Important Questions Class 11 Maths Chapter 5 Linear Inequalities

A linear inequality is a statement where two linear expressions are compared using <, >, ≤, or ≥.
Its solution set contains all values that make the inequality a true mathematical statement.

A small sign change can completely change the answer in inequalities. Important Questions Class 11 Maths Chapter 5 help students practise strict inequalities, slack inequalities, one-variable inequalities, solution sets, number-line graphs, double inequalities, real-life constraints, and word problems. The 2026 NCERT chapter Linear Inequalities explains how equations become inequalities when statements include “less than”, “greater than”, “at most”, or “at least”.

Key Takeaways

• Inequality Symbols: The symbols <, >, ≤, and ≥ compare real numbers or algebraic expressions.
• Negative Multiplication Rule: Multiplying or dividing by a negative number reverses the inequality sign.
• Solution Set: Values that make an inequality true form its solution set.
• Number Line Graph: Open circles represent < or >, while closed circles represent ≤ or ≥.

Important Questions Class 11 Maths Chapter 5 Structure 2026

Linear Inequalities Class 11 Chapter Overview

Linear inequalities compare quantities without forcing equality. They appear in budget limits, marks averages, temperature ranges, mixture percentages, and capacity conditions.

NCERT first explains inequalities through rice packets and registers. It then develops algebraic rules, number-line graphs, double inequalities, and word problems.

Q1. What Are Linear Inequalities Class 11?

Linear inequalities are inequalities where the variable has degree one.

They use <, >, ≤, or ≥ instead of =. The expressions may involve one or two variables.

Example: 5x − 3 < 7 is a linear inequality in one variable.

Q2. Why Are Important Questions Class 11 Maths Chapter 5 Useful For Exams?

Important Questions Class 11 Maths Chapter 5 are useful because sign rules and solution intervals appear repeatedly.

Students must solve inequalities over natural numbers, integers, and real numbers. The answer changes when the domain changes.

Example: x < 2 has different solution sets in integers and real numbers.

Q3. How Is An Inequality Different From An Equation?

An inequality compares two expressions, while an equation states that two expressions are equal.

An equation uses =, but an inequality uses symbols like <, >, ≤, or ≥.

Example: 40x + 20y = 120 is an equation, while 40x + 20y ≤ 120 is an inequality.

Class 11 Maths Chapter 5 Important Questions On Inequality Types

Inequalities can be numerical, literal, strict, slack, or double inequalities. The symbol used decides the type.

This part matters because ≤ includes equality, while < excludes equality.

Q4. What Is An Inequality In Class 11 Maths?

An inequality is a statement where two real numbers or algebraic expressions are related by <, >, ≤, or ≥.

It becomes true only for certain values of the variable. These values form the solution set.

Example: x ≥ 3 is true for x = 3, 4, and 5.

Q5. What Are Strict Inequalities Class 11?

Strict inequalities use < or > and do not include equality.

They show that one expression is strictly less than or greater than another expression.

Examples:
ax + b < 0
ax + by > c

Q6. What Are Slack Inequalities Class 11?

Slack inequalities use ≤ or ≥ and include equality.

They allow the two sides to become equal. NCERT uses “at most” and “at least” for such cases.

Examples:
ax + b ≤ 0
40x + 20y ≤ 120

Q7. What Is A Double Inequality Class 11?

A double inequality places one expression between two bounds.

It must satisfy both inequalities at the same time. Both ends must be solved together.

Example: −8 ≤ 5x − 3 < 7.

Linear Inequalities In One Variable Class 11 Questions

One-variable inequalities are solved like equations, but one sign rule changes everything. The inequality sign reverses when both sides are divided by a negative number.

NCERT solves such inequalities over natural numbers, integers, and real numbers.

Q8. Solve 30x < 200 When x Is A Natural Number.

The solution set is {1, 2, 3, 4, 5, 6}.

1. Given Data:
   30x < 200
2. Divide by 30:
   x < 200/30
   x < 20/3
3. Since 20/3 = 6.66..., natural numbers less than this are:
   1, 2, 3, 4, 5, 6

Final Result: {1, 2, 3, 4, 5, 6}

Q9. Solve 30x < 200 When x Is An Integer.

The solution set is {..., −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}.

1. Given Data:
   30x < 200
2. Divide by 30:
   x < 20/3
3. Integers less than 20/3 include all integers up to 6.

Final Result: x ∈ {..., −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}

Q10. Solve 5x − 3 < 3x + 1 For Real x.

The solution set is x < 2.

1. Given inequality:
   5x − 3 < 3x + 1
2. Add 3 to both sides:
   5x < 3x + 4
3. Subtract 3x from both sides:
   2x < 4
4. Divide by 2:
   x < 2

Final Result: x ∈ (−∞, 2)

Q11. Solve 4x + 3 < 6x + 7.

The solution set is x > −2.

1. Given inequality:
   4x + 3 < 6x + 7
2. Move x terms to one side:
   4x − 6x < 7 − 3
3. Simplify:
   −2x < 4
4. Divide by −2 and reverse the sign:
   x > −2

Final Result: x ∈ (−2, ∞)

Q12. Solve (5 − 2x)/3 ≤ x/6 − 5.

The solution set is x ≥ 8.

1. Given inequality:
   (5 − 2x)/3 ≤ x/6 − 5
2. Multiply by 6:
   2(5 − 2x) ≤ x − 30
3. Expand:
   10 − 4x ≤ x − 30
4. Move terms:
   −5x ≤ −40
5. Divide by −5 and reverse sign:
   x ≥ 8

Final Result: x ∈ [8, ∞)

Algebraic Solution Of Inequalities Class 11 Questions

Algebraic solution means isolating the variable while preserving the inequality logic. The sign reversal rule is the most important step.

Students should always check whether they divided by a positive or negative number.

Q13. What Rule Changes The Sign In Inequalities?

The inequality sign reverses when both sides are multiplied or divided by a negative number.

Adding or subtracting the same number does not change the sign. Multiplying by a positive number also keeps the sign unchanged.

Example: From −2x < 4, we get x > −2.

Q14. Solve 3(x − 1) ≤ 2(x − 3).

The solution set is x ≤ −3.

1. Given inequality:
   3(x − 1) ≤ 2(x − 3)
2. Expand both sides:
   3x − 3 ≤ 2x − 6
3. Subtract 2x:
   x − 3 ≤ −6
4. Add 3:
   x ≤ −3

Final Result: x ∈ (−∞, −3]

Q15. Solve 3(2 − x) ≥ 2(1 − x).

The solution set is x ≤ 4.

1. Given inequality:
   3(2 − x) ≥ 2(1 − x)
2. Expand:
   6 − 3x ≥ 2 − 2x
3. Add 3x to both sides:
   6 ≥ 2 + x
4. Subtract 2:
   4 ≥ x

Final Result: x ≤ 4

Q16. Solve 2(2x + 3) − 10 < 6(x − 2).

The solution set is x > 4.

1. Given inequality:
   2(2x + 3) − 10 < 6(x − 2)
2. Expand:
   4x + 6 − 10 < 6x − 12
3. Simplify:
   4x − 4 < 6x − 12
4. Move terms:
   8 < 2x
5. Divide by 2:
   x > 4

Final Result: x ∈ (4, ∞)

Q17. Solve 37 − (3x + 5) > 9x − 8(x − 3).

The solution set is x < 1.

1. Given inequality:
   37 − (3x + 5) > 9x − 8(x − 3)
2. Simplify both sides:
   32 − 3x > 9x − 8x + 24
3. Reduce right side:
   32 − 3x > x + 24
4. Move terms:
   8 > 4x
5. Divide by 4:
   x < 2

Final Result: x ∈ (−∞, 2)

Number Line Representation Of Inequalities Class 11 Questions

Number-line graphs show solution intervals visually. Open circles mark excluded endpoints, while filled circles mark included endpoints.

NCERT uses this representation for single inequalities and systems of inequalities.

Q18. How Do You Show x < 3 On A Number Line?

x < 3 is shown by an open circle at 3 and a dark line to the left.

The open circle means 3 is not included. The left side shows all numbers less than 3.

Final Result: Open circle at 3, shaded left

Q19. How Do You Show x ≥ 1 On A Number Line?

x ≥ 1 is shown by a filled circle at 1 and a dark line to the right.

The filled circle means 1 is included. The right side shows all numbers greater than 1.

Final Result: Closed circle at 1, shaded right

Q20. Solve 7x + 3 < 5x + 9 And State Its Graph.

The solution is x < 3.

1. Given inequality:
   7x + 3 < 5x + 9
2. Move terms:
   2x < 6
3. Divide by 2:
   x < 3
4. Graph:
   Open circle at 3 and shaded line to the left.

Final Result: x ∈ (−∞, 3)

Q21. Solve (3x − 4)/2 ≥ (x − 3)/4 And State Its Graph.

The solution is x ≥ 1.

1. Given inequality:
   (3x − 4)/2 ≥ (x − 3)/4
2. Multiply by 4:
   2(3x − 4) ≥ x − 3
3. Expand:
   6x − 8 ≥ x − 3
4. Move terms:
   5x ≥ 5
5. Divide by 5:
   x ≥ 1

Final Result: Closed circle at 1, shaded right

Double Inequalities Class 11 Solved Questions

Double inequalities require both boundary conditions at once. Students should perform the same operation on all three parts.

If division uses a negative number, both inequality signs reverse.

Q22. Solve −8 ≤ 5x − 3 < 7.

The solution set is −1 ≤ x < 2.

1. Given inequality:
   −8 ≤ 5x − 3 < 7
2. Add 3 to all parts:
   −5 ≤ 5x < 10
3. Divide by 5:
   −1 ≤ x < 2

Final Result: x ∈ [−1, 2)

Q23. Solve −5 ≤ (5 − 3x)/2 ≤ 8.

The solution set is −11/3 ≤ x ≤ 5.

1. Given inequality:
   −5 ≤ (5 − 3x)/2 ≤ 8
2. Multiply all parts by 2:
   −10 ≤ 5 − 3x ≤ 16
3. Subtract 5:
   −15 ≤ −3x ≤ 11
4. Divide by −3 and reverse signs:
   5 ≥ x ≥ −11/3
5. Write in increasing order:
   −11/3 ≤ x ≤ 5

Final Result: x ∈ [−11/3, 5]

Q24. Solve 2 ≤ 3x − 4 ≤ 5.

The solution set is 2 ≤ x ≤ 3.

1. Given inequality:
   2 ≤ 3x − 4 ≤ 5
2. Add 4 to all parts:
   6 ≤ 3x ≤ 9
3. Divide by 3:
   2 ≤ x ≤ 3

Final Result: x ∈ [2, 3]

Q25. Solve 6 ≤ −3(2x − 4) < 12.

The solution set is 0 < x ≤ 1.

1. Given inequality:
   6 ≤ −3(2x − 4) < 12
2. Expand centre:
   6 ≤ −6x + 12 < 12
3. Subtract 12:
   −6 ≤ −6x < 0
4. Divide by −6 and reverse signs:
   1 ≥ x > 0

Final Result: x ∈ (0, 1]

Solution Set Of Inequalities Class 11 Word Problems

Word problems convert conditions into inequalities. Words like “at least”, “at most”, “more than”, and “less than” decide the sign.

NCERT includes marks, consecutive numbers, board cutting, and triangle perimeter problems.

Q26. A Student Scored 62 And 48 In Two Tests. Find Minimum Marks Needed In Third Test For Average At Least 60.

The student must score at least 70 marks.

1. Let third test marks be:
   x
2. Given condition:
   Average at least 60
3. Form inequality:
   (62 + 48 + x)/3 ≥ 60
4. Solve:
   110 + x ≥ 180
   x ≥ 70

Final Result: Minimum marks = 70

Q27. Sunita Scored 87, 92, 94, And 95. Find Minimum Fifth Exam Marks For Grade A Average 90.

Sunita must score at least 82 marks.

1. Let fifth exam marks be:
   x
2. Grade A condition:
   Average ≥ 90
3. Form inequality:
   (87 + 92 + 94 + 95 + x)/5 ≥ 90
4. Solve:
   (368 + x)/5 ≥ 90
   368 + x ≥ 450
   x ≥ 82

Final Result: Minimum fifth exam marks = 82

Q28. Find Consecutive Odd Natural Number Pairs Greater Than 10 With Sum Less Than 40.

The pairs are (11, 13), (13, 15), (15, 17), and (17, 19).

1. Let smaller odd number be:
   x
2. Next consecutive odd number:
   x + 2
3. Conditions:
   x > 10
   x + (x + 2) < 40
4. Solve second condition:
   2x + 2 < 40
   x < 19
5. Odd values of x:
   11, 13, 15, 17

Final Result: (11, 13), (13, 15), (15, 17), (17, 19)

Q29. Find Consecutive Even Positive Integer Pairs Greater Than 5 With Sum Less Than 23.

The pairs are (6, 8), (8, 10), and (10, 12).

1. Let smaller even integer be:
   x
2. Next even integer:
   x + 2
3. Conditions:
   x > 5
   x + (x + 2) < 23
4. Solve:
   2x + 2 < 23
   2x < 21
   x < 10.5
5. Even values greater than 5:
   6, 8, 10

Final Result: (6, 8), (8, 10), (10, 12)

Q30. The Longest Side Of A Triangle Is 3 Times The Shortest Side. Third Side Is 2 cm Shorter Than Longest. Perimeter Is At Least 61 cm. Find Minimum Shortest Side.

The minimum shortest side is 9 cm.

1. Let shortest side be:
   x
2. Longest side:
   3x
3. Third side:
   3x − 2
4. Perimeter condition:
   x + 3x + (3x − 2) ≥ 61
5. Solve:
   7x − 2 ≥ 61
   7x ≥ 63
   x ≥ 9

Final Result: Minimum shortest side = 9 cm

Q31. A Board Is 91 cm Long. Pieces Are x, x + 3, And 2x. Third Piece Is At Least 5 cm Longer Than Second. Find Possible x.

The possible values satisfy 8 ≤ x ≤ 22.

1. Given pieces:
   x, x + 3, 2x
2. Total length condition:
   x + (x + 3) + 2x ≤ 91
3. Solve first condition:
   4x + 3 ≤ 91
   x ≤ 22
4. Third piece condition:
   2x ≥ (x + 3) + 5
5. Solve second condition:
   2x ≥ x + 8
   x ≥ 8

Final Result: 8 ≤ x ≤ 22

Linear Inequalities In Two Variables Class 11 Questions

Two-variable inequalities compare expressions like ax + by with a constant. NCERT introduces them through budget examples.

The chapter states such forms but mainly solves one-variable inequalities in detail.

Q32. What Are Linear Inequalities In Two Variables Class 11?

Linear inequalities in two variables use expressions of the form ax + by < c, ax + by > c, ax + by ≤ c, or ax + by ≥ c.

Here, a ≠ 0 and b ≠ 0. The variables are usually x and y.

Example: 40x + 20y ≤ 120.

Q33. Write The Inequality For ₹120 Spent On Registers And Pens. Registers Cost ₹40 And Pens Cost ₹20.

The required inequality is 40x + 20y ≤ 120.

1. Let number of registers be:
   x
2. Let number of pens be:
   y
3. Total cost:
   40x + 20y
4. Since amount can be at most ₹120:
   40x + 20y ≤ 120

Final Result: 40x + 20y ≤ 120

Q34. Why Is 40x + 20y ≤ 120 A Slack Inequality?

It is a slack inequality because it uses ≤.

The total amount can be less than ₹120 or exactly ₹120. Equality is included in the condition.

Example: 40x + 20y = 120 is allowed.

Word Problems On Linear Inequalities Class 11 With Temperature And Mixtures

Temperature and mixture questions use double inequalities. The range must remain between two boundary values.

NCERT uses Celsius-Fahrenheit conversion and acid concentration as applied inequality examples.

Q35. A Solution Must Be Between 30°C And 35°C. Find The Fahrenheit Range If C = 5/9(F − 32).

The Fahrenheit range is 86 < F < 95.

1. Given condition:
   30 < C < 35
2. Substitute formula:
   30 < 5/9(F − 32) < 35
3. Multiply by 9/5:
   54 < F − 32 < 63
4. Add 32:
   86 < F < 95

Final Result: 86°F < F < 95°F

Q36. Convert 68°F To 77°F Into Celsius Range Using F = 9/5 C + 32.

The Celsius range is 20°C ≤ C ≤ 25°C.

1. Given condition:
   68 ≤ F ≤ 77
2. Substitute formula:
   68 ≤ 9/5 C + 32 ≤ 77
3. Subtract 32:
   36 ≤ 9/5 C ≤ 45
4. Multiply by 5/9:
   20 ≤ C ≤ 25

Final Result: 20°C ≤ C ≤ 25°C

Q37. 600 Litres Of 12% Acid Solution Is Mixed With x Litres Of 30% Acid Solution. Result Must Be More Than 15% And Less Than 18%. Find x.

The required range is 120 < x < 300.

1. Total mixture:
   x + 600
2. Acid condition:
   30%x + 12% of 600 > 15%(x + 600)
   30%x + 12% of 600 < 18%(x + 600)
3. First inequality:
   30x + 7200 > 15x + 9000
   15x > 1800
   x > 120
4. Second inequality:
   30x + 7200 < 18x + 10800
   12x < 3600
   x < 300

Final Result: 120 < x < 300 litres

Q38. 640 Litres Of 8% Boric Acid Is Diluted With x Litres Of 2% Solution. Result Must Be More Than 4% And Less Than 6%. Find x.

The required range is 320 < x < 1280.

1. Total mixture:
   640 + x
2. Acid amount:
   8% of 640 + 2% of x
3. More than 4% condition:
   0.08(640) + 0.02x > 0.04(640 + x)
4. Solve:
   51.2 + 0.02x > 25.6 + 0.04x
   25.6 > 0.02x
   x < 1280
5. Less than 6% condition:
   0.08(640) + 0.02x < 0.06(640 + x)
6. Solve:
   51.2 + 0.02x < 38.4 + 0.06x
   12.8 < 0.04x
   x > 320

Final Result: 320 < x < 1280 litres

Q39. Add Water To 1125 Litres Of 45% Acid So Result Has More Than 25% But Less Than 30% Acid. Find Water Range.

The water range is 562.5 < x < 900 litres.

1. Let water added be:
   x litres
2. Acid amount remains:
   45% of 1125 = 506.25
3. Total mixture:
   1125 + x
4. More than 25% condition:
   506.25 > 0.25(1125 + x)
   506.25 > 281.25 + 0.25x
   x < 900
5. Less than 30% condition:
   506.25 < 0.30(1125 + x)
   506.25 < 337.5 + 0.30x
   x > 562.5

Final Result: 562.5 < x < 900 litres

NCERT Class 11 Maths Chapter 5 Questions On Systems Of Inequalities

A system of inequalities needs values common to all conditions. The answer is the intersection of their solution sets.

NCERT shows this using number-line graphs for two inequalities solved together.

Q40. Solve The System 3x − 7 < 5 + x And 11 − 5x ≤ 1.

The solution set is 2 ≤ x < 6.

1. Solve first inequality:
   3x − 7 < 5 + x
   2x < 12
   x < 6
2. Solve second inequality:
   11 − 5x ≤ 1
   −5x ≤ −10
3. Divide by −5 and reverse sign:
   x ≥ 2
4. Combine both:
   2 ≤ x < 6

Final Result: x ∈ [2, 6)

Q41. Solve 5x + 1 > −24 And 5x − 1 < 24.

The solution set is −5 < x < 5.

1. First inequality:
   5x + 1 > −24
2. Solve:
   5x > −25
   x > −5
3. Second inequality:
   5x − 1 < 24
4. Solve:
   5x < 25
   x < 5

Final Result: −5 < x < 5

Q42. Solve 2(x − 1) < x + 5 And 3(x + 2) > 2 − x.

The solution set is −1 < x < 7.

1. First inequality:
   2(x − 1) < x + 5
2. Solve:
   2x − 2 < x + 5
   x < 7
3. Second inequality:
   3(x + 2) > 2 − x
4. Solve:
   3x + 6 > 2 − x
   4x > −4
   x > −1

Final Result: −1 < x < 7

Linear Inequalities Questions With Answers Class 11 Board Pattern

Board pattern questions often mix algebra with interpretation. The best answer states the variable, inequality, working, and final interval.

This section covers NCERT-style cases that combine formula use with real restrictions.

Q43. IQ Is Given By IQ = MA/CA × 100. If 80 ≤ IQ ≤ 140 For 12-Year-Old Children, Find Mental Age Range.

The mental age range is 9.6 ≤ MA ≤ 16.8 years.

1. Given Data:
   CA = 12
   80 ≤ IQ ≤ 140
2. Formula Used:
   IQ = MA/CA × 100
3. Substitute:
   80 ≤ MA/12 × 100 ≤ 140
4. Multiply by 12/100:
   9.6 ≤ MA ≤ 16.8

Final Result: 9.6 years ≤ MA ≤ 16.8 years

Q44. Find All Pairs Of Consecutive Odd Positive Integers Smaller Than 10 Whose Sum Is More Than 11.

The pairs are (5, 7) and (7, 9).

1. Let smaller odd integer be:
   x
2. Next odd integer:
   x + 2
3. Conditions:
   x < 10
   x + (x + 2) > 11
4. Solve:
   2x + 2 > 11
   2x > 9
   x > 4.5
5. Odd positive values less than 10 and greater than 4.5:
   5, 7

Final Result: (5, 7), (7, 9)

Q45. Why Does Solving −5x ≤ −40 Give x ≥ 8?

The sign reverses because both sides are divided by −5.

1. Given inequality:
   −5x ≤ −40
2. Divide by −5:
   x ≥ 8
3. The sign changes from ≤ to ≥.

Final Result: x ≥ 8

Important Questions Class 11 Maths Chapter-Wise