Class 9 Mathematics Chapter 5 Prime Time: Complete Important Questions & Answers for CBSE Boards

Chapter 5 carries 8–12 marks in the Class 9 annual examination. The CBSE board emphasizes conceptual understanding over rote learning—especially in number theory. Questions test: (1) identification of primes, composites, and twin primes; (2) application of divisibility tests to solve real-world problems; (3) prime factorisation to find HCF and LCM; (4) understanding the Sieve of Eratosthenes as a historical and algorithmic tool. The 2026-27 pattern expects multi-step reasoning: for example, a 5-mark question might ask students to find HCF and LCM of two numbers using prime factorisation, then verify using the formula HCF × LCM = Product of the two numbers. Short-answer and MCQ sections test divisibility rules (especially 9 and 11), which appear in mental math and data interpretation tasks. Case-study and application-based questions now form 30–40% of the paper, so expect problems like: 'A shopkeeper has 36 red and 48 blue pens. He wants to pack them into identical boxes without mixing colours. What is the maximum number of pens in each box?' This requires HCF reasoning. Preparing systematically through these curated questions ensures you can handle both theoretical definitions and practical problem-solving.

MCQs in CBSE Class 9 test quick recall and conceptual clarity. These five questions cover core definitions and divisibility rules. **Question 1:** Which of the following is a prime number? (A) 1 (B) 51 (C) 97 (D) 99 **Answer:** (C) 97. Explanation: Prime numbers have exactly two factors (1 and itself). 1 has only one factor (itself), so it is neither prime nor composite. 51 = 3 × 17, and 99 = 9 × 11, so both are composite. 97 is divisible only by 1 and 97. **Question 2:** A number is divisible by 9 if: (A) It ends in 0 or 9 (B) The sum of its digits is divisible by 9 (C) It is divisible by 3 (D) The last two digits form a number divisible by 9 **Answer:** (B) The sum of its digits is divisible by 9. Example: 729 → 7+2+9 = 18, which is divisible by 9, so 729 is divisible by 9 (729 ÷ 9 = 81). **Question 3:** Which pair are twin primes? (A) 11 and 13 (B) 13 and 17 (C) 17 and 19 (D) 19 and 23 **Answer:** (A) 11 and 13. Twin primes are pairs of prime numbers that differ by 2. 11 and 13 are both prime with 13 − 11 = 2. (Note: 13 and 17 differ by 4; 17 and 19 differ by 2, but are a separate pair; 19 and 23 differ by 4.) **Question 4:** The HCF of two consecutive even numbers is: (A) 1 (B) 2 (C) 3 (D) 4 **Answer:** (B) 2. Consecutive even numbers like 8 and 10: their factors are {1, 2, 4, 8} and {1, 2, 5, 10}. Common factors: {1, 2}. HCF = 2. Every even number is divisible by 2, so HCF of any two consecutive even numbers is always 2. **Question 5:** A number is divisible by 11 if: (A) Sum of all digits is divisible by 11 (B) The difference between the sum of alternate digits is divisible by 11 (C) It ends in 1 (D) Last digit is 0 or 1 **Answer:** (B) The difference between the sum of alternate digits is divisible by 11. Example: 1331 → Alternate positions: (1+3) − (3+1) = 0, divisible by 11 (1331 = 11 × 121).

These questions require brief explanations and one or two computational steps. They test understanding of definitions and basic applications. **Question 1:** Find the prime factorisation of 84. **Answer:** 84 = 2² × 3 × 7. Working: 84 ÷ 2 = 42; 42 ÷ 2 = 21; 21 ÷ 3 = 7; 7 is prime. So 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7. (1 mark for correct factorisation; 1 mark for method or exponential form.) **Question 2:** Using the Sieve of Eratosthenes, find all prime numbers less than 30. **Answer:** Primes less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Method: Start with 2 (prime), cross all multiples of 2. Then 3 (prime), cross all multiples of 3. Then 5, cross multiples. Continue until √30 ≈ 5.5. Remaining unmarked numbers are prime. (1 mark for correct list; 1 mark for explaining the algorithm.) **Question 3:** Verify that 8 and 15 are co-prime. **Answer:** Co-prime numbers have HCF = 1. Prime factorisation: 8 = 2³ and 15 = 3 × 5. They share no common factor. HCF(8, 15) = 1, so they are co-prime. (1 mark for factorisation; 1 mark for HCF and conclusion.) **Question 4:** Check if 756 is divisible by 6 without division. **Answer:** A number is divisible by 6 if it is divisible by both 2 and 3. 756 is even (ends in 6), so divisible by 2. Sum of digits: 7+5+6 = 18, divisible by 3. So 756 is divisible by 6. (1 mark for checking divisibility by 2 and 3; 1 mark for conclusion.) **Question 5:** Find HCF and LCM of 36 and 48 using prime factorisation. **Answer:** 36 = 2² × 3² and 48 = 2⁴ × 3. HCF = 2² × 3 = 12 (product of lowest powers). LCM = 2⁴ × 3² = 144 (product of highest powers). Verify: HCF × LCM = 12 × 144 = 1728 = 36 × 48 ✓. (1 mark for factorisation; 1 mark for HCF and LCM.)

These questions demand multi-step reasoning and sometimes require combining two concepts. They appear in board papers as short-case-study or application-based questions. **Question 1:** A hall has 102 red chairs and 153 blue chairs. These are to be arranged in rows such that each row has the same number of chairs and no mixing of colours. What is the maximum number of chairs in each row? **Answer:** We need the HCF of 102 and 153. Using Euclidean algorithm: 153 = 102 × 1 + 51; 102 = 51 × 2 + 0. So HCF(102, 153) = 51. The maximum number of chairs in each row is 51. Verification: 102 ÷ 51 = 2 rows of red; 153 ÷ 51 = 3 rows of blue. Total rows = 5, each with 51 chairs. (1 mark for identifying HCF concept; 1 mark for calculation; 1 mark for answer and interpretation.) **Question 2:** Prove that every composite number can be expressed as a product of prime numbers. **Answer:** Assume n is a composite number. By definition, n has at least one factor other than 1 and itself. Let the smallest such factor be p. If p were composite, it would have a factor smaller than p and smaller than n, contradicting our choice of p. So p is prime. Now n = p × q, where q = n/p. If q is prime, we are done. If q is composite, repeat the process for q. Since each quotient is smaller, this process must terminate with a prime. Thus n = p₁ × p₂ × … × pₖ, where each pᵢ is prime. This is the Fundamental Theorem of Arithmetic. (1 mark for logical structure; 1 mark for correct reasoning; 1 mark for clarity of proof.) **Question 3:** Two bells ring at intervals of 12 and 18 minutes. At what interval will they ring together? **Answer:** They ring together at intervals equal to the LCM of 12 and 18. Prime factorisation: 12 = 2² × 3 and 18 = 2 × 3². LCM = 2² × 3² = 36 minutes. So the bells ring together every 36 minutes. Check: 36 ÷ 12 = 3 rings of bell 1; 36 ÷ 18 = 2 rings of bell 2. (1 mark for identifying LCM; 1 mark for factorisation and calculation; 1 mark for interpretation.) **Question 4:** Using divisibility tests, check if 5832 is divisible by 9. Then find its prime factorisation. **Answer:** Divisibility by 9: Sum of digits = 5+8+3+2 = 18, which is divisible by 9. So 5832 is divisible by 9. Prime factorisation: 5832 ÷ 9 = 648; 648 ÷ 9 = 72; 72 ÷ 8 = 9; continuing: 5832 = 2³ × 3⁶. Verify: 8 × 729 = 5832 ✓. (1 mark for divisibility test; 1 mark for correct factorisation; 1 mark for verification.)

These questions integrate multiple concepts—prime factorisation, HCF, LCM, and real-world applications. They require structured working and justify each step. **Question 1:** Three friends, Aji, Bala, and Chitra, visit a gym at intervals of 6, 8, and 12 days respectively. If they all visit today, after how many days will they meet again? Also, how many times will each person visit before they meet again? **Answer:** Step 1: They meet again after LCM(6, 8, 12) days. Prime factorisation: 6 = 2 × 3; 8 = 2³; 12 = 2² × 3. Step 2: LCM = 2³ × 3 = 24 days. Step 3: They will meet again after 24 days. Step 4: Number of visits before they meet: Aji visits every 6 days, so in 24 days: 24 ÷ 6 = 4 visits (including the first). Bala: 24 ÷ 8 = 3 visits. Chitra: 24 ÷ 12 = 2 visits. Final Answer: They meet after 24 days. Aji visits 4 times, Bala 3 times, Chitra 2 times. (1 mark for identifying LCM concept; 1 mark for correct factorisation; 1 mark for LCM calculation; 1 mark for each person's visit count; 1 mark for final interpretation.) **Question 2:** The length, width, and height of a rectangular room are 825 cm, 675 cm, and 450 cm. It is to be packed with cubic boxes of the same size such that no space is left. Find the side of the largest cube that can fit. How many such cubes are needed? **Answer:** Step 1: The side of the largest cube = HCF(825, 675, 450). Step 2: Prime factorisation: 825 = 3 × 5² × 11; 675 = 3³ × 5²; 450 = 2 × 3² × 5². Step 3: HCF = 3 × 5² = 75 cm. Step 4: Volume of room = 825 × 675 × 450 = 250,121,250 cm³. Step 5: Volume of one cube = 75³ = 421,875 cm³. Step 6: Number of cubes = 250,121,250 ÷ 421,875 = 593 (or verify: (825÷75) × (675÷75) × (450÷75) = 11 × 9 × 6 = 594 cubes). Final Answer: Side of largest cube = 75 cm; 594 cubes needed. (1 mark for identifying HCF; 1 mark for factorisation; 1 mark for HCF calculation; 1 mark for volume reasoning; 1 mark for correct count and final answer.) **Question 3:** Using the properties of prime numbers and divisibility, prove that if a number is divisible by both 9 and 11, then it is divisible by 99. Is 1089 divisible by 99? Verify. **Answer:** Step 1: 9 and 11 are coprime (HCF(9,11) = 1) since 9 = 3² and 11 is prime. Step 2: If a number n is divisible by both 9 and 11, then 9|n and 11|n. Step 3: Since 9 and 11 are coprime, their product divides n (by the coprimality theorem). Step 4: 9 × 11 = 99 divides n. Thus, any number divisible by both 9 and 11 is divisible by 99. Step 5: Check 1089: Divisible by 9? Sum of digits = 1+0+8+9 = 18, divisible by 9 ✓. Divisible by 11? (1+8) − (0+9) = 0, divisible by 11 ✓. Step 6: Therefore, 1089 is divisible by 99. Verify: 1089 ÷ 99 = 11 ✓. Final Answer: Proof established; 1089 is divisible by 99. (1 mark each for proof structure, coprimality reasoning, application to 1089, divisibility checks, and verification.)

**Question:** A shopkeeper wants to arrange 60 mangoes, 84 apples, and 108 oranges into identical gift boxes such that each fruit type is equally distributed and no fruit is left over. (a) What is the maximum number of gift boxes? (b) How many of each fruit will be in one box? (c) If mangoes cost ₹5, apples ₹3, and oranges ₹4, what is the cost of one gift box? **Full Solution:** Step 1 (Part a): Find the HCF of 60, 84, and 108 to determine the maximum number of boxes. Prime factorisation: - 60 = 2² × 3 × 5 - 84 = 2² × 3 × 7 - 108 = 2² × 3³ HCF = 2² × 3 = 12. Maximum number of boxes = 12. (2 marks for correct factorisation and HCF.) Step 2 (Part b): Distribute each fruit equally among 12 boxes. - Mangoes per box = 60 ÷ 12 = 5 - Apples per box = 84 ÷ 12 = 7 - Oranges per box = 108 ÷ 12 = 9 Each box contains 5 mangoes, 7 apples, and 9 oranges. (1 mark for correct distribution.) Step 3 (Part c): Calculate cost of one box. Cost = (5 × ₹5) + (7 × ₹3) + (9 × ₹4) = ₹25 + ₹21 + ₹36 = ₹82 Cost of one gift box = ₹82. (2 marks for arithmetic and final answer.) **Total: 5 marks.** This question integrates HCF, divisibility, and real-world problem-solving—typical of modern CBSE case-study sections.

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Before your board exam or pre-board test, verify you can: ✓ Identify primes and composites instantly (list: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, …). ✓ Apply divisibility tests (2: even; 3 & 9: digit sum; 5: ends in 0 or 5; 10: ends in 0; 11: alternate digit difference). ✓ Find prime factorisation using repeated division. ✓ Compute HCF using either prime factorisation (lowest powers) or Euclidean algorithm. ✓ Compute LCM using prime factorisation (highest powers) and verify HCF × LCM = n₁ × n₂. ✓ Solve real-world problems (arrangement, bells, gym visits, packaging) by recognizing HCF or LCM cues. ✓ Explain the Sieve of Eratosthenes as a method and identify twin primes. ✓ Prove statements like 'composite numbers are products of primes' or 'coprime × coprime concept'. These 18 questions cover all six marks guaranteed in your annual exam. Practise them thrice—first for understanding, second for speed, third for confidence under timed conditions.