Class 9 Mathematics Chapter 2: Linear Equations in One Variable — Previous Year Questions with Full Solutions

Students who solve previous year questions score 15–20% higher on Chapter 2 than those who only memorize formulas. Here's why: (1) **Pattern recognition.** CBSE examiners repeat structural patterns. A 3-mark 'solve and verify' question using fractions appears almost every year. Solving it once primes your brain to spot it instantly in the exam. (2) **Time management.** Real papers teach you which steps to show for full marks. A 5-mark word problem on age or distance expects setup → equation formation → solving → verification. Skip any step, lose marks. (3) **Confidence under pressure.** Facing a question you've seen before (or its variant) in the exam hall is psychologically powerful. You solve faster, make fewer errors. (4) **Identifying your weaknesses.** Past papers reveal whether your gap is in translating words into equations (most common) or in solving equations with variables on both sides. Theory alone won't show this. CBSE Class 9 Mathematics tests deep understanding, not rote learning. Previous year questions are the closest mirror to what's coming.

One-mark questions typically test definition recall and simple substitution. Here are five patterns that repeat every year: **Q1: Which of the following is a linear equation in one variable?** (a) 2x + y = 3 (b) x² + x = 2 (c) 3x − 5 = 7 (d) 2x + 3y − z = 0 **Answer: (c) 3x − 5 = 7** Explanation: A linear equation in one variable has exactly one variable with power 1. Option (c) has only x, with degree 1. (a) has two variables, (b) has degree 2, (d) has three variables. **Q2: Solve: 5x = 20** **Answer: x = 4** Direct division. Examiners test whether you can isolate the variable in 10 seconds. **Q3: If 2x − 3 = 7, what is the value of x?** **Answer: x = 5** Setup: 2x = 10 → x = 5. Two-step equation, always appears. **Q4: Which value of x satisfies the equation x/2 = 3?** **Answer: x = 6** Fractional coefficient. Common in 1-mark slots as a quick check of fraction handling. **Q5: Is x = 2 a solution of 3x − 1 = 5?** **Answer: Yes** (Verification: 3(2) − 1 = 6 − 1 = 5 ✓) One-mark questions often test 'verify this value' as a speed check. Takes 15 seconds if you substitute carefully.

Three-mark questions expect clear step-by-step working and often involve fractions, variables on both sides, or simple word problems. **Q1: Solve 2x − 5 = 3x + 2. Verify your answer.** **Solution:** Step 1: Move variable terms to left, constants to right. 2x − 3x = 2 + 5 Step 2: Simplify. −x = 7 Step 3: Divide by −1 (or multiply by −1). x = −7 Step 4: Verify by substituting x = −7 into both sides: LHS: 2(−7) − 5 = −14 − 5 = −19 RHS: 3(−7) + 2 = −21 + 2 = −19 LHS = RHS ✓ **Answer: x = −7** **Q2: Solve: (2x + 1)/3 = (3x − 2)/2** **Solution:** Step 1: Cross-multiply. 2(2x + 1) = 3(3x − 2) Step 2: Expand. 4x + 2 = 9x − 6 Step 3: Collect terms. 4x − 9x = −6 − 2 −5x = −8 Step 4: Divide. x = 8/5 or 1.6 **Answer: x = 8/5** **Q3: A number is 5 more than another. If their sum is 23, find the numbers.** **Solution:** Step 1: Define variable. Let the smaller number = x. Then larger = x + 5. Step 2: Form equation. x + (x + 5) = 23 Step 3: Solve. 2x + 5 = 23 → 2x = 18 → x = 9 Step 4: Find both numbers. Smaller = 9, Larger = 9 + 5 = 14 Step 5: Verify. 9 + 14 = 23 ✓ **Answer: The numbers are 9 and 14** **Q4: Solve 3(x − 2) = 2(x + 1) − 4** **Solution:** Step 1: Expand both sides. 3x − 6 = 2x + 2 − 4 3x − 6 = 2x − 2 Step 2: Collect terms. 3x − 2x = −2 + 6 x = 4 **Answer: x = 4** **Q5: Solve (x − 1)/2 + (x − 2)/3 = 1** **Solution:** Step 1: Find LCM of 2 and 3 = 6. Multiply throughout by 6. 6 · (x − 1)/2 + 6 · (x − 2)/3 = 6 · 1 3(x − 1) + 2(x − 2) = 6 Step 2: Expand. 3x − 3 + 2x − 4 = 6 5x − 7 = 6 Step 3: Solve. 5x = 13 x = 13/5 **Answer: x = 13/5**

Five-mark questions demand full problem setup, systematic solving, and explicit verification or contextual conclusion. These typically feature word problems or multi-step equations. **Q1: The sum of three consecutive integers is 48. Find the integers.** **Full Solution:** Step 1: Define variables. Let the three consecutive integers be x, x + 1, and x + 2. Step 2: Form equation from the given condition. x + (x + 1) + (x + 2) = 48 Step 3: Simplify. 3x + 3 = 48 Step 4: Solve. 3x = 45 x = 15 Step 5: Find all three integers. First integer = 15, Second = 16, Third = 17 Step 6: Verify. 15 + 16 + 17 = 48 ✓ **Answer: The three consecutive integers are 15, 16, and 17.** **Q2: Ravi's father is 3 times as old as Ravi. After 12 years, his father's age will be twice his age. Find their present ages.** **Full Solution:** Step 1: Define variables. Let Ravi's present age = x years Father's present age = 3x years Step 2: Set up equation using 'after 12 years' condition. After 12 years: Ravi's age = x + 12, Father's age = 3x + 12 Given: 3x + 12 = 2(x + 12) Step 3: Expand. 3x + 12 = 2x + 24 Step 4: Solve. 3x − 2x = 24 − 12 x = 12 Step 5: Find both ages. Ravi's present age = 12 years Father's present age = 3 × 12 = 36 years Step 6: Verify. After 12 years: Ravi = 24, Father = 48. Is 48 = 2 × 24? Yes ✓ **Answer: Ravi is 12 years old and his father is 36 years old.** **Q3: A rectangle's length is 7 cm more than its width. Its perimeter is 54 cm. Find its dimensions and area.** **Full Solution:** Step 1: Define variables. Let width = x cm Then length = (x + 7) cm Step 2: Form equation using perimeter formula. Perimeter = 2(length + width) 54 = 2[(x + 7) + x] Step 3: Simplify. 54 = 2(2x + 7) 54 = 4x + 14 Step 4: Solve. 40 = 4x x = 10 Step 5: Find dimensions. Width = 10 cm Length = 10 + 7 = 17 cm Step 6: Calculate area. Area = length × width = 17 × 10 = 170 cm² Step 7: Verify perimeter. 2(17 + 10) = 2 × 27 = 54 cm ✓ **Answer: Width is 10 cm, length is 17 cm, and area is 170 cm².**

The 2026–27 CBSE framework emphasizes competency-based learning and case-study integration. For Linear Equations in One Variable, expect these shifts: (1) **Real-world context dominance.** Simple 'solve 2x = 10' questions are nearly extinct. Equations now embed in budgeting scenarios, recipe scaling, distance-time problems, or business breakeven analysis. A 3-mark question might now read: 'A shop sells notebooks at ₹8 each. If fixed costs are ₹200 monthly, how many notebooks must it sell to break even at ₹10 revenue per unit?' (2) **Multi-step literacy.** Expect 4–5 consecutive steps in a single problem. A word problem now demands: form equation, interpret the solution in context, decide validity (e.g., 'can age be negative?'), and propose an alternative scenario. (3) **Technology integration hints.** Questions may reference spreadsheet use or graphical interpretation. Example: 'Use an online tool to verify your equation solution graphically.' (4) **Emphasis on equation comparison.** Instead of 'solve one equation,' you might see 'compare two linear equations and decide which represents the more efficient process.' (5) **Higher reasoning levels.** CBSE's focus on Bloom's taxonomy means fewer 'recall' and 'apply' questions; more 'analyse' and 'evaluate' questions. Start solving equations with explicit reasoning statements: 'I chose to move variable terms first because…' Practicing with authentic word problems on cbsetutor.ai preps you for this shift today.

**Time allocation for a 2-hour exam (full paper with all chapters):** Allocate roughly 35–40 minutes to Linear Equations if it carries 8–10 marks. Here's the order: **Minute 1–2: Read all questions.** Don't solve yet. Circle or underline equation keywords: 'solve,' 'verify,' 'form equation,' 'find ages,' 'dimensions.' This primes your brain. **Minute 3–5: Solve all 1-mark questions first.** These are confidence-builders and typically take 1 minute each. You'll finish 4–5 of them in 2–3 minutes if you're familiar with patterns. Verify by substitution if unsure. **Minute 6–20: Tackle 3-mark questions.** Take 4–5 minutes per question. Write three lines minimum: (1) define variables, (2) form equation, (3) solve and simplify. Always include verification if the question asks or if space permits. Skip step-by-step algebra only if 100% sure. **Minute 21–35: Solve 5-mark questions.** Invest 7 minutes per question. Write all six steps: define variables, form equation, expand/simplify, solve, interpret answer, verify. Word problem answers need context statements: 'The ages are…' not just 'x = 12.' **Minute 36–40: Review and recheck.** Scan for arithmetic errors (especially in fraction/cross-multiplication) and missing negative signs. Rework any equation where LHS ≠ RHS after verification. **Key exam-hall tactics:** (1) If stuck on an equation, clear your working and restart—careless errors compound. (2) For word problems, always define your variable in one sentence: 'Let age = x' not just 'Let x = age.' Examiners award marks for clarity. (3) If solving a fractional equation, clear denominators by multiplying by LCM immediately. Don't leave fractions in steps. (4) Negative solutions are valid; don't panic if x = −7. Verify to ensure it satisfies the equation. (5) In word problems, reject solutions that make no physical sense (e.g., negative age, zero quantity where quantity must be positive).

Before attempting past papers, ensure you've memorized these core concepts (no calculator needed): **Definition:** A linear equation in one variable is an equation of the form ax + b = 0 (or ax + b = c), where a ≠ 0 and x is the variable. **Standard solving steps:** 1. Expand all brackets using distributive property. 2. Collect all variable terms on one side (usually left). 3. Collect all constant terms on the other side. 4. Divide by the coefficient of the variable. 5. Simplify to get x = (some number). 6. Verify by substituting back into the original equation. **Equations with variables on both sides:** Move all x terms to the left (or right—choose once), all numbers to the opposite side. Example: 3x − 2 = x + 4 becomes 3x − x = 4 + 2, so 2x = 6, thus x = 3. **Equations with fractions:** Multiply entire equation by the LCM of all denominators to eliminate fractions before solving. Example: x/2 + x/3 = 5. LCM(2,3) = 6. Multiply by 6: 3x + 2x = 30, so 5x = 30, x = 6. **Word problem keywords:** - 'More than' / 'Less than' → Addition / Subtraction - 'Times' → Multiplication - 'Consecutive integers' → x, x+1, x+2, … - 'Ratio' → Use multipliers (e.g., if ratio is 2:3, write as 2x and 3x) - 'After n years' / 'n years ago' → Add/subtract n from present value **Red flags in your working:** - Forgetting to apply operations to both sides. - Arithmetic slip (e.g., 3 − 5 = −2, not 2). - Not simplifying negative signs correctly. - Skipping verification step—always verify, especially under exam stress. Start a 3-day free trial at cbsetutor.ai to access interactive drills on every sub-topic of Chapter 2 and get instant feedback on your solutions.