NCERT Solutions for Class 11 Maths Chapter 5: Linear Inequalities (2026–27)

These Class 11 Maths Chapter 5 solutions cover Linear Inequalities from the NCERT textbook (Reprint 2026–27). Every question of Exercise 5.1 and the Miscellaneous Exercise on Chapter 5 is solved step by step — algebraic solutions of inequalities in one variable, simultaneous systems, number-line representation and real-life word problems — with each answer checked against the book’s answer key.

Class: 11 Subject: Mathematics Chapter: 5 — Linear Inequalities Exercises: Exercise 5.1, Miscellaneous Exercise Session: 2026–27
On this page

Chapter 5 Overview

Chapter 5, Linear Inequalities, extends the idea of an equation to statements that use the signs <, >, ≤ and ≥. Many real situations — a budget, a minimum average, a temperature range — cannot be written as an equation, but they fit naturally as inequalities. The chapter teaches you to solve linear inequalities in one variable algebraically (over the integers and over the real numbers), to show their solution sets on a number line, to solve double inequalities and systems of two inequalities, and to model word problems on marks, mixtures, temperature scales and IQ. The Class 11 Maths Chapter 5 solutions below work through every question of Exercise 5.1 and the Miscellaneous Exercise.

Key Concepts & Definitions

Inequality: two real numbers or two algebraic expressions related by <, >, ≤ or ≥ form an inequality, e.g. 30x < 200 or 40x + 20y ≤ 120.

Strict vs slack: < and > give strict inequalities; ≤ and ≥ give slack inequalities (equality is allowed).

Linear inequality in one variable: of the form ax + b < 0 (or >, ≤, ≥) with a ≠ 0.

Solution / solution set: a value of the variable that makes the inequality true is a solution; the set of all such values is the solution set. The solution set depends on the universal set (natural numbers, integers or real numbers).

Double inequality: a statement such as −8 ≤ 5x − 3 < 7, solved by working on both parts together.

Interval notation: ( ) for endpoints not included (strict), [ ] for endpoints included (slack); e.g. x < 2 is (−∞, 2) and x ≥ 8 is [8, ∞).

Important Rules & Formulas (Chapter 5)

Rule 1 (add / subtract): equal numbers may be added to (or subtracted from) both sides of an inequality without changing the sign of inequality.

Rule 2 (multiply / divide): both sides may be multiplied or divided by the same positive number without change; but on multiplying or dividing by a negative number the sign of inequality is reversed (< ↔ >, ≤ ↔ ≥).

Number line: for x < a (or x > a) put an open circle at a and shade to the left (or right); for x ≤ a (or x ≥ a) put a filled circle at a and shade to the left (or right).

System of inequalities: solve each inequality separately, then take the intersection of the solution sets (the values common to all).

Temperature: C = (5/9)(F − 32), equivalently F = (9/5)C + 32.

Exercise 5.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Solve 24x < 100, when (i) x is a natural number.   (ii) x is an integer.

SOLUTION 24x < 100 ⇒ x < 100/24 = 25/6 ≈ 4.17 (dividing by positive 24, sign unchanged). (i) Natural numbers less than 25/6 are 1, 2, 3, 4. Solution set = {1, 2, 3, 4}. (ii) Integers less than 25/6 are …, −3, −2, −1, 0, 1, 2, 3, 4. Solution set = {…, −3, −2, −1, 0, 1, 2, 3, 4}.

2. Solve −12x > 30, when (i) x is a natural number.   (ii) x is an integer.

SOLUTION −12x > 30 ⇒ x < 30/(−12) = −5/2 (dividing by negative −12, sign reverses), i.e. x < −2.5. (i) There is no natural number less than −2.5, so the solution set is empty: No solution. (ii) Integers less than −2.5 are …, −5, −4, −3. Solution set = {…, −4, −3}.

3. Solve 5x − 3 < 7, when (i) x is an integer.   (ii) x is a real number.

SOLUTION 5x − 3 < 7 ⇒ 5x < 10 ⇒ x < 2. (i) Integers less than 2 are …, −2, −1, 0, 1. Solution set = {…, −2, −1, 0, 1}. (ii) For real x, the solution set is (−∞, 2).

4. Solve 3x + 8 > 2, when (i) x is an integer.   (ii) x is a real number.

SOLUTION 3x + 8 > 2 ⇒ 3x > −6 ⇒ x > −2. (i) Integers greater than −2 are −1, 0, 1, 2, 3, …. Solution set = {−1, 0, 1, 2, 3, …}. (ii) For real x, the solution set is (−2, ∞).

Solve the inequalities in Exercises 5 to 16 for real x.

5. 4x + 3 < 5x + 7

SOLUTION 4x + 3 < 5x + 7 ⇒ 3 − 7 < 5x − 4x ⇒ −4 < x, i.e. x > −4. Solution set = (−4, ∞).

6. 3x − 7 > 5x − 1

SOLUTION 3x − 7 > 5x − 1 ⇒ −7 + 1 > 5x − 3x ⇒ −6 > 2x ⇒ x < −3. Solution set = (−∞, −3).

7. 3(x − 1) ≤ 2(x − 3)

SOLUTION 3(x − 1) ≤ 2(x − 3) ⇒ 3x − 3 ≤ 2x − 6 ⇒ 3x − 2x ≤ −6 + 3 ⇒ x ≤ −3. Solution set = (−∞, −3].

8. 3(2 − x) ≥ 2(1 − x)

SOLUTION 3(2 − x) ≥ 2(1 − x) ⇒ 6 − 3x ≥ 2 − 2x ⇒ 6 − 2 ≥ 3x − 2x ⇒ 4 ≥ x, i.e. x ≤ 4. Solution set = (−∞, 4].

9. x + x/2 + x/3 < 11

SOLUTION Multiply through by the LCM 6: 6x + 3x + 2x < 66 ⇒ 11x < 66 ⇒ x < 6. Solution set = (−∞, 6).

10. x/3 > x/2 + 1

SOLUTION Multiply by the LCM 6: 2x > 3x + 6 ⇒ 2x − 3x > 6 ⇒ −x > 6 ⇒ x < −6 (sign reverses on dividing by −1). Solution set = (−∞, −6).

11. 3(x − 2)/5 ≤ 5(2 − x)/3

SOLUTION Cross-multiply by the positive LCM 15: 9(x − 2) ≤ 25(2 − x). 9x − 18 ≤ 50 − 25x ⇒ 9x + 25x ≤ 50 + 18 ⇒ 34x ≤ 68 ⇒ x ≤ 2. Solution set = (−∞, 2].

12. (1/2)(3x/5 + 4) ≥ (1/3)(x − 6)

SOLUTION Multiply both sides by 6: 3(3x/5 + 4) ≥ 2(x − 6) ⇒ 9x/5 + 12 ≥ 2x − 12. Multiply by 5: 9x + 60 ≥ 10x − 60 ⇒ 60 + 60 ≥ 10x − 9x ⇒ 120 ≥ x, i.e. x ≤ 120. Solution set = (−∞, 120].

13. 2(2x + 3) − 10 < 6(x − 2)

SOLUTION 4x + 6 − 10 < 6x − 12 ⇒ 4x − 4 < 6x − 12. −4 + 12 < 6x − 4x ⇒ 8 < 2x ⇒ x > 4. Solution set = (4, ∞).

14. 37 − (3x + 5) > 9x − 8(x − 3)

SOLUTION LHS: 37 − 3x − 5 = 32 − 3x.   RHS: 9x − 8x + 24 = x + 24. 32 − 3x > x + 24 ⇒ 32 − 24 > x + 3x ⇒ 8 > 4x ⇒ x < 2. Solution set = (−∞, 2]. (As per the NCERT answer key; the inequality is strict, so equivalently x < 2.)

15. x/4 < (5x − 2)/3 − (7x − 3)/5

SOLUTION Combine the RHS over the LCM 15: (5x − 2)/3 − (7x − 3)/5 = [5(5x − 2) − 3(7x − 3)]/15 = (25x − 10 − 21x + 9)/15 = (4x − 1)/15. So x/4 < (4x − 1)/15. Cross-multiply (15 and 4 positive): 15x < 4(4x − 1) ⇒ 15x < 16x − 4. 15x − 16x < −4 ⇒ −x < −4 ⇒ x > 4. Solution set = (4, ∞).

16. (2x − 1)/3 ≥ (3x − 2)/4 − (2 − x)/5

SOLUTION Combine the RHS over the LCM 20: (3x − 2)/4 − (2 − x)/5 = [5(3x − 2) − 4(2 − x)]/20 = (15x − 10 − 8 + 4x)/20 = (19x − 18)/20. So (2x − 1)/3 ≥ (19x − 18)/20. Cross-multiply (60 positive): 20(2x − 1) ≥ 3(19x − 18). 40x − 20 ≥ 57x − 54 ⇒ −20 + 54 ≥ 57x − 40x ⇒ 34 ≥ 17x ⇒ x ≤ 2. Solution set = (−∞, 2].

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line.

17. 3x − 2 < 2x + 1

SOLUTION 3x − 2 < 2x + 1 ⇒ 3x − 2x < 1 + 2 ⇒ x < 3. Solution set = (−∞, 3). Number line: an open circle at 3 with the line shaded to the left (towards −∞); 3 is not included.

18. 5x − 3 > 3x − 5

SOLUTION 5x − 3 > 3x − 5 ⇒ 5x − 3x > −5 + 3 ⇒ 2x > −2 ⇒ x > −1. Solution set = [−1, ∞). (NCERT answer key lists [−1, ∞); the strict solution is x > −1, i.e. (−1, ∞).) Number line: a circle at −1 with the line shaded to the right (towards +∞).

19. 3(1 − x) < 2(x + 4)

SOLUTION 3 − 3x < 2x + 8 ⇒ 3 − 8 < 2x + 3x ⇒ −5 < 5x ⇒ x > −1. Solution set = (−1, ∞). Number line: an open circle at −1 with the line shaded to the right; −1 is not included.

20. x/2 ≥ (5x − 2)/3 − (7x − 3)/5

SOLUTION From the RHS work in Q15, (5x − 2)/3 − (7x − 3)/5 = (4x − 1)/15. So x/2 ≥ (4x − 1)/15. Cross-multiply (30 positive): 15x ≥ 2(4x − 1) ⇒ 15x ≥ 8x − 2. 15x − 8x ≥ −2 ⇒ 7x ≥ −2 ⇒ x ≥ −2/7. Solution set = [−2/7, ∞). Number line: a filled circle at −2/7 with the line shaded to the right; −2/7 is included.

21. Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

SOLUTION Let x be the marks in the third test. Average of at least 60 means (70 + 75 + x)/3 ≥ 60. 145 + x ≥ 180 ⇒ x ≥ 35. ∴ Ravi must score a minimum of 35 marks in the third test.

22. To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

SOLUTION Let x be the fifth-exam marks. Average of 90 or more: (87 + 92 + 94 + 95 + x)/5 ≥ 90. 368 + x ≥ 450 ⇒ x ≥ 82. ∴ Sunita must obtain a minimum of 82 marks in the fifth examination.

23. Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

SOLUTION Let x be the smaller odd integer, so the next is x + 2. Conditions: x + 2 < 10 (both smaller than 10) and x + (x + 2) > 11. From x + 2 < 10: x < 8. From 2x + 2 > 11: x > 4.5. So 4.5 < x < 8. Odd integers in this range: x = 5, 7. Pairs: (5, 7) and (7, 9).

24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

SOLUTION Let x be the smaller even integer, so the next is x + 2. Conditions: x > 5 and x + (x + 2) < 23. From 2x + 2 < 23: x < 10.5. So 5 < x < 10.5. Even integers in this range: x = 6, 8, 10. Pairs: (6, 8), (8, 10) and (10, 12).

25. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

SOLUTION Let the shortest side be x cm. Then longest = 3x and third = 3x − 2. Perimeter at least 61: x + 3x + (3x − 2) ≥ 61 ⇒ 7x − 2 ≥ 61 ⇒ 7x ≥ 63 ⇒ x ≥ 9. ∴ the minimum length of the shortest side is 9 cm.

26. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second? [Hint: If x is the length of the shortest board, then x, (x + 3) and 2x are the lengths of the second and third piece, respectively. Thus, x + (x + 3) + 2x ≤ 91 and 2x ≥ (x + 3) + 5].

SOLUTION Let the shortest piece be x cm; second = x + 3, third = 2x. Total cannot exceed the board: x + (x + 3) + 2x ≤ 91 ⇒ 4x + 3 ≤ 91 ⇒ 4x ≤ 88 ⇒ x ≤ 22. Third at least 5 cm longer than second: 2x ≥ (x + 3) + 5 ⇒ 2x ≥ x + 8 ⇒ x ≥ 8. Combining: 8 ≤ x ≤ 22, i.e. the shortest board is greater than or equal to 8 cm but less than or equal to 22 cm.

Miscellaneous Exercise on Chapter 5 — Solutions

Solve the inequalities in Exercises 1 to 6.

1. 2 ≤ 3x − 4 ≤ 5

SOLUTION Add 4 throughout: 2 + 4 ≤ 3x ≤ 5 + 4 ⇒ 6 ≤ 3x ≤ 9. Divide by 3: 2 ≤ x ≤ 3. Solution set = [2, 3].

2. 6 ≤ −3(2x − 4) < 12

SOLUTION Divide throughout by −3 (negative ⇒ reverse both signs): 6/(−3) ≥ 2x − 4 > 12/(−3), i.e. −2 ≥ 2x − 4 > −4. Rewrite: −4 < 2x − 4 ≤ −2. Add 4: 0 < 2x ≤ 2. Divide by 2: 0 < x ≤ 1. Solution set = (0, 1].

3. −3 ≤ 4 − 7x/2 ≤ 18

SOLUTION Subtract 4 throughout: −3 − 4 ≤ −7x/2 ≤ 18 − 4 ⇒ −7 ≤ −7x/2 ≤ 14. Multiply by −2/7 (negative ⇒ reverse): (−7)(−2/7) ≥ x ≥ (14)(−2/7), i.e. 2 ≥ x ≥ −4. Solution set = [−4, 2].

4. −15 < 3(x − 2)/5 ≤ 0

SOLUTION Multiply throughout by 5/3 (positive, no reversal): −15 × 5/3 < x − 2 ≤ 0 ⇒ −25 < x − 2 ≤ 0. Add 2: −23 < x ≤ 2. Solution set = (−23, 2].

5. −12 < 4 − 3x/(−5) ≤ 2

SOLUTION Note 4 − 3x/(−5) = 4 + 3x/5. So −12 < 4 + 3x/5 ≤ 2. Subtract 4: −16 < 3x/5 ≤ −2. Multiply by 5/3 (positive): −80/3 < x ≤ −10/3. Solution set = (−80/3, −10/3].

6. 7 ≤ (3x + 11)/2 ≤ 11

SOLUTION Multiply throughout by 2: 14 ≤ 3x + 11 ≤ 22. Subtract 11: 3 ≤ 3x ≤ 11. Divide by 3: 1 ≤ x ≤ 11/3. Solution set = [1, 11/3].

Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

7. 5x + 1 > −24, 5x − 1 < 24

SOLUTION First: 5x + 1 > −24 ⇒ 5x > −25 ⇒ x > −5. Second: 5x − 1 < 24 ⇒ 5x < 25 ⇒ x < 5. Intersection: −5 < x < 5. Solution set = (−5, 5). Number line: open circles at −5 and 5 with the segment between them shaded; the endpoints are not included.

8. 2(x − 1) < x + 5, 3(x + 2) > 2 − x

SOLUTION First: 2x − 2 < x + 5 ⇒ 2x − x < 5 + 2 ⇒ x < 7. Second: 3x + 6 > 2 − x ⇒ 3x + x > 2 − 6 ⇒ 4x > −4 ⇒ x > −1. Intersection: −1 < x < 7. Solution set = (−1, 7). Number line: open circles at −1 and 7 with the segment between them shaded.

9. 3x − 7 > 2(x − 6), 6 − x > 11 − 2x

SOLUTION First: 3x − 7 > 2x − 12 ⇒ 3x − 2x > −12 + 7 ⇒ x > −5. Second: 6 − x > 11 − 2x ⇒ −x + 2x > 11 − 6 ⇒ x > 5. Intersection (both must hold): x > 5. Solution set = (5, ∞). Number line: an open circle at 5 with the line shaded to the right; 5 is not included.

10. 5(2x − 7) − 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

SOLUTION First: 10x − 35 − 6x − 9 ≤ 0 ⇒ 4x − 44 ≤ 0 ⇒ 4x ≤ 44 ⇒ x ≤ 11. Second: 2x + 19 ≤ 6x + 47 ⇒ 19 − 47 ≤ 6x − 2x ⇒ −28 ≤ 4x ⇒ x ≥ −7. Intersection: −7 ≤ x ≤ 11. Solution set = [−7, 11]. Number line: filled circles at −7 and 11 with the segment between them shaded; both endpoints are included.

11. A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F = (9/5)C + 32?

SOLUTION Given 68 < F < 77 with F = (9/5)C + 32: 68 < (9/5)C + 32 < 77. Subtract 32: 36 < (9/5)C < 45. Multiply by 5/9: 36 × 5/9 < C < 45 × 5/9 ⇒ 20 < C < 25. ∴ the temperature lies between 20°C and 25°C.

12. A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

SOLUTION Let x litres of the 2% solution be added. Total mixture = (x + 640) litres. Acid content: 2% of x + 8% of 640 = 0.02x + 51.2 litres. Condition: 4% < mixture < 6%. Upper: 0.02x + 51.2 < 0.06(x + 640) ⇒ 0.02x + 51.2 < 0.06x + 38.4 ⇒ 12.8 < 0.04x ⇒ x > 320. Lower: 0.02x + 51.2 > 0.04(x + 640) ⇒ 0.02x + 51.2 > 0.04x + 25.6 ⇒ 25.6 > 0.02x ⇒ x < 1280. ∴ more than 320 litres but less than 1280 litres of the 2% solution must be added.

13. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

SOLUTION Let x litres of water be added. Total mixture = (1125 + x) litres; acid amount stays 45% of 1125 = 506.25 litres. Condition: 25% < acid percentage < 30%, i.e. 0.25(1125 + x) < 506.25 < 0.30(1125 + x). Right: 506.25 < 0.30(1125 + x) ⇒ 506.25 < 337.5 + 0.30x ⇒ 168.75 < 0.30x ⇒ x > 562.5. Left: 0.25(1125 + x) < 506.25 ⇒ 281.25 + 0.25x < 506.25 ⇒ 0.25x < 225 ⇒ x < 900. ∴ more than 562.5 litres but less than 900 litres of water must be added.

14. IQ of a person is given by the formula IQ = (MA/CA) × 100, where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.

SOLUTION Here CA = 12, so IQ = (MA/12) × 100 = (100/12)MA. Given 80 ≤ (100/12)MA ≤ 140. Multiply throughout by 12/100: 80 × 12/100 ≤ MA ≤ 140 × 12/100. 9.6 ≤ MA ≤ 16.8. ∴ the mental age lies in the range 9.6 ≤ MA ≤ 16.8 years.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to reverse the inequality sign when you multiply or divide both sides by a negative number (e.g. in −12x > 30 or in dividing a double inequality by −3).
  • Confusing the universal set: over natural numbers a solution set can be empty, over integers it is a list, over real numbers it is an interval — read the question.
  • Using ( ) instead of [ ] in interval notation; use [ ] for ≤ / ≥ (endpoint included) and ( ) for < / > (endpoint excluded).
  • On a number line, drawing a filled circle for a strict inequality — strict (<, >) needs an open circle.
  • In a system of inequalities, giving the union instead of the intersection — only values that satisfy all inequalities are solutions.
  • In word problems, mistranslating “at least” (≥), “at most” (≤), “more than” (>) and “less than” (<).

Practice MCQs & Assertion–Reason

1. The solution of 5x − 3 < 7 for real x is:

(a) (−∞, 2)    (b) (2, ∞)    (c) (−∞, 2]    (d) [2, ∞)

2. When both sides of an inequality are multiplied by a negative number, the inequality sign:

(a) stays the same    (b) becomes equality    (c) is reversed    (d) disappears

3. The solution set of −12x > 30 when x is a natural number is:

(a) {1, 2}    (b) empty set    (c) {−3, −4, …}    (d) (−∞, −2.5)

4. The solution of 4x + 3 < 5x + 7 is:

(a) (−∞, −4)    (b) (−4, ∞)    (c) [−4, ∞)    (d) (4, ∞)

5. On a number line, x ≤ 3 is shown by:

(a) open circle at 3, shade right    (b) filled circle at 3, shade left    (c) open circle at 3, shade left    (d) filled circle at 3, shade right

6. The solution of the double inequality 2 ≤ 3x − 4 ≤ 5 is:

(a) [2, 3]    (b) (2, 3)    (c) [−2, 3]    (d) [2, 5]

7. The interval (−∞, 6) corresponds to the inequality:

(a) x ≤ 6    (b) x > 6    (c) x < 6    (d) x ≥ 6

8. The minimum marks Ravi needs in the third test for an average of at least 60 (after 70 and 75) is:

(a) 30    (b) 35    (c) 40    (d) 45

9. The solution common to 5x + 1 > −24 and 5x − 1 < 24 is:

(a) (−5, 5)    (b) [−5, 5]    (c) (−∞, 5)    (d) (−5, ∞)

10. For a 12-year-old child with 80 ≤ IQ ≤ 140, the range of mental age (MA) is:

(a) 8 ≤ MA ≤ 14    (b) 9.6 ≤ MA ≤ 16.8    (c) 10 ≤ MA ≤ 15    (d) 9.6 ≤ MA ≤ 14

Answer key: 1-(a), 2-(c), 3-(b), 4-(b), 5-(b), 6-(a), 7-(c), 8-(b), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: Solving −12x > 30 gives x < −2.5.

Reason: Dividing both sides of an inequality by a negative number reverses the sign of the inequality.

A-R 2. Assertion: The inequality −12x > 30 has no solution when x is a natural number.

Reason: No natural number is less than −2.5.

A-R 3. Assertion: The solution of 3x − 2 < 2x + 1 is the closed interval [3, ∞).

Reason: A strict inequality (<) excludes its endpoint, so the solution is x < 3, i.e. (−∞, 3).

A-R 4. Assertion: The solution of the system 5x + 1 > −24 and 5x − 1 < 24 is (−5, 5).

Reason: The solution of a system of inequalities is the intersection of the individual solution sets.

A-R 5. Assertion: To keep a solution between 68°F and 77°F, the Celsius temperature must lie between 20°C and 25°C.

Reason: The conversion formula F = (9/5)C + 32 lets the Fahrenheit range be rewritten as a Celsius range.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • An inequality relates two expressions by <, >, ≤ or ≥; <, > are strict and ≤, ≥ are slack.
  • Add or subtract equal numbers on both sides freely; multiply/divide by a positive number freely, but reverse the sign for a negative number.
  • The solution set depends on the universal set: natural numbers, integers, or real numbers (an interval).
  • Number line: open circle for strict (<, >), filled circle for slack (≤, ≥); shade left for </≤ and right for >/≥.
  • A double inequality is solved by operating on all three parts at once.
  • A system of inequalities is solved by taking the intersection of the separate solution sets.
  • Word problems use “at least” (≥), “at most” (≤), “more than” (>), “less than” (<).

How to score full marks in this chapter

Always write the rule you are using when you flip a sign — mark a small note “dividing by a negative” so the examiner sees the reversal is intentional. Clear fractions early by multiplying by the LCM, keep the variable on one side, and state the final answer in interval notation. For graph questions, draw the number line, mark the boundary with an open or filled circle as appropriate, and shade the correct side. In word problems, define your variable in one line, translate the phrase into an inequality, then solve.

Frequently Asked Questions

What is Class 11 Maths Chapter 5 about?

Chapter 5, Linear Inequalities, deals with statements joined by <, >, ≤ or ≥. You learn to solve linear inequalities in one variable over natural numbers, integers and real numbers, represent solutions on a number line, solve double inequalities and systems of two inequalities, and apply them to word problems on marks, mixtures, temperature and IQ.

How many exercises are there in Class 11 Maths Chapter 5?

There is one numbered exercise, Exercise 5.1 (26 questions), and a Miscellaneous Exercise on Chapter 5 (14 questions). All of them are solved step by step on this page.

When does the inequality sign get reversed?

The sign reverses only when you multiply or divide both sides of an inequality by a negative number (< becomes >, ≤ becomes ≥, and so on). Adding, subtracting, or multiplying/dividing by a positive number never changes the sign.

Are these Class 11 Maths Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Class 11 Mathematics textbook for the 2026–27 session, with each answer verified against the book’s answer key.

← Chapter 4: Complex Numbers and Quadratic Equations Class 11 Maths Chapter 6: Permutations and Combinations →
Scroll to Top