 ≤ 13 is
    (a) −4 ≤ x ≤ 5     (b) −2 ≤ x ≤ 4     (c) −8 ≤ x ≤ 16     (d) −11 ≤ x ≤ 10

Sol: (b) −2 ≤ x ≤ 4
−11 ≤ 4x − 3 ≤ 13
Add 3 throughout:
−8 ≤ 4x ≤ 16
Divide by 4:
−2 ≤ x ≤ 4

Answer: (b) −2 ≤ x ≤ 4

10. If |x − 3|/(x − 3) > 0, then
    (a) x > 1     (b) x > 2     (c) x > −3     (d) x > 3

Sol: (d) x > 3
If x > 3, then |x − 3| = x − 3.
Therefore,

|x − 3|/(x − 3) = 1 > 0

If x < 3, then |x − 3| = −(x − 3).
Therefore,

|x − 3|/(x − 3) = −1 < 0

Hence the inequality is satisfied only when x > 3.

Answer: (d) x > 3

— : End of Self Evaluation on Linear Inequations in One Variable ICSE Class 10 Maths OP Malhotra (2026-27) :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

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