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Class 6 Maths: Prime Time, 8 practice questions with answers

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These are 8 practice questions on Chapter 5, Prime Time, of the new NCERT Class 6 Maths textbook, Ganita Prakash. Each has a worked answer and the textbook page it comes from. They ask you to test a claim about common multiples, spot where a sieve was stopped too early, decide whether 1 is prime, build a co-prime pair of composite numbers, find where a prime factorisation stopped too soon, correct a wrong co-prime check, reason about divisibility by 4 and by 8, and see why the last digit alone cannot decide divisibility by 4. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help, and every answer was checked against the textbook page it cites, 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: Mathematics, Textbook for Grade 6, Chapter 5, printed pages 107 to 128):

Multiples, primes and the sieve

Question 1. Dev says, "A common multiple of two numbers is always bigger than both of them." Test his claim with 34 and 68. Is he right?

Answer: no. The multiples of 34 are 34, 68, 102, … and the multiples of 68 start with 68, …, so 68 is a common multiple of 34 and 68 (68 = 34 × 2 and 68 = 68 × 1). It is not bigger than 68, one of the two numbers. A common multiple is at least as big as the larger number, and it can be equal to it, so the first common multiple here, 68, is even less than the product 34 × 68. (Common multiples, page 107; a number is a factor and a multiple of itself, page 109; the first common multiple against the product, page 116.)

Question 2. Reena is using the sieve on the numbers from 1 to 50. She has crossed out 1 and the multiples of 2, 3 and 5 that come after them, and she says, "Every number still standing is prime." Find her mistake. Which number gives her away?

Answer: 49 is still standing but it is composite; she should go on to 7. After crossing out 1 and the later multiples of 2, 3 and 5, the numbers left are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 and 49. The next uncrossed number after 5 is 7, so she must circle 7 and cross out its later multiples: 14, 21, 28, 35 and 42 are already crossed out, and 49 is the new one. Since 49 = 7 × 7 it has a factor other than 1 and itself, so it is composite. The method says to continue until every number is either circled or crossed out. (The sieve, page 113.)

Question 3. Nandini says, "The number 1 is a prime, because its only factor is 1, and primes have only 1 and the number itself as factors." Find her mistake.

Answer: a prime must have exactly two factors; 1 has only one. A prime number has two factors, 1 and itself, and these are different numbers. The number 1 has just one factor, so it is not prime. It is not composite either, because composite numbers have more than two factors. So 1 is neither prime nor composite. It also has no prime factorisation, since it is not divisible by any prime number. (Primes and composites, pages 112 and 113; 1 has no prime factorisation, page 118.)

Co-primes and prime factorisation

Question 4. Find two composite numbers that are co-prime to each other, and show that they are co-prime using prime factorisations.

Answer: for example, 34 and 77. 34 = 2 × 17 and 77 = 7 × 11, so both are composite. The prime factors of 34 are 2 and 17, and the prime factors of 77 are 7 and 11. They have no common prime factor, so 34 and 77 are co-prime. Many other pairs work; check yours the same way. (Co-prime numbers and the prime-factorisation check, pages 116 and 120 to 121; summary, page 128.)

Question 5. Raju writes the prime factorisation of 294 as 2 × 3 × 49. Find his mistake and give the correct prime factorisation.

Answer: 49 is not a prime; the factorisation is 2 × 3 × 7 × 7. Raju stopped too soon: 49 is composite because 49 = 7 × 7, so it can be broken down further. Replacing 49 gives 294 = 2 × 3 × 7 × 7, and now every factor is prime. Check: 2 × 3 = 6, 6 × 7 = 42 and 42 × 7 = 294. The idea is to keep breaking composite numbers into factors till only primes are left. (Prime factorisation, page 118.)

Question 6. Sam finds that 138 = 2 × 3 × 23. He says, "138 is even and is a multiple of 3, so no number from 2 to 10 can be co-prime to 138." Find his mistake. Which numbers from 2 to 10 are co-prime to 138?

Answer: 5 and 7 are co-prime to 138. Two numbers are co-prime when they have no common prime factor. The prime factors of 138 are 2, 3 and 23. The numbers 2, 4, 6, 8 and 10 share the prime factor 2 with 138, and 3, 6 and 9 share the prime factor 3, so none of these is co-prime to 138. But 5 and 7 are primes that are not among 2, 3 and 23, so they have no common prime factor with 138. Sam checked only the primes 2 and 3 and forgot the numbers made of other primes. (Using prime factorisation to check co-primes, pages 120 and 121; summary, page 128.)

Divisibility tests

Question 7. Meena says, "A number that is divisible by 8 is always divisible by 4, but a number divisible by 4 need not be divisible by 8." Is she right? Explain with prime factorisations and give one example for each part.

Answer: yes, she is right. 8 = 2 × 2 × 2 and 4 = 2 × 2. If a number is divisible by another, the prime factorisation of the second is included in the first. A multiple of 8 has three 2s in its factorisation, so it also has the two 2s of 4, and is divisible by 4. For example, 424 = 2 × 2 × 2 × 53, so 424 is divisible by 8 (424 = 8 × 53) and by 4 (424 = 4 × 106). The reverse fails: 164 = 2 × 2 × 41 has only two 2s, so it is divisible by 4 (164 = 4 × 41) but not by 8, because 8 needs three 2s. The test for 8 agrees: 164 ÷ 8 leaves a remainder of 4. (Divisibility from prime factorisations, pages 121 and 122; the tests for 4 and 8, pages 124 and 125.)

Question 8. Mira says, "3,926 is divisible by 4, because it ends in 6 and 6 is an even digit." Find her mistake. Is 3,926 divisible by 4?

Answer: the last digit cannot decide divisibility by 4; 3,926 is not divisible by 4. Numbers with the same last digit can behave differently: 12 and 22 both end in 2, but 12 is a multiple of 4 and 22 is not. To test for 4, look at the number formed by the last two digits. Here the last two digits form 26, and 26 = 4 × 6 + 2, so it is not divisible by 4, and neither is 3,926 (3,926 = 4 × 981 + 2). Mira's reasoning would work for 2: a number ending in 6 is divisible by 2. (Divisibility by 2 and by 4, pages 123 and 124.)

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Sources

All opened and read on 22 October 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 106.

The facts are from the textbook pages above. The questions and worked answers are our own. This guide is general information; follow your teacher and the textbook.