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Linear Inequations (In One Variable) — ICSE Class 10 Maths Important Questions

13 ICSE Class 10 Maths practice questions on Linear Inequations (In One Variable), each with a full model answer, covering 5 topics from the chapter in the question formats used in the exam.

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13
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5
Topics
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Key concepts
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With answers

Linear Inequations (In One Variable) — ICSE Class 10 Maths Important Questions

Inequalities on the Number Line

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Quick answer

Reliable-scoring ICSE Linear Inequations questions ask you to solve an inequation in one variable over a stated replacement set (natural numbers, integers or real numbers), write the solution set, and represent it on a number line. Remember to reverse the inequality sign when multiplying or dividing by a negative number. Combined (double) inequations with fractions are asked almost every year.

About Linear Inequations (In One Variable)

Within the ICSE Class 10 Maths chapter Linear Inequations you solve inequalities in one variable, taking care to reverse the sign when multiplying or dividing by a negative number. The solution depends on the replacement set — natural numbers, whole numbers, integers or real numbers — and you write the solution set and show it on a number line.

Solving linear inequationsReplacement set and solution setRule for multiplying or dividing by a negativeCombined (double) inequationsNumber-line representation

Key concepts & formulas

Sign-reversal rule

You may add or subtract any quantity, and multiply or divide by any positive number, without changing the sign. Multiplying or dividing both sides by a negative number reverses the inequality: from −x>3-x>3-x>3 you get x<−3x<-3x<-3.

Replacement and solution sets

The replacement set is the set from which xxx may be chosen (e.g. N,W,Z,R\mathbb{N},\mathbb{W},\mathbb{Z},\mathbb{R}N,W,Z,R). The solution set is the subset that satisfies the inequation — it may be finite (for integers) or an interval (for real numbers).

Number line

On the number line a filled (solid) dot means the end value is included (≤,≥\le,\ge≤,≥) and an open dot means it is excluded (<,><,><,>). For real solutions the whole segment is shaded; for integers only the marked points are shown.

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Important questions with answers

Try each on paper first, then reveal the model answer to check your method.

Multiple-choice questions (1 mark)

Q1MCQEasy1 mark

If xxx is a natural number and x<4x<4x<4, the solution set is:

  1. (a)

    {1,2,3}\{1,2,3\}\1,2,3\

  2. (b)

    {0,1,2,3}\{0,1,2,3\}\0,1,2,3\

  3. (c)

    {1,2,3,4}\{1,2,3,4\}\1,2,3,4\

  4. (d)

    {0,1,2,3,4}\{0,1,2,3,4\}\0,1,2,3,4\

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Answer: (a) {1,2,3}\{1,2,3\}\1,2,3\.

Natural numbers start at 111, and x<4x<4x<4 excludes 444, so x∈{1,2,3}.x\in\{1,2,3\}.x\1,2,3\.

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Q2MCQEasy1 mark

The solution of 2x>62x>62x>6, where x∈Rx\in\mathbb{R}xR, is:

  1. (a)

    x>3x>3x>3

  2. (b)

    x<3x<3x<3

  3. (c)

    x≥3x\ge3x≥3

  4. (d)

    x>6x>6x>6

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Answer: (a) x>3x>3x>3.

Dividing both sides by the positive number 222 keeps the sign: x>3.x>3.x>3.

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Q3MCQModerate1 mark

The solution of −3x≥9-3x\ge9-3x≥9, where x∈Rx\in\mathbb{R}xR, is:

  1. (a)

    x≤−3x\le-3x≤-3

  2. (b)

    x≥−3x\ge-3x≥-3

  3. (c)

    x≤3x\le3x≤3

  4. (d)

    x≥3x\ge3x≥3

Show model answer

Answer: (a) x≤−3x\le-3x≤-3.

Dividing by −3-3-3 reverses the sign: x≤9−3=−3.x\le\dfrac{9}{-3}=-3.x≤9/-3=-3.

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Q4MCQHOTS1 mark

The solution set of 3≤2x−1<73\le2x-1<73≤2x-1<7, where x∈Zx\in\mathbb{Z}xZ, is:

  1. (a)

    {2,3}\{2,3\}\2,3\

  2. (b)

    {2,3,4}\{2,3,4\}\2,3,4\

  3. (c)

    {3}\{3\}\3\

  4. (d)

    {2}\{2\}\2\

Show model answer

Answer: (a) {2,3}\{2,3\}\2,3\.

3≤2x−1⇒4≤2x⇒x≥23\le2x-1\Rightarrow4\le2x\Rightarrow x\ge23≤2x-14≤2x x≥2; and 2x−1<7⇒2x<8⇒x<42x-1<7\Rightarrow2x<8\Rightarrow x<42x-1<72x<8 x<4. So 2≤x<42\le x<42≤ x<4, giving integers {2,3}.\{2,3\}.\2,3\.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): Multiplying both sides of −x>3-x>3-x>3 by −1-1-1 gives x>−3x>-3x>-3.

Reason (R): When both sides of an inequation are multiplied by a negative number, the inequality sign reverses.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (d) R is true, but A is false: multiplying −x>3-x>3-x>3 by −1-1-1 reverses the sign to give x<−3x<-3x<-3, not x>−3x>-3x>-3.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Solve the inequation 5x−3<3x+75x-3<3x+75x-3<3x+7, where x∈Rx\in\mathbb{R}xR.

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5x−3<3x+75x-3<3x+75x-3<3x+7

5x−3x<7+35x-3x<7+35x-3x<7+3

2x<10⇒x<5.2x<10\Rightarrow x<5.2x<10 x<5.

Solution set ={x:x<5, x∈R}.=\{x:x<5,\ x\in\mathbb{R}\}.=\x:x<5, xR\.

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Q7Very ShortModerate2 marks

Given the replacement set {−2,−1,0,1,2,3}\{-2,-1,0,1,2,3\}\-2,-1,0,1,2,3\, find the solution set of 2x−1<32x-1<32x-1<3.

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2x−1<3⇒2x<4⇒x<2.2x-1<3\Rightarrow2x<4\Rightarrow x<2.2x-1<32x<4 x<2.

From the replacement set, the values less than 222 are −2,−1,0,1-2,-1,0,1-2,-1,0,1.

Solution set ={−2,−1,0,1}.=\{-2,-1,0,1\}.=\-2,-1,0,1\.

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Short answer questions (3 marks)

Q8Short AnswerEasy3 marks

Solve −2≤x+3<5-2\le x+3<5-2≤ x+3<5, where x∈Rx\in\mathbb{R}xR, and represent the solution on a number line.

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Subtract 333 throughout: −2−3≤x<5−3-2-3\le x<5-3-2-3≤ x<5-3, i.e. −5≤x<2.-5\le x<2.-5≤ x<2.

Solution set ={x:−5≤x<2, x∈R}=\{x:-5\le x<2,\ x\in\mathbb{R}\}=\x:-5≤ x<2, xR\ — a solid dot at −5-5-5 (included) and an open dot at 222 (excluded).

ICSE Class 10 Maths — Linear Inequations (In One Variable): Solve -2\le x+3<5, where x\in\mathbb{R}, and represent the solution on a number line.
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Q9Short AnswerModerate3 marks

Solve x2+5≤x3+6\dfrac{x}{2}+5\le\dfrac{x}{3}+6x/2+5/3+6, where x∈Rx\in\mathbb{R}xR.

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Multiply every term by 666 (the LCM):

6×x2+6×5≤6×x3+6×66\times\dfrac{x}{2}+6\times5\le6\times\dfrac{x}{3}+6\times66×x/2+6×5≤6×x/3+6×6

3x+30≤2x+363x+30\le2x+363x+30≤2x+36

3x−2x≤36−30⇒x≤6.3x-2x\le36-30\Rightarrow x\le6.3x-2x≤36-30 x≤6.

Solution set ={x:x≤6, x∈R}.=\{x:x\le6,\ x\in\mathbb{R}\}.=\x:x≤6, xR\.

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Q10Short AnswerHOTS3 marks

If xxx is an integer and −5<2x−1≤5-5<2x-1\le5-5<2x-1≤5, find the solution set and state its greatest value.

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−5<2x−1⇒−4<2x⇒x>−2.-5<2x-1\Rightarrow-4<2x\Rightarrow x>-2.-5<2x-1-4<2x x>-2.

2x−1≤5⇒2x≤6⇒x≤3.2x-1\le5\Rightarrow2x\le6\Rightarrow x\le3.2x-1≤52x≤6 x≤3.

So −2<x≤3-2<x\le3-2<x≤3. For integers, the solution set ={−1,0,1,2,3}=\{-1,0,1,2,3\}=\-1,0,1,2,3\, and its greatest value is 333.

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Long answer questions (5 marks)

Q11Long AnswerModerate5 marks

Solve the pair of inequations 3x−2>2(x−1)3x-2>2(x-1)3x-2>2(x-1) and 2x+5≤152x+5\le152x+5≤15 simultaneously, where x∈Rx\in\mathbb{R}xR, and represent the solution on a number line.

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First inequation: 3x−2>2x−2⇒3x−2x>−2+2⇒x>0.3x-2>2x-2\Rightarrow3x-2x>-2+2\Rightarrow x>0.3x-2>2x-23x-2x>-2+2 x>0.

Second inequation: 2x+5≤15⇒2x≤10⇒x≤5.2x+5\le15\Rightarrow2x\le10\Rightarrow x\le5.2x+5≤152x≤10 x≤5.

Common solution: 0<x≤5.0<x\le5.0<x≤5.

Solution set ={x:0<x≤5, x∈R}=\{x:0<x\le5,\ x\in\mathbb{R}\}=\x:0<x≤5, xR\ — open dot at 000, solid dot at 555.

ICSE Class 10 Maths — Linear Inequations (In One Variable): Solve the pair of inequations 3x-22(x-1) and 2x+5\le15 simultaneously, where x\in\mathbb{R}, and represent the solution
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Q12Long AnswerHOTS5 marks

Solve the following inequation, write the solution set and represent it on a number line: −3(x−7)≥15−7x>x+13-3(x-7)\ge15-7x>\dfrac{x+1}{3}-3(x-7)≥15-7x>x+1/3, where x∈Rx\in\mathbb{R}xR.

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Split into two inequations.

Left part: −3(x−7)≥15−7x⇒−3x+21≥15−7x⇒−3x+7x≥15−21⇒4x≥−6⇒x≥−32.-3(x-7)\ge15-7x\Rightarrow-3x+21\ge15-7x\Rightarrow-3x+7x\ge15-21\Rightarrow4x\ge-6\Rightarrow x\ge-\dfrac{3}{2}.-3(x-7)≥15-7x-3x+21≥15-7x-3x+7x≥15-214x≥-6 x≥-3/2.

Right part: 15−7x>x+1315-7x>\dfrac{x+1}{3}15-7x>x+1/3. Multiply by 333: 45−21x>x+1⇒45−1>x+21x⇒44>22x⇒x<2.45-21x>x+1\Rightarrow45-1>x+21x\Rightarrow44>22x\Rightarrow x<2.45-21x>x+145-1>x+21x44>22x x<2.

Combining: −32≤x<2.-\dfrac{3}{2}\le x<2.-3/2≤ x<2.

Solution set ={x:−1.5≤x<2, x∈R}=\left\{x:-1.5\le x<2,\ x\in\mathbb{R}\right\}=\x:-1.5≤ x<2, xR\ — solid dot at −1.5-1.5-1.5, open dot at 222.

ICSE Class 10 Maths — Linear Inequations (In One Variable): Solve the following inequation, write the solution set and represent it on a number line: -3(x-7)\ge15-7x\dfrac{x+1}{3},
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Case-based questions (4 marks)

Q13Case-basedModerate4 marks

The replacement set for the variable xxx is the set of integers Z\mathbb{Z}Z. Consider the inequation −4≤3x+2<11-4\le3x+2<11-4≤3x+2<11.

(i) Solve the inequation for xxx.

(ii) Write the solution set.

(iii) State the number of elements in the solution set.

(iv) Represent the solution on a number line.

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(i) −4≤3x+2⇒−6≤3x⇒x≥−2-4\le3x+2\Rightarrow-6\le3x\Rightarrow x\ge-2-4≤3x+2-6≤3x x≥-2; and 3x+2<11⇒3x<9⇒x<33x+2<11\Rightarrow3x<9\Rightarrow x<33x+2<113x<9 x<3. So −2≤x<3.-2\le x<3.-2≤ x<3.

(ii) For integers, solution set ={−2,−1,0,1,2}.=\{-2,-1,0,1,2\}.=\-2,-1,0,1,2\.

(iii) It has 555 elements.

(iv) Mark the five integer points −2-2-2 to 222 with solid dots.

ICSE Class 10 Maths — Linear Inequations (In One Variable): The replacement set for the variable x is the set of integers \mathbb{Z}. Consider the inequation -4\le3x+2<11. (i) Solv
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  • Do these Linear Inequations (In One Variable) questions follow the latest ICSE syllabus?
    Yes — they are aligned to the CISCE 2026–27 syllabus for ICSE Class 10 Maths, so nothing here is outside the current course.

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