1
Solve 24x < 100 when (i) x is a natural number (ii) x is an integer.
Solution
x < 100/24 = 25/6 ≈ 4.17. (i) {1, 2, 3, 4}. (ii) {…, −3, −2, −1, 0, 1, 2, 3, 4}.
(i) {1, 2, 3, 4}; (ii) {…, −3, −2, −1, 0, 1, 2, 3, 4}
NCERT Solutions · Class 11 Maths · Chapter 5
Chapter 5: Linear Inequalities (Algebra)
Step-by-step solutions to all exercises of Chapter 5, Linear Inequalities (NCERT Class 11 Maths, 2026-27): solving linear inequalities in one variable, graphing on the number line, word problems and systems of inequalities. All 40 questions are answered, with the key answer highlighted.
NCERT textbook Chapter 5 (PDF)
Exam revision? Chapter 5 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 14 chapters
Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-maths/chapter-5-linear-inequalities
Adding or subtracting the same number keeps the sign; multiplying or dividing by a negative number reverses it. Show < and > with open circles, ≤ and ≥ with filled circles.
Exercise 5.1
2
Solve −12x > 30 when (i) x is natural (ii) x is an integer.
Solution
Divide by −12 (reverse): x < −30/12 = −5/2. (i) No natural number is negative: no solution. (ii) {…, −5, −4, −3}.
(i) no solution; (ii) {…, −5, −4, −3}
3
Solve 5x − 3 < 7 when (i) x is an integer (ii) x is real.
Solution
5x < 10 ⇒ x < 2. (i) {…, −2, −1, 0, 1}. (ii) (−∞, 2).
(i) {…, −2, −1, 0, 1}; (ii) (−∞, 2)
4
Solve 3x + 8 > 2 when (i) x is an integer (ii) x is real.
Solution
3x > −6 ⇒ x > −2. (i) {−1, 0, 1, 2, …}. (ii) (−2, ∞).
(i) {−1, 0, 1, 2, …}; (ii) (−2, ∞)
5
Solve for real x: 4x + 3 < 5x + 7
Solution
3 − 7 < 5x − 4x ⇒ x > −4, (−4, ∞).
x > −4
6
Solve for real x: 3x − 7 > 5x − 1
Solution
−7 + 1 > 5x − 3x ⇒ −6 > 2x ⇒ x < −3.
x < −3
7
Solve for real x: 3(x − 1) ≤ 2(x − 3)
Solution
3x − 3 ≤ 2x − 6 ⇒ x ≤ −3, (−∞, −3].
x ≤ −3
8
Solve for real x: 3(2 − x) ≥ 2(1 − x)
Solution
6 − 3x ≥ 2 − 2x ⇒ 4 ≥ x: (−∞, 4].
(−∞, 4]
9
Solve for real x: x + x/2 + x/3 < 11
Solution
11x/6 < 11 ⇒ x < 6.
x < 6
10
Solve for real x: x/3 > x/2 + 1
Solution
×6: 2x > 3x + 6 ⇒ −x > 6 ⇒ x < −6.
x < −6
11
Solve for real x: 3(x − 2)/5 ≤ 5(2 − x)/3
Solution
×15: 9(x − 2) ≤ 25(2 − x) ⇒ 9x − 18 ≤ 50 − 25x ⇒ 34x ≤ 68 ⇒ x ≤ 2.
x ≤ 2
12
Solve for real x: ½(3x/5 + 4) ≥ ⅓(x − 6)
Solution
×30: 15(3x/5 + 4) ≥ 10(x − 6) ⇒ 9x + 60 ≥ 10x − 60 ⇒ x ≤ 120.
x ≤ 120
13
Solve for real x: 2(2x + 3) − 10 < 6(x − 2)
Solution
4x + 6 − 10 < 6x − 12 ⇒ 8 < 2x ⇒ x > 4.
x > 4
14
Solve for real x: 37 − (3x + 5) ≥ 9x − 8(x − 3)
Solution
32 − 3x ≥ x + 24 ⇒ 8 ≥ 4x ⇒ x ≤ 2.
x ≤ 2
15
Solve for real x: x/4 < (5x − 2)/3 − (7x − 3)/5
Solution
×60: 15x < 20(5x − 2) − 12(7x − 3) = 16x − 4 ⇒ x > 4.
x > 4
16
Solve for real x: (2x − 1)/3 ≥ (3x − 2)/4 − (2 − x)/5.
Solution
×60: 20(2x − 1) ≥ 15(3x − 2) − 12(2 − x) ⇒ 40x − 20 ≥ 57x − 54 ⇒ 34 ≥ 17x ⇒ x ≤ 2.
x ≤ 2
17
Solve and show on the number line: 3x − 2 < 2x + 1
Solution
x < 3: open circle at 3, line to the left.
x < 3
18
Solve and show on the number line: 5x − 3 ≥ 3x − 5
Solution
2x ≥ −2 ⇒ x ≥ −1: filled circle at −1, line to the right.
x ≥ −1
19
Solve and show on the number line: 3(1 − x) < 2(x + 4)
Solution
3 − 3x < 2x + 8 ⇒ −5 < 5x ⇒ x > −1: open circle at −1, line to the right.
x > −1
20
Solve and show on the number line: x/2 ≥ (5x − 2)/3 − (7x − 3)/5.
Solution
×30: 15x ≥ 10(5x − 2) − 6(7x − 3) = 8x − 2 ⇒ 7x ≥ −2 ⇒ x ≥ −2/7: filled circle at −2/7, line to the right.
x ≥ −2/7
21
Ravi obtained 70 and 75 marks in two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60.
Solution
(70 + 75 + x)/3 ≥ 60 ⇒ 145 + x ≥ 180 ⇒ x ≥ 35: at least 35 marks.
at least 35 marks
22
For grade A an average of 90 or more in five examinations is needed. Sunita's marks in four are 87, 92, 94 and 95. Find the minimum marks she must get in the fifth.
Solution
(87 + 92 + 94 + 95 + x)/5 ≥ 90 ⇒ 368 + x ≥ 450 ⇒ x ≥ 82: at least 82 marks.
at least 82 marks
23
Find all pairs of consecutive odd positive integers, both smaller than 10, with sum more than 11.
Solution
x, x + 2 odd: x + 2 < 10 ⇒ x < 8; 2x + 2 > 11 ⇒ x > 4.5. So x = 5, 7: (5, 7), (7, 9).
(5, 7), (7, 9)
24
Find all pairs of consecutive even positive integers, both larger than 5, with sum less than 23.
Solution
x, x + 2 even: x > 5; 2x + 2 < 23 ⇒ x < 10.5. So x = 6, 8, 10: (6, 8), (8, 10), (10, 12).
(6, 8), (8, 10), (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. If the perimeter is at least 61 cm, find the minimum length of the shortest side.
Solution
Sides x, 3x, 3x − 2: x + 3x + 3x − 2 ≥ 61 ⇒ 7x ≥ 63 ⇒ x ≥ 9: the shortest side is at least 9 cm.
at least 9 cm
26
Three lengths from a 91 cm board: the second 3 cm longer than the shortest, the third twice the shortest. What lengths of the shortest piece are possible if the third is to be at least 5 cm longer than the second?
Solution
Pieces x, x + 3, 2x: x + (x + 3) + 2x ≤ 91 ⇒ 4x ≤ 88 ⇒ x ≤ 22; 2x ≥ (x + 3) + 5 ⇒ x ≥ 8. So 8 cm ≤ shortest ≤ 22 cm.
8 cm ≤ shortest ≤ 22 cm
Miscellaneous Exercise on Chapter 5
1
Solve: 2 ≤ 3x − 4 ≤ 5
Solution
+4: 6 ≤ 3x ≤ 9 ⇒ 2 ≤ x ≤ 3, [2, 3].
2 ≤ x ≤ 3
2
Solve: 6 ≤ −3(2x − 4) < 12
Solution
÷(−3) (reverse): −2 ≥ 2x − 4 > −4 ⇒ 2 ≥ 2x > 0 ⇒ 0 < x ≤ 1, (0, 1].
0 < x ≤ 1
3
Solve: −3 ≤ 4 − 7x/2 ≤ 18
Solution
−4: −7 ≤ −7x/2 ≤ 14; ×(−2/7) (reverse): 2 ≥ x ≥ −4: [−4, 2].
[−4, 2]
4
Solve: −15 < 3(x − 2)/5 ≤ 0
Solution
×5/3: −25 < x − 2 ≤ 0 ⇒ −23 < x ≤ 2.
−23 < x ≤ 2
5
Solve: −12 < 4 − 3x/(−5) ≤ 2
Solution
4 − 3x/(−5) = 4 + 3x/5: −16 < 3x/5 ≤ −2 ⇒ −80/3 < x ≤ −10/3.
−80/3 < x ≤ −10/3
6
Solve: 7 ≤ (3x + 11)/2 ≤ 11.
Solution
×2: 14 ≤ 3x + 11 ≤ 22 ⇒ 3 ≤ 3x ≤ 11 ⇒ 1 ≤ x ≤ 11/3.
1 ≤ x ≤ 11/3
7
Solve and show on the number line: 5x + 1 > −24, 5x − 1 < 24
Solution
x > −5 and x < 5: (−5, 5), open circles at ±5.
(−5, 5)
8
Solve and show on the number line: 2(x − 1) < x + 5, 3(x + 2) > 2 − x
Solution
x < 7 and 4x > −4 ⇒ x > −1: (−1, 7).
(−1, 7)
9
Solve and show on the number line: 3x − 7 > 2(x − 6), 6 − x > 11 − 2x
Solution
x > −5 and x > 5: common part x > 5, (5, ∞).
x > 5
10
Solve and show on the number line: 5(2x − 7) − 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47.
Solution
4x − 44 ≤ 0 ⇒ x ≤ 11; −28 ≤ 4x ⇒ x ≥ −7: [−7, 11], filled circles at −7 and 11.
[−7, 11]
11
A solution is to be kept between 68 °F and 77 °F. What is the range in °C if F = 9C/5 + 32?
Solution
68 < 9C/5 + 32 < 77 ⇒ 36 < 9C/5 < 45 ⇒ 20 < C < 25: between 20 °C and 25 °C.
20 < C < 25
12
640 L of 8% boric acid is diluted with 2% boric acid; the mixture must be more than 4% but less than 6%. How many litres of the 2% solution must be added?
Solution
Add x L: acid = 0.08 × 640 + 0.02x = 51.2 + 0.02x; total 640 + x. 51.2 + 0.02x > 0.04(640 + x) ⇒ 25.6 > 0.02x ⇒ x < 1280; 51.2 + 0.02x < 0.06(640 + x) ⇒ 12.8 < 0.04x ⇒ x > 320. More than 320 L but less than 1280 L.
More than 320 L but less than 1280 L
13
How many litres of water must be added to 1125 L of a 45% acid solution so that the mixture contains more than 25% but less than 30% acid?
Solution
Acid = 0.45 × 1125 = 506.25 L; add x L water: 0.25(1125 + x) < 506.25 < 0.30(1125 + x). Left: 1125 + x < 2025 ⇒ x < 900; right: 1125 + x > 1687.5 ⇒ x > 562.5. More than 562.5 L but less than 900 L.
More than 562.5 L but less than 900 L
14
IQ = (MA/CA) × 100. If 80 ≤ IQ ≤ 140 for 12-year-old children, find the range of their mental age.
Solution
80 ≤ (MA/12) × 100 ≤ 140 ⇒ 960 ≤ 100 MA ≤ 1680 ⇒ 9.6 ≤ MA ≤ 16.8 years.
9.6 ≤ MA ≤ 16.8
Done with the answers?Now revise the whole chapter: Class 11 Mathematics notes
- For NCERT Mathematics Class 11 (2026-27 edition) - all 14 chapters
- Typed concept notes with every formula, theorem and method, with worked examples
- Every NCERT exercise and miscellaneous question solved step by step
₹199₹24920% off · all 14 chapters · instant PDF
More free NCERT solutions for Class 11Class 11 Physics · 14 chaptersClass 11 Chemistry · 9 chaptersClass 11 Biology · 19 chaptersAll classes →
Found a mistake or need help with a question? Message us on WhatsApp. Free study tips and new solutions: join our WhatsApp Channel.