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Class 7 Maths: Finding Common Ground, 8 practice questions with answers

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These are 8 practice questions on Chapter 3, Finding Common Ground, of the new NCERT Class 7 Maths textbook, Ganita Prakash, Part 2. Each has a worked answer and the textbook page it comes from. They cover the longest equal pieces two ropes can be cut into, what a prime factorisation shows about odd factors and multiples, the HCF and the LCM by prime factors, the link between HCF, LCM and the product, what happens to the HCF when only one number is doubled, the division ladder for both at once, and a sharing problem. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the chapter's own examples is reused, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 7, Part 2, Chapter 3, printed pages 47 to 66):

HCF

Question 1. Ravi has two ropes, one 85 m long and the other 51 m long. He wants to cut both into pieces that all have the same length, a whole number of metres each, with no rope left over, and he wants each piece to be as long as possible. How long should each piece be, and how many pieces will he get in all?

Answer: 17 m pieces, 8 of them. The length of a piece must divide both 85 and 51 exactly, so it is a common factor. The factors of 85 are 1, 5, 17, 85 and the factors of 51 are 1, 3, 17, 51, so the common factors are 1 and 17. The longest possible piece is the biggest common factor, the HCF: 17 m. The 85 m rope gives 85 ÷ 17 = 5 pieces and the 51 m rope gives 51 ÷ 17 = 3 pieces, so Ravi gets 8 pieces in all. (Common factors and the HCF, pages 47 and 48.)

Question 2. 1260 = 2 × 2 × 3 × 3 × 5 × 7. Using only this prime factorisation, answer: (a) What is the largest odd factor of 1260? (b) What is the smallest whole number that 1260 must be multiplied by to give a multiple of 8? (c) And to give a multiple of 25?

Answer: (a) 315; (b) 2; (c) 5. (a) An odd factor can have no 2 in it, so the largest one keeps every other prime: 3 × 3 × 5 × 7 = 315. Check: 315 × 4 = 1260. (b) 8 = 2 × 2 × 2 needs three 2s, and 1260 has only two, so one more 2 is needed: multiply by 2 (1260 × 2 = 2520 = 8 × 315). (c) 25 = 5 × 5 needs two 5s, and 1260 has only one, so multiply by 5 (1260 × 5 = 6300 = 25 × 252). (Factors from the prime factorisation, pages 49 to 51.)

Question 3. Find the HCF of 168 and 252 using prime factorisation.

Answer: 84. 168 = 2 × 2 × 2 × 3 × 7 and 252 = 2 × 2 × 3 × 3 × 7. The common primes are 2, 3 and 7. Take each the smallest number of times it appears: 2 appears three times in 168 but only twice in 252, so take two 2s; 3 appears once in 168, so take one 3; 7 appears once in each. HCF = 2 × 2 × 3 × 7 = 84. Check: 168 = 84 × 2 and 252 = 84 × 3, and 2 and 3 have no common factor other than 1. (HCF using prime factorisation, pages 52 to 54.)

LCM

Question 4. One lighthouse flashes every 18 seconds and another every 32 seconds. They flash together now. After how many seconds will they next flash together? Use prime factorisation.

Answer: after 288 seconds. The time must be a common multiple of 18 and 32, and the next time together is the LCM. 18 = 2 × 3 × 3 and 32 = 2 × 2 × 2 × 2 × 2. Take each prime the largest number of times it appears: five 2s (from 32) and two 3s (from 18). LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 32 × 9 = 288. No smaller number is a multiple of both. (The LCM and LCM by prime factorisation, pages 55 to 58.)

Question 5. For 98 and 70, find the HCF and the LCM, and check that HCF × LCM equals 98 × 70.

Answer: HCF 14, LCM 490, and 14 × 490 = 6860 = 98 × 70. 98 = 2 × 7 × 7 and 70 = 2 × 5 × 7. The HCF takes the common primes the fewest times: 2 × 7 = 14. The LCM takes every prime the most times: 2 × 5 × 7 × 7 = 490. Then 14 × 490 = 6860, and 98 × 70 = 6860 too. The common primes (2 and 7) go into the HCF and the remaining ones (5 and the other 7) go into the LCM, so between them they use every prime of both numbers exactly once. (HCF × LCM = product, pages 62 and 63.)

Patterns and procedures

Question 6. The HCF of two numbers is 13. Only ONE of the two numbers is doubled; the other stays the same. Does the HCF always double? Give one pair where it does and one pair where it does not, and explain the difference.

Answer: no; for example 13 and 26 (it doubles) and 26 and 39 (it does not). Write each number as its prime factors. Doubling one number adds a 2 only to that number's factors, and the 2 joins the HCF only if the other number also has a 2 to spare. (1) 13 and 26 have HCF 13. Doubling 13 gives 26 and 26, whose HCF is 26: it doubled, because 26 = 2 × 13 already had a 2 left over. (2) 26 and 39 have HCF 13. Doubling 26 gives 52 = 2 × 2 × 13 and 39 = 3 × 13, whose HCF is still 13: 39 has no 2, so the new 2 cannot be shared. (HCF from prime factorisations, pages 52 to 54.)

Question 7. Use the division ladder to find both the HCF and the LCM of 104 and 156.

Answer: HCF 52, LCM 312. Divide both numbers by a common prime at each step: by 2 to get 52 and 78, by 2 again to get 26 and 39, by 13 to get 2 and 3. Now 2 and 3 have no common factor, so stop. The HCF is the product of the divisors: 2 × 2 × 13 = 52. The LCM is that times the two numbers left at the bottom: 52 × 2 × 3 = 312. Check: 52 × 312 = 16224 = 104 × 156. (The procedure for the HCF and the LCM together, pages 60 to 62.)

Question 8. A box of sweets can be shared equally, with none left over, among 8 children, or among 12 children, or among 18 children. The box has more than 160 sweets. What is the smallest number of sweets it could have?

Answer: 216. The number of sweets must be a common multiple of 8, 12 and 18. 8 = 2 × 2 × 2, 12 = 2 × 2 × 3 and 18 = 2 × 3 × 3, so the LCM is 2 × 2 × 2 × 3 × 3 = 72. Every common multiple is a multiple of 72: 72, 144, 216, … The first one above 160 is 216. (Common multiples and the LCM of more than two numbers, pages 55 to 58.)

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Sources

All opened and read on 1 November 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 46 for this chapter.

The rules are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.