Class 11 Maths NCERT Solutions Chapter 5 Ex 5.1 – Linear Inequalities in One Variable | Boundless Maths
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Ex 5.1 Class 11 Maths NCERT Solutions · Chapter 5

Class 11 Maths NCERT Solutions Chapter 5 Ex 5.1 – Linear Inequalities in One Variable

Free, step-by-step Class 11 Maths NCERT Solutions for Chapter 5 Ex 5.1 — all 26 questions solved, covering algebraic solutions of linear inequalities, number-line graphs, and real-world word problems.

Solving an inequality works almost exactly like solving an equation — add, subtract, multiply, or divide both sides — with one rule that changes everything: multiplying or dividing by a negative number flips the inequality sign. Questions 1–4 restrict the answer to natural numbers or integers before moving to real numbers, Questions 5–16 are pure algebraic practice, Questions 17–20 add a number-line graph on top of the algebra, and Questions 21–26 translate real situations — test averages, consecutive integers, triangle sides, cut lengths of a board — into an inequality and solve it.

26Questions
Easy–MediumDifficulty Mix
2026-27CBSE Syllabus

Class 11 Maths NCERT Solutions Chapter 5 Ex 5.1 — All 26 Questions

1

Solve 24x\lt100, when:
(i) x is a natural number.
(ii) x is an integer.

Easy +
Solution

Given 24x\lt100. Dividing both sides by 24 (positive, sign unchanged):

x\lt\dfrac{100}{24}=\dfrac{25}{6}

(i) The natural numbers less than \tfrac{25}{6}\approx4.17 are 1, 2, 3, 4.

Solution set: {1, 2, 3, 4}

(ii) The integers less than \tfrac{25}{6} continue indefinitely in the negative direction.

Solution set: {..., −3, −2, −1, 0, 1, 2, 3, 4}
2

Solve -12x\gt30, when:
(i) x is a natural number.
(ii) x is an integer.

Easy +
Solution

Given -12x\gt30. Dividing both sides by −12 (negative, so the sign flips):

x\lt\dfrac{30}{-12}=-\dfrac{5}{2}

(i) Natural numbers are always positive, so none can be less than -\tfrac{5}{2}.

Solution set: φ (no natural number satisfies this)

(ii) The integers less than -\tfrac{5}{2}=-2.5 are −3, −4, −5, ... continuing indefinitely.

Solution set: {..., −5, −4, −3}
3

Solve 5x-3\lt7, when:
(i) x is an integer.
(ii) x is a real number.

Easy +
Solution

Given 5x-3\lt7.

5x\lt10\ \Rightarrow\ x\lt2

(i) The integers less than 2 are ..., −2, −1, 0, 1.

Solution set: {..., −2, −1, 0, 1}

(ii) Every real number less than 2 is a solution.

Solution set: (−∞, 2)
4

Solve 3x+8\gt2, when:
(i) x is an integer.
(ii) x is a real number.

Easy +
Solution

Given 3x+8\gt2.

3x\gt-6\ \Rightarrow\ x\gt-2

(i) The integers greater than −2 are −1, 0, 1, 2, ... continuing indefinitely.

Solution set: {−1, 0, 1, 2, ...}

(ii) Every real number greater than −2 is a solution.

Solution set: (−2, ∞)
5

Solve 4x+3\lt5x+7 for real x.

Easy +
Solution

4x+3\lt5x+7

4x-5x\lt7-3

-x\lt4\ \Rightarrow\ x\gt-4   (dividing by −1 flips the sign)

Solution set: (−4, ∞)
6

Solve 3x-7\gt5x-1 for real x.

Easy +
Solution

3x-7\gt5x-1

3x-5x\gt-1+7

-2x\gt6\ \Rightarrow\ x\lt-3

Solution set: (−∞, −3)
7

Solve 3(x-1)\le2(x-3) for real x.

Easy +
Solution

3(x-1)\le2(x-3)

3x-3\le2x-6

3x-2x\le-6+3\ \Rightarrow\ x\le-3

Solution set: (−∞, −3]
8

Solve 3(2-x)\ge2(1-x) for real x.

Easy +
Solution

3(2-x)\ge2(1-x)

6-3x\ge2-2x

6-2\ge-2x+3x\ \Rightarrow\ 4\ge x

Solution set: (−∞, 4]
9

Solve x+\dfrac{x}{2}+\dfrac{x}{3}\lt11 for real x.

Easy +
Solution

Multiplying every term by 6 (the LCM of 1, 2, 3) to clear fractions:

6x+3x+2x\lt66

11x\lt66\ \Rightarrow\ x\lt6

Solution set: (−∞, 6)
10

Solve \dfrac{x}{3}\gt\dfrac{x}{2}+1 for real x.

Easy +
Solution

Multiplying every term by 6:

2x\gt3x+6

2x-3x\gt6\ \Rightarrow\ -x\gt6\ \Rightarrow\ x\lt-6

Solution set: (−∞, −6)
11

Solve \dfrac{3(x-2)}{5}\le\dfrac{5(2-x)}{3} for real x.

Medium +
Solution

Multiplying both sides by 15 (the LCM of 5 and 3):

9(x-2)\le25(2-x)

9x-18\le50-25x

9x+25x\le50+18

34x\le68\ \Rightarrow\ x\le2

Solution set: (−∞, 2]
12

Solve \dfrac{1}{2}\left(\dfrac{3x}{5}+4\right)\ge\dfrac{1}{3}(x-6) for real x.

Medium +
Solution

Multiplying both sides by 30 (the LCM of 2, 5, 3):

15\left(\dfrac{3x}{5}+4\right)\ge10(x-6)

9x+60\ge10x-60

60+60\ge10x-9x

120\ge x, i.e. x\le120

Solution set: (−∞, 120]
13

Solve 2(2x+3)-10\lt6(x-2) for real x.

Easy +
Solution

2(2x+3)-10\lt6(x-2)

4x+6-10\lt6x-12

4x-4\lt6x-12

-4+12\lt6x-4x\ \Rightarrow\ 8\lt2x\ \Rightarrow\ x\gt4

Solution set: (4, ∞)
14

Solve 37-(3x+5)\ge9x-8(x-3) for real x.

Medium +
Solution

37-(3x+5)\ge9x-8(x-3)

37-3x-5\ge9x-8x+24

32-3x\ge x+24

32-24\ge x+3x\ \Rightarrow\ 8\ge4x\ \Rightarrow\ x\le2

Solution set: (−∞, 2]
15

Solve \dfrac{x}{4}\lt\dfrac{5x-2}{3}-\dfrac{7x-3}{5} for real x.

Medium +
Solution

Multiplying every term by 60 (the LCM of 4, 3, 5):

15x\lt20(5x-2)-12(7x-3)

15x\lt100x-40-84x+36

15x\lt16x-4

15x-16x\lt-4\ \Rightarrow\ -x\lt-4\ \Rightarrow\ x\gt4

Solution set: (4, ∞)
16

Solve \dfrac{2x-1}{3}\ge\dfrac{3x-2}{4}-\dfrac{2-x}{5} for real x.

Medium +
Solution

Multiplying every term by 60 (the LCM of 3, 4, 5):

20(2x-1)\ge15(3x-2)-12(2-x)

40x-20\ge45x-30-24+12x

40x-20\ge57x-54

-20+54\ge57x-40x\ \Rightarrow\ 34\ge17x\ \Rightarrow\ x\le2

Solution set: (−∞, 2]
17

Solve 3x-2\lt2x+1 and show the graph of the solution on a number line.

Medium +
Solution

3x-2\lt2x+1

3x-2x\lt1+2\ \Rightarrow\ x\lt3

−2−10 123 4
x < 3 — open circle at 3 (not included), bold line extends left toward −∞.
Solution set: (−∞, 3)
18

Solve 5x-3\ge3x-5 and show the graph of the solution on a number line.

Medium +
Solution

5x-3\ge3x-5

5x-3x\ge-5+3\ \Rightarrow\ 2x\ge-2\ \Rightarrow\ x\ge-1

−3−2−1 012 3
x ≥ −1 — filled circle at −1 (included), bold line extends right toward +∞.
Solution set: [−1, ∞)
19

Solve 3(1-x)\lt2(x+4) and show the graph of the solution on a number line.

Medium +
Solution

3(1-x)\lt2(x+4)

3-3x\lt2x+8

3-8\lt2x+3x\ \Rightarrow\ -5\lt5x\ \Rightarrow\ -1\lt x, i.e. x\gt-1

−3−2−1 012 3
x > −1 — open circle at −1 (not included), bold line extends right toward +∞.
Solution set: (−1, ∞)
20

Solve \dfrac{x}{2}\ge\dfrac{5x-2}{3}-\dfrac{7x-3}{5} and show the graph of the solution on a number line.

Medium +
Solution

Multiplying every term by 30 (the LCM of 2, 3, 5):

15x\ge10(5x-2)-6(7x-3)

15x\ge50x-20-42x+18

15x\ge8x-2

15x-8x\ge-2\ \Rightarrow\ 7x\ge-2\ \Rightarrow\ x\ge-\dfrac{2}{7}

−2−10 123 −2/7
x ≥ −2/7 — filled circle just left of 0 (included), bold line extends right toward +∞.
Solution set: [−2/7, ∞)
21

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

Easy +
Solution

Let x be the marks Ravi obtains in the third test.

The average of the three tests must be at least 60:

\dfrac{70+75+x}{3}\ge60

145+x\ge180

x\ge35

Ravi must score at least 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 the first four examinations are 87, 92, 94 and 95, find the minimum marks that Sunita must obtain in the fifth examination to get grade 'A' in the course.

Easy +
Solution

Let x be the marks Sunita obtains in the fifth examination.

The average of all five examinations must be at least 90:

\dfrac{87+92+94+95+x}{5}\ge90

368+x\ge450

x\ge82

Sunita must score at least 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.

Medium +
Solution

Let x be the smaller of the two consecutive odd positive integers, so the other is x+2.

Both integers must be smaller than 10:

x+2\lt10\ \Rightarrow\ x\lt8   ...(1)

Their sum must be more than 11:

x+(x+2)\gt11\ \Rightarrow\ 2x+2\gt11\ \Rightarrow\ x\gt4.5   ...(2)

Combining (1) and (2): 4.5\lt x\lt8. Since x is an odd positive integer, x=5 or x=7.

The possible pairs are (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.

Medium +
Solution

Let x be the smaller of the two consecutive even positive integers, so the other is x+2.

Both integers must be larger than 5:

x\gt5   ...(1)

Their sum must be less than 23:

x+(x+2)\lt23\ \Rightarrow\ 2x+2\lt23\ \Rightarrow\ x\lt10.5   ...(2)

Combining (1) and (2): 5\lt x\lt10.5. Since x is an even positive integer, x=6,8 or 10.

The possible pairs are (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.

Medium +
Solution

Let x cm be the length of the shortest side.

Then the longest side =3x cm, and the third side =3x-2 cm.

The perimeter must be at least 61 cm:

x+3x+(3x-2)\ge61

7x-2\ge61

7x\ge63\ \Rightarrow\ x\ge9

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?

Hard +
Solution

Let x cm be the length of the shortest board.

Then the second piece =(x+3) cm, and the third piece =2x cm.

Since all three pieces come from a 91 cm board, their total cannot exceed 91 cm:

x+(x+3)+2x\le91

4x+3\le91\ \Rightarrow\ 4x\le88\ \Rightarrow\ x\le22   ...(1)

The third piece must be at least 5 cm longer than the second:

2x\ge(x+3)+5

2x\ge x+8\ \Rightarrow\ x\ge8   ...(2)

Combining (1) and (2):

The shortest board must be between 8 cm and 22 cm, i.e., 8 ≤ x ≤ 22.

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Common Questions

Class 11 Maths NCERT Solutions Chapter 5 Ex 5.1 — FAQs

How many questions are there in Exercise 5.1?
Exercise 5.1 has 26 questions — the first 4 restrict solutions to natural numbers or integers, Questions 5–16 solve for real x algebraically, Questions 17–20 also ask for a number-line graph, and Questions 21–26 are real-world word problems modelled as inequalities.
What is the key rule to remember when solving inequalities?
You can add or subtract the same number from both sides without changing the inequality sign, and you can multiply or divide both sides by a positive number without changing the sign. But multiplying or dividing both sides by a negative number reverses the inequality sign — this is the single most common mistake in this exercise.
Where can I find the official NCERT textbook for this chapter?
Linear Inequalities is Chapter 5 of the NCERT Class 11 Mathematics textbook, published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the exercise exactly as it appears there.

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