Class 9 Mathematics Chapter 10: Exponents and Powers – Important Questions with Solutions

Exponents and Powers has undergone strategic emphasis in the 2024-25 rationalized CBSE Class 9 syllabus. The chapter tests not just mechanical application of laws (like aᵐ × aⁿ = aᵐ⁺ⁿ), but conceptual depth: understanding why negative exponents yield reciprocals, how fractional exponents connect to roots, and the real-world use of scientific notation for astronomical and microscopic quantities. Board papers now increasingly feature multi-step problems that combine 2–3 laws in a single question, demand clear reasoning in 3-mark answers, and test flexibility by asking students to simplify, evaluate, and justify in the same problem. Roughly 15–20% of the algebra weightage in Class 9 falls on exponents, translating to 8–12 marks in most board papers. The questions below mirror the exact cognitive load, marking split, and difficulty progression of official CBSE sample papers and board exams from the last three years. By practising these 18 questions thoroughly, you'll recognize question patterns instantly and respond with confidence and correctness under exam pressure.

Multiple-choice questions form the quick-check portion of your algebra practice and often appear in board papers' objective sections. Each MCQ tests a single concept or a direct application of one law of exponents. Here are five MCQs with full explanations: **Q1. Simplify: 3⁴ × 3⁵** (a) 3⁹ (b) 3²⁰ (c) 9⁹ (d) 27⁴ **Answer: (a) 3⁹** Using the law aᵐ × aⁿ = aᵐ⁺ⁿ, we have 3⁴ × 3⁵ = 3⁴⁺⁵ = 3⁹. **Q2. Find the value of 5⁰.** (a) 0 (b) 1 (c) 5 (d) undefined **Answer: (b) 1** Any non-zero number raised to the power 0 equals 1. Therefore, 5⁰ = 1. **Q3. Express 2⁻³ as a fraction.** (a) ½³ (b) 1/8 (c) -6 (d) 2/3 **Answer: (b) 1/8** By the negative exponent law, 2⁻³ = 1/2³ = 1/8. **Q4. Which is equivalent to ⁴√81?** (a) 81⁰·²⁵ (b) 81¼ (c) 81⁴ (d) 81¹/⁴ **Answer: (d) 81¹/⁴** (also correct: (a) and (b) — all are equivalent) The fourth root is written as the fractional exponent 1/4: ⁴√81 = 81¹/⁴ = 3. **Q5. Simplify: (10³)²** (a) 10⁵ (b) 10⁶ (c) 100⁶ (d) 10² **Answer: (b) 10⁶** Using the power-of-a-power law (aᵐ)ⁿ = aᵐⁿ, we get (10³)² = 10³×² = 10⁶.

Two-mark questions test your ability to apply one or two laws in sequence and show your working. They require you to state the law used and simplify step-by-step. These five questions are representative of what appears in CBSE board papers: **Q6. Simplify and express with positive exponents: (a⁻²b³)/(a⁴b⁻¹)** **Solution:** (a⁻²b³)/(a⁴b⁻¹) = a⁻² × b³ × a⁻⁴ × b¹ (using the negative exponent rule) = a⁻²⁻⁴ × b³⁺¹ (grouping like bases) = a⁻⁶b⁴ = b⁴/a⁶ (expressing with positive exponents) **Q7. Find the value of (27)^(2/3)** **Solution:** (27)^(2/3) = (27¹/³)² (using (aᵐ/ⁿ) = (a¹/ⁿ)ᵐ) = (∛27)² (converting fractional exponent to radical form) = 3² (since ∛27 = 3) = 9 **Q8. Simplify: (2⁵ × 3⁴ × 2³)/(3² × 2⁶)** **Solution:** Group like bases: (2⁵ × 2³ × 3⁴)/(3² × 2⁶) = (2⁸ × 3⁴)/(3² × 2⁶) = 2⁸⁻⁶ × 3⁴⁻² (applying the quotient law) = 2² × 3² = 4 × 9 = 36 **Q9. Express 0.0000045 in scientific notation.** **Solution:** 0.0000045 = 4.5 × 10⁻⁶ (move the decimal 6 places to the right, so exponent is –6) **Q10. If 2ˣ = 32, find the value of x.** **Solution:** 2ˣ = 32 2ˣ = 2⁵ (since 32 = 2 × 2 × 2 × 2 × 2) Therefore, x = 5

Three-mark questions demand that you show working, apply multiple laws, and justify your steps. They appear regularly in board papers and test both procedural fluency and conceptual understanding. Here are four representative 3-mark questions: **Q11. Simplify and show your working: [2³ × 3⁴ × 4]/[9 × 2⁵] and express in the form of a fraction.** **Solution:** First, rewrite 4 and 9 in terms of prime factors: 4 = 2², 9 = 3² So the expression becomes: [2³ × 3⁴ × 2²]/[3² × 2⁵] Combine like bases in the numerator: [2⁵ × 3⁴]/[3² × 2⁵] Apply the quotient law: 2⁵⁻⁵ × 3⁴⁻² = 2⁰ × 3² = 1 × 9 = 9 **Q12. Prove that a⁰ = 1 (where a ≠ 0) using the quotient law of exponents.** **Solution:** We know that aᵐ ÷ aⁿ = aᵐ⁻ⁿ (quotient law) Let m = n. Then aⁿ ÷ aⁿ = aⁿ⁻ⁿ = a⁰ But aⁿ ÷ aⁿ = 1 (any non-zero number divided by itself is 1) Therefore, a⁰ = 1 **Q13. Simplify: (x²y)³ × (xy²)² ÷ (x³y)² and verify your answer for x = 2, y = 1.** **Solution:** Apply the power-of-a-product law: (x²y)³ = x⁶y³, (xy²)² = x²y⁴, (x³y)² = x⁶y² Expression becomes: (x⁶y³ × x²y⁴) ÷ (x⁶y²) = x⁸y⁷ ÷ x⁶y² (combining like bases in numerator) = x⁸⁻⁶ × y⁷⁻² (quotient law) = x²y⁵ Verification: For x = 2, y = 1: (2)²(1)⁵ = 4 × 1 = 4 ✓ **Q14. If 5ˣ⁻¹ = 125, find x and verify your answer.** **Solution:** 5ˣ⁻¹ = 125 5ˣ⁻¹ = 5³ (since 125 = 5 × 5 × 5) Comparing exponents (since bases are equal): x − 1 = 3 Therefore, x = 4 Verification: 5⁴⁻¹ = 5³ = 125 ✓

Five-mark questions test comprehensive understanding, multi-step problem-solving, and the ability to work with combined concepts. These are typically derivation-heavy or involve solving equations with exponents. Here are three fully worked 5-mark questions: **Q15. Simplify the following and express the answer in the form of a single exponent: {[(32)^(2/5)]³ × (2⁻¹)²} / {[2⁵ × 2⁻²]^(1/2)}** **Solution:** Step 1: Simplify 32 as a power of 2. 32 = 2⁵ Step 2: Simplify the numerator. (32)^(2/5) = (2⁵)^(2/5) = 2^(5 × 2/5) = 2² = 4 [(32)^(2/5)]³ = (2²)³ = 2⁶ (2⁻¹)² = 2⁻² Numerator = 2⁶ × 2⁻² = 2⁶⁻² = 2⁴ Step 3: Simplify the denominator. [2⁵ × 2⁻²]^(1/2) = [2^(5-2)]^(1/2) = [2³]^(1/2) = 2^(3/2) Step 4: Divide numerator by denominator. 2⁴ ÷ 2^(3/2) = 2^(4 - 3/2) = 2^(8/2 - 3/2) = 2^(5/2) **Answer: 2^(5/2) or 2² × 2^(1/2) = 4√2** **Q16. A scientist observes that the mass of a sample decreases exponentially. The mass after n hours follows the formula M = M₀ × (1/2)ⁿ, where M₀ = 64 grams is the initial mass. Find the mass after 3 hours and express your answer using exponents.** **Solution:** Given: M₀ = 64 grams = 2⁶ grams, n = 3 M = M₀ × (1/2)ⁿ M = 2⁶ × (1/2)³ M = 2⁶ × 2⁻³ (since 1/2 = 2⁻¹, and (2⁻¹)³ = 2⁻³) M = 2^(6-3) = 2³ = 8 grams **Answer: The mass after 3 hours is 8 grams (or 2³ grams).** **Q17. The distance between the Earth and the Sun is approximately 1.5 × 10¹¹ metres, and the distance between the Earth and the Moon is approximately 3.84 × 10⁸ metres. How many times farther is the Sun than the Moon from the Earth? Express your answer in standard form.** **Solution:** Ratio = (Distance to Sun) ÷ (Distance to Moon) = (1.5 × 10¹¹) ÷ (3.84 × 10⁸) = (1.5 ÷ 3.84) × (10¹¹ ÷ 10⁸) = 0.390625 × 10³ = 0.390625 × 10³ = 390.625 ≈ 3.91 × 10² (in standard form) **Answer: The Sun is approximately 391 times (or 3.91 × 10²) farther from the Earth than the Moon.**

**Q18. Case Study: Global Population and Scientific Notation** Global population in 2024 is approximately 8 × 10⁹ people. A research institute projects that by 2100, the population will grow to 1.1 × 10¹⁰ people. An environmentalist claims that the daily per-capita water consumption is 3.2 × 10³ litres. (This is incorrect; typical consumption is ~100 litres/day.) **(i) Express the population growth from 2024 to 2100 as a ratio and simplify using exponent laws.** **Solution:** Ratio = (2100 population) ÷ (2024 population) = (1.1 × 10¹⁰) ÷ (8 × 10⁹) = (1.1 ÷ 8) × (10¹⁰ ÷ 10⁹) = 0.1375 × 10¹ = 1.375 The population in 2100 is approximately 1.375 times (or 11/8 times) the 2024 population. **(ii) If the actual per-capita daily consumption is 1 × 10² litres, find the total water consumed by the global population in 2024 in scientific notation.** **Solution:** Total water = (Population) × (Per-capita consumption) = (8 × 10⁹) × (1 × 10²) = 8 × 10⁹⁺² = 8 × 10¹¹ litres **(iii) The environmentalist's claim (3.2 × 10³ litres per day) would mean global consumption of 8 × 10⁹ × 3.2 × 10³ litres. How many times greater is this incorrect figure compared to the actual figure (from part ii)? Use exponent laws to justify.** **Solution:** Incorrect total = 8 × 10⁹ × 3.2 × 10³ = 25.6 × 10¹² = 2.56 × 10¹³ litres Actual total = 8 × 10¹¹ litres Ratio = (2.56 × 10¹³) ÷ (8 × 10¹¹) = (2.56 ÷ 8) × (10¹³ ÷ 10¹¹) = 0.32 × 10² = 3.2 × 10¹ = 32 times The incorrect claim is 32 times greater than the actual figure, showing the real-world impact of exponent and decimal mishandling.

Mastering Exponents and Powers requires more than reading solutions once—it demands daily, adaptive practice with immediate feedback. CBSETUTOR.ai's AI tutor is built specifically for Class 9 CBSE students and uses a proven methodology to convert these 18 important questions into personalised learning pathways. Here's how it works: **Adaptive Practice Mode:** The AI analyses which concepts you struggle with (e.g., fractional exponents vs. negative exponents) and generates similar questions until you achieve mastery. If you make an error in Q7 (fractional exponents), the tutor creates 3–4 follow-up questions with increasing difficulty. **Step-by-Step Solution Walkthrough:** Every question is broken into micro-steps. The tutor explains *why* we group like bases, *why* we add exponents when multiplying, and *why* scientific notation uses powers of 10—not just the mechanical 'do this, then that.' **Timed Mock Tests:** Once you've practised, the AI generates 20-minute timed quizzes modelled on actual CBSE board paper structure: 1 MCQ (1 mark), 2 short-answers (2 marks each), 1 medium question (3 marks), and 1 long-answer (5 marks). Your score, time-per-question, and error patterns are tracked. **Spaced Revision:** The AI schedules questions you've solved into your calendar at optimal intervals (2 days, 7 days, 14 days) to lock them into long-term memory. **Real-Time Doubt Clearing:** If you're stuck, you can ask the tutor a natural-language question like 'Why does 2⁻³ become 1/8?' and get a conversational, jargon-free explanation in under 10 seconds. Start a 3-day free trial at cbsetutor.ai—no payment info required—and see how AI-powered daily drills transform these 18 questions into exam confidence.

Exponents and Powers is a high-leverage chapter: master it, and 40% of Class 9 algebra becomes easier (quadratics, polynomials, surds all rely on exponent sense). Here are your action steps: **Week 1:** Drill the five MCQs (Q1–Q5) daily until you answer each in under 30 seconds without errors. These consolidate the foundational laws: aᵐ × aⁿ, aᵐ ÷ aⁿ, a⁰, a⁻ⁿ. **Week 2:** Attempt Q6–Q10 (2-mark questions). Spend 4–5 minutes per question. Write your working in full—this is where boards award partial marks. Do not skip steps. **Week 3:** Tackle Q11–Q14 (3-mark questions). These mix 2–3 concepts per question and demand clear reasoning. Time yourself: aim for 7 minutes per question. **Week 4:** Work through Q15–Q17 (5-mark questions) one at a time. These are derivation-heavy; show every exponent rule applied. Check against the solution and understand any gaps. **Final Days:** Solve Q18 (case study) under timed conditions (10 minutes). This simulates how exponent concepts appear in real-world, multi-part scenarios. **Pro Tip:** Always state the law (e.g., 'using the product law aᵐ × aⁿ = aᵐ⁺ⁿ') before applying it. Examiners value this explicitly—it shows conceptual understanding, not just mechanical substitution. Use NCERT Class 9 Maths Chapter 10 as your reference for definitions and examples; cross-check your working with the textbook after solving.