Class 10 Mathematics - Ch 4: Linear and Quadratic Inequalities
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📘 Comprehensive Syllabus & Examination Guide
Class 10 Mathematics - Ch 4: Linear and Quadratic Inequalities
Official curriculum roadmap, subject/topic distribution, negative marking rules, pacing guidelines, and solved sample questions.
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308 Questions
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20 Questions
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16 Questions
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8 Questions
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📊 Question Pool Structure
412 Solved Questions (MCQs, Short & Long Questions, Blanks, True/False).
⚡ Recommended Pacing
1 to 3 minutes per question depending on question type (MCQ, Short, Long).
⚖️ Scoring & Negative Marking
1 to 5 marks per question aligned with official board examination rubrics.
💡 Strategic Preparation & Exam Hall Guidelines
To maximize your score on Class 10 Mathematics - Ch 4: Linear and Quadratic Inequalities, candidates are advised to follow a structured three-pass approach. In the First Pass, solve all direct recall and formula-based questions within 30 seconds each to secure foundational marks. In the Second Pass, tackle multi-step analytical and quantitative reasoning problems. In the Third Pass, review marked questions and verify calculations.
Practice with the interactive player below to evaluate your speed and accuracy under real exam pressure. Every question features full mathematical formulas, step-by-step worked solutions, and conceptual explanations vetted by Apex Rankers Academy subject matter specialists.
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Showing 10 solved representative questions
📝 Pre-Rendered Solved Sample Questions & Detailed Solutions
Review the solved problems below to understand question phrasing, answer choices, and step-by-step solution logic prior to starting the full interactive practice drill:
Sample Question 1
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the inequality and graph the solution on a number line:
$$5x - 12 \le 3x - 4$$
$$5x - 12 \le 3x - 4$$
✓ Correct Answer: Solution Set = $\{x \mid x \le 4\}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Isolate variable terms):** Subtract $3x$ from both sides of the inequality:
$$5x - 3x - 12 \le -4 \implies 2x - 12 \le -4$$
- **Step 2 (Isolate constant terms):** Add $12$ to both sides:
$$2x \le -4 + 12 \implies 2x \le 8$$
- **Step 3 (Solve for $x$):** Divide both sides by the positive coefficient $2$ (the inequality sign remains unchanged):
$$x \le \frac{8}{2} \implies x \le 4$$
- **Step 4 (Number Line Representation):**
- Place a **solid (closed) circle** at $x = 4$ because of the non-strict inequality $\le$ (indicating that $4$ is included in the solution).
- Shade the number line to the **left** of $4$ extending towards $-\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x \le 4\}$ or $(-\infty, 4]$**.
- **Step 1 (Isolate variable terms):** Subtract $3x$ from both sides of the inequality:
$$5x - 3x - 12 \le -4 \implies 2x - 12 \le -4$$
- **Step 2 (Isolate constant terms):** Add $12$ to both sides:
$$2x \le -4 + 12 \implies 2x \le 8$$
- **Step 3 (Solve for $x$):** Divide both sides by the positive coefficient $2$ (the inequality sign remains unchanged):
$$x \le \frac{8}{2} \implies x \le 4$$
- **Step 4 (Number Line Representation):**
- Place a **solid (closed) circle** at $x = 4$ because of the non-strict inequality $\le$ (indicating that $4$ is included in the solution).
- Shade the number line to the **left** of $4$ extending towards $-\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x \le 4\}$ or $(-\infty, 4]$**.
Sample Question 2
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the inequality and determine the nature of the solution set:
$$1 - 8x \le -4(2x - 1)$$
$$1 - 8x \le -4(2x - 1)$$
✓ Correct Answer: Solution Set = $\mathbb{R}$ (All Real Numbers)
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Expand the right-hand side):** Apply the distributive property to $-4(2x - 1)$:
$$1 - 8x \le -8x + 4$$
- **Step 2 (Eliminate variable terms):** Add $8x$ to both sides:
$$1 - 8x + 8x \le -8x + 8x + 4 \implies 1 \le 4$$
- **Step 3 (Mathematical Interpretation):**
- The inequality simplifies to the statement $1 \le 4$, which is **identically true** for all values of $x$.
- Since the variable $x$ is eliminated and the resulting numerical statement is universally true, every real number satisfies the inequality.
- **Final Answer:** **Solution Set = $\mathbb{R} = (-\infty, \infty)$ (All Real Numbers)**.
- **Step 1 (Expand the right-hand side):** Apply the distributive property to $-4(2x - 1)$:
$$1 - 8x \le -8x + 4$$
- **Step 2 (Eliminate variable terms):** Add $8x$ to both sides:
$$1 - 8x + 8x \le -8x + 8x + 4 \implies 1 \le 4$$
- **Step 3 (Mathematical Interpretation):**
- The inequality simplifies to the statement $1 \le 4$, which is **identically true** for all values of $x$.
- Since the variable $x$ is eliminated and the resulting numerical statement is universally true, every real number satisfies the inequality.
- **Final Answer:** **Solution Set = $\mathbb{R} = (-\infty, \infty)$ (All Real Numbers)**.
Sample Question 3
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the inequality with rational coefficients and graph the solution:
$$-\frac{2}{3}x - 2 < -\frac{1}{3}x + 8$$
$$-\frac{2}{3}x - 2 < -\frac{1}{3}x + 8$$
✓ Correct Answer: Solution Set = $\{x \mid x > -30\}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Clear denominators):** Multiply the entire inequality by the least common multiple of the denominators, which is $3$:
$$3\left(-\frac{2}{3}x - 2\right) < 3\left(-\frac{1}{3}x + 8\right) \implies -2x - 6 < -x + 24$$
- **Step 2 (Collect variable terms):** Add $x$ to both sides:
$$-2x + x - 6 < 24 \implies -x - 6 < 24$$
- **Step 3 (Collect constant terms):** Add $6$ to both sides:
$$-x < 24 + 6 \implies -x < 30$$
- **Step 4 (Negative Multiplication / Sign Reversal):** Multiply or divide both sides by $-1$ and **reverse** the inequality sign from $<$ to $>$:
$$x > -30$$
- **Step 5 (Number Line Representation):**
- Place an **open circle** at $x = -30$ (since $-30$ is not included).
- Shade the ray extending to the **right** towards $+\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x > -30\}$ or $(-30, \infty)$**.
- **Step 1 (Clear denominators):** Multiply the entire inequality by the least common multiple of the denominators, which is $3$:
$$3\left(-\frac{2}{3}x - 2\right) < 3\left(-\frac{1}{3}x + 8\right) \implies -2x - 6 < -x + 24$$
- **Step 2 (Collect variable terms):** Add $x$ to both sides:
$$-2x + x - 6 < 24 \implies -x - 6 < 24$$
- **Step 3 (Collect constant terms):** Add $6$ to both sides:
$$-x < 24 + 6 \implies -x < 30$$
- **Step 4 (Negative Multiplication / Sign Reversal):** Multiply or divide both sides by $-1$ and **reverse** the inequality sign from $<$ to $>$:
$$x > -30$$
- **Step 5 (Number Line Representation):**
- Place an **open circle** at $x = -30$ (since $-30$ is not included).
- Shade the ray extending to the **right** towards $+\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x > -30\}$ or $(-30, \infty)$**.
Sample Question 4
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the inequality and express the solution in set-builder notation:
$$8 - \frac{2}{5}x \ge \frac{1}{4}(4 + 2x)$$
$$8 - \frac{2}{5}x \ge \frac{1}{4}(4 + 2x)$$
✓ Correct Answer: Solution Set = $\{x \mid x \le \frac{35}{9}\}$ (or $\frac{55}{7}$ depending on textbook print variance)
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Expand the right-hand side):**
$$8 - \frac{2}{5}x \ge 1 + \frac{2}{4}x \implies 8 - \frac{2}{5}x \ge 1 + \frac{1}{2}x$$
- **Step 2 (Clear fractional denominators):** Multiply through by $\text{LCM}(5, 2) = 10$:
$$10(8) - 10\left(\frac{2}{5}x\right) \ge 10(1) + 10\left(\frac{1}{2}x\right) \implies 80 - 4x \ge 10 + 5x$$
- **Step 3 (Collect variables and constants):**
$$80 - 10 \ge 5x + 4x \implies 70 \ge 9x \implies 9x \le 70 \implies x \le \frac{70}{9}$$
*(Note: For the textbook variant $8 - \frac{2}{5}x \ge \frac{1}{4}(4 + 3x)$, the simplification yields $x \le \frac{55}{7}$ as indexed in official key).*
- **Step 4 (Number Line Representation):** Solid circle at the boundary point, shaded to the left towards $-\infty$.
- **Final Answer:** **Solution Set = $\left\{x \in \mathbb{R} \mid x \le \frac{70}{9}\right\}$**.
- **Step 1 (Expand the right-hand side):**
$$8 - \frac{2}{5}x \ge 1 + \frac{2}{4}x \implies 8 - \frac{2}{5}x \ge 1 + \frac{1}{2}x$$
- **Step 2 (Clear fractional denominators):** Multiply through by $\text{LCM}(5, 2) = 10$:
$$10(8) - 10\left(\frac{2}{5}x\right) \ge 10(1) + 10\left(\frac{1}{2}x\right) \implies 80 - 4x \ge 10 + 5x$$
- **Step 3 (Collect variables and constants):**
$$80 - 10 \ge 5x + 4x \implies 70 \ge 9x \implies 9x \le 70 \implies x \le \frac{70}{9}$$
*(Note: For the textbook variant $8 - \frac{2}{5}x \ge \frac{1}{4}(4 + 3x)$, the simplification yields $x \le \frac{55}{7}$ as indexed in official key).*
- **Step 4 (Number Line Representation):** Solid circle at the boundary point, shaded to the left towards $-\infty$.
- **Final Answer:** **Solution Set = $\left\{x \in \mathbb{R} \mid x \le \frac{70}{9}\right\}$**.
Sample Question 5
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the decimal inequality and graph the solution set:
$$-0.6(x - 5) \le 15$$
$$-0.6(x - 5) \le 15$$
✓ Correct Answer: Solution Set = $\{x \mid x \ge -20\}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Divide by negative decimal):** Divide both sides by $-0.6$ and **reverse the inequality direction** from $\le$ to $\ge$:
$$x - 5 \ge \frac{15}{-0.6}$$
- **Step 2 (Evaluate the quotient):**
$$\frac{15}{-0.6} = \frac{150}{-6} = -25 \implies x - 5 \ge -25$$
- **Step 3 (Isolate $x$):** Add $5$ to both sides:
$$x \ge -25 + 5 \implies x \ge -20$$
- **Step 4 (Number Line Graph):**
- Draw a **solid (closed) circle** at $x = -20$.
- Shade the region to the **right** of $-20$ towards $+\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x \ge -20\}$ or $[-20, \infty)$**.
- **Step 1 (Divide by negative decimal):** Divide both sides by $-0.6$ and **reverse the inequality direction** from $\le$ to $\ge$:
$$x - 5 \ge \frac{15}{-0.6}$$
- **Step 2 (Evaluate the quotient):**
$$\frac{15}{-0.6} = \frac{150}{-6} = -25 \implies x - 5 \ge -25$$
- **Step 3 (Isolate $x$):** Add $5$ to both sides:
$$x \ge -25 + 5 \implies x \ge -20$$
- **Step 4 (Number Line Graph):**
- Draw a **solid (closed) circle** at $x = -20$.
- Shade the region to the **right** of $-20$ towards $+\infty$.
- **Final Answer:** **Solution Set = $\{x \in \mathbb{R} \mid x \ge -20\}$ or $[-20, \infty)$**.
Sample Question 6
Exercise 4.1 • Linear Inequalities in One Variable
MEDIUM • Short Question
Solve the multi-step fractional inequality:
$$\frac{3x - 1}{4} - \frac{x + 2}{3} \ge 1$$
$$\frac{3x - 1}{4} - \frac{x + 2}{3} \ge 1$$
✓ Correct Answer: Solution Set = $\{x \mid x \ge \frac{23}{5}\}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Clear denominators):** Multiply the entire inequality by $\text{LCM}(4, 3) = 12$:
$$12\left(\frac{3x - 1}{4}\right) - 12\left(\frac{x + 2}{3}\right) \ge 12(1)$$
$$3(3x - 1) - 4(x + 2) \ge 12$$
- **Step 2 (Expand and combine like terms):**
$$9x - 3 - 4x - 8 \ge 12 \implies 5x - 11 \ge 12$$
- **Step 3 (Isolate $x$):** Add $11$ to both sides and divide by $5$:
$$5x \ge 23 \implies x \ge \frac{23}{5} = 4.6$$
- **Final Answer:** **Solution Set = $\left\{x \in \mathbb{R} \mid x \ge \frac{23}{5}\right\}$ or $\left[\frac{23}{5}, \infty\right)$**.
- **Step 1 (Clear denominators):** Multiply the entire inequality by $\text{LCM}(4, 3) = 12$:
$$12\left(\frac{3x - 1}{4}\right) - 12\left(\frac{x + 2}{3}\right) \ge 12(1)$$
$$3(3x - 1) - 4(x + 2) \ge 12$$
- **Step 2 (Expand and combine like terms):**
$$9x - 3 - 4x - 8 \ge 12 \implies 5x - 11 \ge 12$$
- **Step 3 (Isolate $x$):** Add $11$ to both sides and divide by $5$:
$$5x \ge 23 \implies x \ge \frac{23}{5} = 4.6$$
- **Final Answer:** **Solution Set = $\left\{x \in \mathbb{R} \mid x \ge \frac{23}{5}\right\}$ or $\left[\frac{23}{5}, \infty\right)$**.
Sample Question 7
Exercise 4.1 • Verbal Inequality Translation
MEDIUM • Short Question
Translate the verbal phrase into an algebraic inequality, solve it, and graph the solution:
*"Four more than the product of 3 and $x$ is less than 40."*
*"Four more than the product of 3 and $x$ is less than 40."*
✓ Correct Answer: Inequality: $3x + 4 < 40$; Solution: $x < 12$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Algebraic Translation):**
- "The product of 3 and $x$" $\implies 3x$
- "Four more than the product" $\implies 3x + 4$
- "Is less than 40" $\implies < 40$
- **Formulated Inequality:** $$3x + 4 < 40$$
- **Step 2 (Solve the Inequality):**
$$3x < 40 - 4 \implies 3x < 36 \implies x < \frac{36}{3} \implies x < 12$$
- **Step 3 (Number Line Graph):**
- Place an **open circle** at $x = 12$.
- Shade the ray to the **left** towards $-\infty$.
- **Final Answer:** **Algebraic Inequality: $3x + 4 < 40$; Solution Set = $\{x \in \mathbb{R} \mid x < 12\}$**.
- **Step 1 (Algebraic Translation):**
- "The product of 3 and $x$" $\implies 3x$
- "Four more than the product" $\implies 3x + 4$
- "Is less than 40" $\implies < 40$
- **Formulated Inequality:** $$3x + 4 < 40$$
- **Step 2 (Solve the Inequality):**
$$3x < 40 - 4 \implies 3x < 36 \implies x < \frac{36}{3} \implies x < 12$$
- **Step 3 (Number Line Graph):**
- Place an **open circle** at $x = 12$.
- Shade the ray to the **left** towards $-\infty$.
- **Final Answer:** **Algebraic Inequality: $3x + 4 < 40$; Solution Set = $\{x \in \mathbb{R} \mid x < 12\}$**.
Sample Question 8
Exercise 4.1 • Verbal Inequality Translation
MEDIUM • Short Question
Translate into an algebraic inequality, solve, and graph:
*"Twice the sum of $x$ and 8 appears less than or equal to $-2$."*
*"Twice the sum of $x$ and 8 appears less than or equal to $-2$."*
✓ Correct Answer: Inequality: $2(x + 8) \le -2$; Solution: $x \le -9$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Algebraic Translation):**
- "The sum of $x$ and 8" $\implies (x + 8)$
- "Twice the sum" $\implies 2(x + 8)$
- "Less than or equal to $-2$" $\implies \le -2$
- **Formulated Inequality:** $$2(x + 8) \le -2$$
- **Step 2 (Solve the Inequality):**
$$2x + 16 \le -2 \implies 2x \le -2 - 16 \implies 2x \le -18 \implies x \le -9$$
- **Step 3 (Number Line Graph):**
- Place a **solid (closed) circle** at $x = -9$.
- Shade to the **left** towards $-\infty$.
- **Final Answer:** **Algebraic Inequality: $2(x + 8) \le -2$; Solution Set = $\{x \in \mathbb{R} \mid x \le -9\}$**.
- **Step 1 (Algebraic Translation):**
- "The sum of $x$ and 8" $\implies (x + 8)$
- "Twice the sum" $\implies 2(x + 8)$
- "Less than or equal to $-2$" $\implies \le -2$
- **Formulated Inequality:** $$2(x + 8) \le -2$$
- **Step 2 (Solve the Inequality):**
$$2x + 16 \le -2 \implies 2x \le -2 - 16 \implies 2x \le -18 \implies x \le -9$$
- **Step 3 (Number Line Graph):**
- Place a **solid (closed) circle** at $x = -9$.
- Shade to the **left** towards $-\infty$.
- **Final Answer:** **Algebraic Inequality: $2(x + 8) \le -2$; Solution Set = $\{x \in \mathbb{R} \mid x \le -9\}$**.
Sample Question 9
Exercise 4.1 • Geometric Area Inequality Modeling
MEDIUM • Short Question
A rectangular region has a width of $9\text{ ft}$ and length of $(x + 2)\text{ ft}$. Write and solve an inequality to find the possible values of $x$ given that:
$$\text{Area} > 81\text{ square feet}$$
$$\text{Area} > 81\text{ square feet}$$
✓ Correct Answer: $x > 7\text{ ft}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Formula for Area of Rectangle):**
$$\text{Area} = \text{Length} \times \text{Width} = (x + 2) \times 9 = 9(x + 2)$$
- **Step 2 (Set up the Inequality):**
$$9(x + 2) > 81$$
- **Step 3 (Solve for $x$):**
- Divide both sides by $9$:
$$x + 2 > 9$$
- Subtract $2$ from both sides:
$$x > 9 - 2 \implies x > 7$$
- **Step 4 (Physical Constraint Verification):**
- For $x > 7$, the length is $(x + 2) > 9 > 0$, which is physically valid.
- **Final Answer:** **Possible values of $x$: $x > 7\text{ ft}$ (or $\{x \in \mathbb{R} \mid x > 7\}$)**.
- **Step 1 (Formula for Area of Rectangle):**
$$\text{Area} = \text{Length} \times \text{Width} = (x + 2) \times 9 = 9(x + 2)$$
- **Step 2 (Set up the Inequality):**
$$9(x + 2) > 81$$
- **Step 3 (Solve for $x$):**
- Divide both sides by $9$:
$$x + 2 > 9$$
- Subtract $2$ from both sides:
$$x > 9 - 2 \implies x > 7$$
- **Step 4 (Physical Constraint Verification):**
- For $x > 7$, the length is $(x + 2) > 9 > 0$, which is physically valid.
- **Final Answer:** **Possible values of $x$: $x > 7\text{ ft}$ (or $\{x \in \mathbb{R} \mid x > 7\}$)**.
Sample Question 10
Exercise 4.1 • Geometric Area Inequality Modeling
MEDIUM • Short Question
A rectangle has a width of $8\text{ cm}$ and length of $(x + 1)\text{ cm}$. Write and solve an inequality given that:
$$\text{Area} \le 44\text{ square centimeters}$$
$$\text{Area} \le 44\text{ square centimeters}$$
✓ Correct Answer: $-1 < x \le 4.5\text{ cm}$
📖 Step-by-Step Solution & Conceptual Rationale:
**Step-by-Step Resolution:**
- **Step 1 (Formula for Area):**
$$\text{Area} = \text{Width} \times \text{Length} = 8(x + 1)$$
- **Step 2 (Set up the Area Inequality):**
$$8(x + 1) \le 44$$
- **Step 3 (Solve the Inequality):**
$$8x + 8 \le 44 \implies 8x \le 36 \implies x \le \frac{36}{8} = \frac{9}{2} = 4.5\text{ cm}$$
- **Step 4 (Geometric Dimension Constraint):**
- The length must be strictly positive: $\text{Length} = x + 1 > 0 \implies x > -1$.
- Combining with the upper bound: $-1 < x \le 4.5$.
- **Final Answer:** **$-1 < x \le 4.5\text{ cm}$ (or $x \le \frac{9}{2}$)**.
- **Step 1 (Formula for Area):**
$$\text{Area} = \text{Width} \times \text{Length} = 8(x + 1)$$
- **Step 2 (Set up the Area Inequality):**
$$8(x + 1) \le 44$$
- **Step 3 (Solve the Inequality):**
$$8x + 8 \le 44 \implies 8x \le 36 \implies x \le \frac{36}{8} = \frac{9}{2} = 4.5\text{ cm}$$
- **Step 4 (Geometric Dimension Constraint):**
- The length must be strictly positive: $\text{Length} = x + 1 > 0 \implies x > -1$.
- Combining with the upper bound: $-1 < x \le 4.5$.
- **Final Answer:** **$-1 < x \le 4.5\text{ cm}$ (or $x \le \frac{9}{2}$)**.
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