Linear Inequalities — Class 11 Mathematics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Linear Inequalities" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Mathematics chapter "Linear Inequalities" — 7 important questions with detailed answers for CBSE board…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 11 Mathematics — Linear Inequalities. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Linear Inequalities — All 7 Questions Solved
Q1. Solve the linear inequality 3x - 5 > 7 and represent the solution on a number line.
Step 1: Isolate the variable term.
3x - 5 > 7
3x > 7 + 5
3x > 12
Step 2: Divide both sides by 3.
x > 12/3
x > 4
Step 3: Write the solution set.
Solution = {x ∈ ℝ : x > 4} or in interval notation: (4, ∞)
Step 4: Represent on a number line.
←────○────────────→
0 4 5 6 7
Open circle at 4, arrow pointing right to indicate all numbers greater than 4.
Step 5: Verify with test points.
Test x = 5: 3(5) - 5 = 15 - 5 = 10 > 7 ✓
Test x = 4: 3(4) - 5 = 12 - 5 = 7 ≯ 7 ✗ (boundary point not included)
Test x = 3: 3(3) - 5 = 9 - 5 = 4 ≯ 7 ✗
Final Answer: x > 4 or x ∈ (4, ∞)
Q2. Solve the compound inequality -3 ≤ 2x + 1 < 5.
Step 1: Break into two separate inequalities.
-3 ≤ 2x + 1 AND 2x + 1 < 5
Step 2: Solve the left inequality -3 ≤ 2x + 1.
-3 ≤ 2x + 1
-3 - 1 ≤ 2x
-4 ≤ 2x
-2 ≤ x
Or: x ≥ -2
Step 3: Solve the right inequality 2x + 1 < 5.
2x + 1 < 5
2x < 5 - 1
2x < 4
x < 2
Step 4: Combine the solutions.
x ≥ -2 AND x < 2
Solution: -2 ≤ x < 2
Step 5: Write in interval notation.
[-2, 2)
Step 6: Verify with test points.
Test x = -2: -3 ≤ 2(-2) + 1 = -3 ≤ -3 ✓ and -3 < 5 ✓
Test x = 0: -3 ≤ 2(0) + 1 = -3 ≤ 1 ✓ and 1 < 5 ✓
Test x = 2: -3 ≤ 2(2) + 1 = -3 ≤ 5 ✓ but 5 ≮ 5 ✗
Final Answer: -2 ≤ x < 2 or x ∈ [-2, 2)
Q3. Solve the absolute value inequality |x - 3| ≤ 2.
Step 1: Recall the definition of absolute value inequality.
|A| ≤ B (where B ≥ 0) is equivalent to -B ≤ A ≤ B
Step 2: Apply the definition.
|x - 3| ≤ 2
-2 ≤ x - 3 ≤ 2
Step 3: Add 3 to all parts.
-2 + 3 ≤ x - 3 + 3 ≤ 2 + 3
1 ≤ x ≤ 5
Step 4: Write the solution set.
Solution: [1, 5]
Step 5: Verify with test points.
Test x = 1: |1 - 3| = |-2| = 2 ≤ 2 ✓
Test x = 3: |3 - 3| = 0 ≤ 2 ✓
Test x = 5: |5 - 3| = 2 ≤ 2 ✓
Test x = 0: |0 - 3| = 3 ≤ 2 ✗
Test x = 6: |6 - 3| = 3 ≤ 2 ✗
Final Answer: 1 ≤ x ≤ 5 or x ∈ [1, 5]
Q4. Solve the system of linear inequalities: 2x + y ≤ 8 x - y ≥ -2 x ≥ 0, y ≥ 0
Step 1: Identify the inequalities.
(1) 2x + y ≤ 8
(2) x - y ≥ -2, or x ≥ y - 2
(3) x ≥ 0 (non-negative x)
(4) y ≥ 0 (non-negative y)
Step 2: Find boundary lines by converting to equations.
Line 1: 2x + y = 8 ⟹ y = 8 - 2x
Line 2: x - y = -2 ⟹ y = x + 2
Line 3: x = 0 (y-axis)
Line 4: y = 0 (x-axis)
Step 3: Find key intersection points.
Intersection of 2x + y = 8 and x - y = -2:
2x + y = 8
x - y = -2
Adding: 3x = 6 ⟹ x = 2, y = 4
Point: (2, 4)
Intersection of 2x + y = 8 and x = 0:
2(0) + y = 8 ⟹ y = 8
Point: (0, 8)
Intersection of 2x + y = 8 and y = 0:
2x + 0 = 8 ⟹ x = 4
Point: (4, 0)
Intersection of x - y = -2 and y = 0:
x - 0 = -2 ⟹ x = -2 (outside x ≥ 0)
Intersection of x - y = -2 and x = 0:
0 - y = -2 ⟹ y = 2
Point: (0, 2)
Origin: (0, 0)
Step 4: Identify the feasible region vertices.
Vertices of the feasible region: (0, 0), (4, 0), (2, 4), (0, 2)
Step 5: Verify each vertex satisfies all inequalities.
(0, 0): 2(0) + 0 = 0 ≤ 8 ✓, 0 - 0 = 0 ≥ -2 ✓, x ≥ 0 ✓, y ≥ 0 ✓
(4, 0): 2(4) + 0 = 8 ≤ 8 ✓, 4 - 0 = 4 ≥ -2 ✓, x ≥ 0 ✓, y ≥ 0 ✓
(2, 4): 2(2) + 4 = 8 ≤ 8 ✓, 2 - 4 = -2 ≥ -2 ✓, x ≥ 0 ✓, y ≥ 0 ✓
(0, 2): 2(0) + 2 = 2 ≤ 8 ✓, 0 - 2 = -2 ≥ -2 ✓, x ≥ 0 ✓, y ≥ 0 ✓
Final Answer: The feasible region is a quadrilateral with vertices (0,0), (4,0), (2,4), and (0,2).
Q5. Solve the quadratic inequality x² - 5x + 6 > 0.
Step 1: Factor the quadratic expression.
x² - 5x + 6 = (x - 2)(x - 3)
Step 2: Find the zeros (critical points).
(x - 2)(x - 3) = 0
x = 2 or x = 3
Step 3: Determine the sign of (x - 2)(x - 3) in each interval.
Test intervals: (-∞, 2), (2, 3), (3, ∞)
Test x = 0 (from interval (-∞, 2)):
(0 - 2)(0 - 3) = (-2)(-3) = 6 > 0 ✓
Test x = 2.5 (from interval (2, 3)):
(2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 ✗
Test x = 4 (from interval (3, ∞)):
(4 - 2)(4 - 3) = (2)(1) = 2 > 0 ✓
Step 4: Write the solution.
The expression is positive in intervals (-∞, 2) and (3, ∞).
Solution: x < 2 or x > 3
Step 5: Write in interval notation.
(-∞, 2) ∪ (3, ∞)
Step 6: Verify at boundary points.
At x = 2: (2 - 2)(2 - 3) = 0 ≯ 0 ✗ (not included)
At x = 3: (3 - 2)(3 - 3) = 0 ≯ 0 ✗ (not included)
Final Answer: x < 2 or x > 3, or x ∈ (-∞, 2) ∪ (3, ∞)
Q6. Solve the inequality (x + 1)/(x - 2) > 0.
Step 1: Find critical points where the expression equals zero or is undefined.
Numerator: x + 1 = 0 ⟹ x = -1
Denominator: x - 2 = 0 ⟹ x = 2 (undefined, asymptote)
Step 2: Create a sign chart with intervals.
Intervals: (-∞, -1), (-1, 2), (2, ∞)
Step 3: Test the sign in each interval.
Test x = -2 (from (-∞, -1)):
(-2 + 1) / (-2 - 2) = (-1) / (-4) = 1/4 > 0 ✓
Test x = 0 (from (-1, 2)):
(0 + 1) / (0 - 2) = 1 / (-2) = -1/2 < 0 ✗
Test x = 3 (from (2, ∞)):
(3 + 1) / (3 - 2) = 4 / 1 = 4 > 0 ✓
Step 4: Determine where the expression is positive.
Positive in intervals: (-∞, -1) and (2, ∞)
Step 5: Check boundary points.
At x = -1: (-1 + 1) / (-1 - 2) = 0 / (-3) = 0 ≯ 0 ✗ (not included)
At x = 2: undefined ✗ (not included)
Final Answer: x < -1 or x > 2, or x ∈ (-∞, -1) ∪ (2, ∞)
Q7. A manufacturer produces items with a cost of 5 rupees per item and a fixed cost of 1000 rupees. The selling price is 8 rupees per item. How many items must be produced and sold to make a profit of at least 4000 rupees?
Step 1: Define variables.
Let x = number of items produced and sold
Step 2: Calculate total cost.
Total Cost = Fixed Cost + Variable Cost
TC = 1000 + 5x
Step 3: Calculate total revenue.
Total Revenue = Selling Price × Quantity
TR = 8x
Step 4: Set up the profit inequality.
Profit = Revenue - Cost ≥ 4000
8x - (1000 + 5x) ≥ 4000
Step 5: Simplify.
8x - 1000 - 5x ≥ 4000
3x - 1000 ≥ 4000
3x ≥ 5000
x ≥ 5000/3
x ≥ 1666.67
Step 6: Interpret the result.
Since x must be a whole number (number of items):
x ≥ 1667
Step 7: Verify.
For x = 1667:
Profit = 8(1667) - (1000 + 5(1667)) = 13336 - 9335 = 4001 ≥ 4000 ✓
For x = 1666:
Profit = 8(1666) - (1000 + 5(1666)) = 13328 - 9330 = 3998 < 4000 ✗
Final Answer: At least 1667 items must be produced and sold.
Linear Inequalities — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. When multiplying both sides of a linear inequality by a negative number, what must be done to the inequality sign?
- A. It remains the same.
- B. It flips direction.
- C. It changes to an equality.
- D. It becomes an absolute value.
Answer: B. It flips direction. — A fundamental rule of inequalities states that multiplying or dividing by a negative number reverses the direction of the inequality sign.
MCQ 2. Solve the inequality `2x - 3 < 7`.
- A. `x < 5`
- B. `x > 5`
- C. `x < 2`
- D. `x > 2`
Answer: A. `x < 5` — Add 3 to both sides to get `2x < 10`, then divide by 2 to get `x < 5`.
MCQ 3. Which interval represents the solution `x >= -2`?
- A. `(-2, infinity)`
- B. `[-2, infinity)`
- C. `(-infinity, -2)`
- D. `(-infinity, -2]`
Answer: B. `[-2, infinity)` — The square bracket `[` indicates that -2 is included in the solution set, and `infinity)` is always represented with a parenthesis.
MCQ 4. Which of the following points satisfies the inequality `3x + 2y > 10`?
- A. `(1, 2)`
- B. `(0, 5)`
- C. `(3, 1)`
- D. `(2, 3)`
Answer: D. `(2, 3)` — Substitute the coordinates `(2, 3)` into the inequality: `3(2) + 2(3) = 6 + 6 = 12`. Since `12 > 10`, the point `(2, 3)` satisfies the inequality.
MCQ 5. Find the solution set for the system of inequalities: `x + 3 > 5` and `2x - 1 < 9`.
- A. `(2, 5)`
- B. `(2, 4)`
- C. `(3, 5)`
- D. `(4, 5)`
Answer: A. `(2, 5)` — Solving `x + 3 > 5` gives `x > 2`. Solving `2x - 1 < 9` gives `2x < 10`, so `x < 5`. The intersection of `x > 2` and `x < 5` is `(2, 5)`.
MCQ 6. What is the solution to `|2x - 4| < 6`?
- A. `x < 5`
- B. `-1 < x < 5`
- C. `x > -1`
- D. `-5 < x < 1`
Answer: B. `-1 < x < 5` — The inequality `|2x - 4| < 6` can be written as `-6 < 2x - 4 < 6`. Adding 4 to all parts gives `-2 < 2x < 10`. Dividing by 2 gives `-1 < x < 5`.
MCQ 7. How is the solution `x <= 4` represented on a number line?
- A. An open circle at 4, arrow pointing right.
- B. A closed circle at 4, arrow pointing right.
- C. An open circle at 4, arrow pointing left.
- D. A closed circle at 4, arrow pointing left.
Answer: D. A closed circle at 4, arrow pointing left. — A closed circle (solid dot) indicates that 4 is included in the solution, and the arrow pointing left indicates all numbers less than or equal to 4.
MCQ 8. Solve `(x/3) + 1 >= (x/2) - 1`.
- A. `x <= 12`
- B. `x >= 12`
- C. `x <= 6`
- D. `x >= 6`
Answer: A. `x <= 12` — Multiply all terms by the LCD (6) to clear fractions: `2x + 6 >= 3x - 6`. Rearrange the terms to get `12 >= x`, or `x <= 12`.
MCQ 9. A student needs an average of at least 80 marks in 5 tests. If her marks in the first four tests are 75, 82, 78, and 85, what is the minimum mark she must obtain in the fifth test?
- A. 70
- B. 80
- C. 82
- D. 85
Answer: B. 80 — Let `x` be the mark in the fifth test. The average is `(75 + 82 + 78 + 85 + x) / 5 >= 80`. This simplifies to `(320 + x) / 5 >= 80`, which leads to `320 + x >= 400`, so `x >= 80`.
MCQ 10. Which of the following is a strict inequality?
- A. `x >= 5`
- B. `y <= 2`
- C. `2x + 3y > 7`
- D. `a + b = c`
Answer: C. `2x + 3y > 7` — Strict inequalities use `>` or `<` signs, meaning the boundary values are not included in the solution set.
MCQ 11. When graphing the inequality `y > 2x - 1`, what kind of line is used for the boundary, and which region is shaded?
- A. Solid line, region above.
- B. Solid line, region below.
- C. Dashed line, region above.
- D. Dashed line, region below.
Answer: C. Dashed line, region above. — The `>` sign indicates a strict inequality, so the boundary line is dashed. For `y > mx + c`, the region above the line is shaded.
MCQ 12. Which point lies in the feasible region defined by `x >= 0`, `y >= 0`, `x + y <= 5`, and `x + 2y >= 4`?
- A. `(0, 0)`
- B. `(1, 3)`
- C. `(5, 1)`
- D. `(0, 1)`
Answer: B. `(1, 3)` — Test `(1, 3)`: `1 >= 0` (True), `3 >= 0` (True), `1 + 3 <= 5` (True, `4 <= 5`), `1 + 2(3) >= 4` (True, `7 >= 4`). All conditions are met by `(1,3)`.
MCQ 13. Solve `5x - 7 > 3x + 1`.
- A. `x < 4`
- B. `x > 4`
- C. `x < -4`
- D. `x > -4`
Answer: B. `x > 4` — Subtract `3x` from both sides: `2x - 7 > 1`. Add `7` to both sides: `2x > 8`. Divide by `2`: `x > 4`.
MCQ 14. What is the solution to `|3x + 2| >= 5`?
- A. `x >= 1` or `x <= -7/3`
- B. `-7/3 <= x <= 1`
- C. `x <= 1`
- D. `x >= -7/3`
Answer: A. `x >= 1` or `x <= -7/3` — For `|X| >= a`, the solution is `X >= a` or `X <= -a`. So, `3x + 2 >= 5` (giving `x >= 1`) or `3x + 2 <= -5` (giving `x <= -7/3`).
MCQ 15. The inequality `x < -3` is represented by which interval notation?
- A. `[-3, infinity)`
- B. `(-3, infinity)`
- C. `(-infinity, -3]`
- D. `(-infinity, -3)`
Answer: D. `(-infinity, -3)` — `x < -3` means all numbers strictly less than -3. This is represented by an open interval `(-infinity, -3)`.
MCQ 16. If a car rental costs ₹200 per day plus ₹15 per km, what is the maximum distance (in km) you can travel in a day if your budget is ₹500?
- A. 10 km
- B. 20 km
- C. 30 km
- D. 40 km
Answer: B. 20 km — Let `d` be the distance in km. The total cost is `200 + 15d`. Setting up the inequality `200 + 15d <= 500` gives `15d <= 300`, so `d <= 20` km.
MCQ 17. Which of the following statements is true for the inequality `2x + 5y < 10`?
- A. The line `2x + 5y = 10` is part of the solution.
- B. The origin `(0,0)` satisfies the inequality.
- C. Points on the line `2x + 5y = 10` are included in the solution.
- D. The region above the line `2x + 5y = 10` is always the solution.
Answer: B. The origin `(0,0)` satisfies the inequality. — Substitute `(0,0)` into `2x + 5y < 10`: `2(0) + 5(0) < 10` simplifies to `0 < 10`, which is a true statement.
MCQ 18. Solve `-5 <= (2x - 1) / 3 <= 1`.
- A. `[-7, 2]`
- B. `[-14, 2]`
- C. `[2, 7]`
- D. `[-7, -2]`
Answer: A. `[-7, 2]` — Multiply all parts by 3: `-15 <= 2x - 1 <= 3`. Add 1 to all parts: `-14 <= 2x <= 4`. Divide all parts by 2: `-7 <= x <= 2`.
MCQ 19. Which system of inequalities represents the shaded region in the first quadrant bounded by the lines `x + y = 3` and `y = x` and the x-axis?
- A. `x >= 0, y >= 0, x + y <= 3, y >= x`
- B. `x >= 0, y >= 0, x + y >= 3, y <= x`
- C. `x >= 0, y >= 0, x + y <= 3, y <= x`
- D. `x >= 0, y >= 0, x + y >= 3, y >= x`
Answer: C. `x >= 0, y >= 0, x + y <= 3, y <= x` — The region is in the first quadrant (`x >= 0, y >= 0`). It is below `x + y = 3` (`x + y <= 3`) and below `y = x` (`y <= x`).
MCQ 20. The symbol `less than or equal to` is represented as:
- A. `<`
- B. `>`
- C. `<=`
- D. `>=`
Answer: C. `<=` — The symbol `<=` denotes 'less than or equal to', including the boundary value.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.