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Class 7 Maths: Number Play, 8 practice questions with answers

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These are 8 practice questions on Chapter 6, Number Play, of the new NCERT Class 7 Maths textbook, Ganita Prakash, Part 1. Each has a worked answer and the textbook page it comes from. They cover the parity of products, four consecutive numbers, the row sums of a 3 × 3 grid, why n × n + n is always even, a basketball score, a sequence on the Virahāṅka rule, rhythms of 1-beat and 3-beat sounds, and a multiplication cryptarithm. 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 questions retells a worked example, activity or exercise from the chapter. 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 1, Chapter 6, printed pages 127 to 145):

Parity

Question 1. Without multiplying, decide whether each product is odd or even, and explain: (a) 13 × 15 × 17; (b) 21 × 35 × 48 × 9; (c) the product of any ten odd numbers. (d) What has to be true of a list of numbers for their product to be odd?

Answer: (a) odd; (b) even; (c) odd; (d) every number in the list must be odd. Think of a product as a grid, rows times columns: if one factor is even, the rows (or each row) can be split into pairs, so the whole product is even; odd × odd is odd. (a) 13 × 15 is odd, and that odd number × 17 is again odd. (b) 48 is even, so the whole product is even, whatever the other numbers are. (c) Multiplying odd numbers one at a time always gives odd × odd = odd, so ten of them give an odd product. (d) A single even factor makes the product even, so the product is odd only when all the numbers are odd. (Parity of the squares in a grid, pages 131 and 132; parity of products, page 145.)

Question 2. (a) Take any four consecutive numbers, such as 6, 7, 8, 9. Is their sum odd or even, and is it always so? (b) Write the sum of n, n + 1, n + 2 and n + 3 as an expression. (c) Can four consecutive numbers add up to 50? To 52?

Answer: (a) always even; (b) 4n + 6; (c) 50 yes (11, 12, 13, 14); 52 no. (a) Four consecutive numbers make two pairs of consecutive numbers, and each such pair adds up to an odd number. Odd + odd is even, so the sum is always even: 6 + 7 + 8 + 9 = 30. (b) n + (n + 1) + (n + 2) + (n + 3) = 4n + 6. (c) 4n + 6 = 50 gives 4n = 44 and n = 11, so 11 + 12 + 13 + 14 = 50. For 52, 4n = 46, but 46 is not a multiple of 4, so no four consecutive numbers add up to 52, even though 52 is even. (Consecutive numbers and parity, pages 130 and 131; expressions, page 132.)

Question 3. The numbers 2 to 10 are placed in a 3 × 3 grid, each once, and the sum of each row is written beside it. (a) What must the three row sums add up to? (b) Can the row sums be 17, 18 and 20? (c) Can they be 16, 18 and 20? If so, give the three rows.

Answer: (a) 54; (b) no; (c) yes, for example rows {2, 4, 10}, {3, 6, 9}, {5, 7, 8}. (a) The three rows together use every number once, so the row sums add up to 2 + 3 + … + 10 = 54. (b) 17 + 18 + 20 = 55, not 54, so no filling gives those row sums. (c) 16 + 18 + 20 = 54, so it is possible: 2 + 4 + 10 = 16, 3 + 6 + 9 = 18 and 5 + 7 + 8 = 20. (Why the row sums of a 3 × 3 grid always add up to the same total, page 134.)

Grids and formulas

Question 4. (a) Work out n × n + n for n = 1, 2, 3, 4 and 5. What do you notice about the parity? (b) Explain why it is always even, using the fact that n × n + n = n × (n + 1).

Answer: (a) 2, 6, 12, 20, 30: all even; (b) n and n + 1 are consecutive, so one of them is even. (a) 1 + 1 = 2, 4 + 2 = 6, 9 + 3 = 12, 16 + 4 = 20 and 25 + 5 = 30: all even. (b) n × (n + 1) multiplies two consecutive numbers, and of any two consecutive numbers one is even and the other odd. A product with an even factor is even, so n × n + n is always even, whatever n is. (Consecutive numbers, page 130; parity of expressions, page 132; parity of products, pages 131 and 145.)

Question 5. A basketball team scores only with 2-point and 3-point baskets. In one match it scores 47 points. (a) Was the number of 3-point baskets odd or even? Why? (b) Give one possible number of each kind of basket.

Answer: (a) odd; (b) for example, 13 three-pointers and 4 two-pointers. (a) The 2-point baskets always give an even number of points, since 2 times any number is even. 47 is odd, and even + odd is odd, so the points from 3-point baskets must be odd. 3 times a number is odd only when that number is odd, so the number of 3-point baskets was odd. (b) 13 × 3 = 39 and 47 − 39 = 8 = 4 × 2, so 13 three-pointers and 4 two-pointers work (so do, for example, 1 and 22, or 15 and 1). (Parity of sums, pages 129 to 131; even numbers as 2 times a number, page 132.)

Sequences, rhythms and a cryptarithm

Question 6. A sequence follows the Virahāṅka rule (each number is the sum of the two before it) but starts with 2 and 5. (a) Write the next five numbers. (b) Which is the first number in the sequence that is greater than 100, and in which place does it come? (c) If the sequence started with 2 and 6 instead, what would its 5th number be?

Answer: (a) 7, 12, 19, 31, 50; (b) 131, the 9th number; (c) 22. (a) 2 + 5 = 7, 5 + 7 = 12, 7 + 12 = 19, 12 + 19 = 31, 19 + 31 = 50. (b) Carry on with the rule: 31 + 50 = 81, the 8th number, still below 100; then 50 + 81 = 131, the 9th number and the first above 100. (c) 2, 6, then 2 + 6 = 8, 6 + 8 = 14, 8 + 14 = 22. Only the two numbers just before are needed each time. (The rule for the Virahāṅka numbers, pages 141 and 142.)

Question 7. In a new rhythm game, a short sound takes 1 beat and a very long sound takes 3 beats. (a) List all the rhythms of 4 beats. (b) Explain why the number of n-beat rhythms is the number of (n − 1)-beat rhythms plus the number of (n − 3)-beat rhythms. (c) Using 1, 1 and 2 as the numbers of 1-, 2- and 3-beat rhythms, find how many 7-beat rhythms there are.

Answer: (a) 1+1+1+1, 1+3 and 3+1; (b) every rhythm starts with a 1-beat or a 3-beat sound; (c) 9. (a) The 4-beat rhythms are 1 + 1 + 1 + 1, 1 + 3 and 3 + 1: 3 of them. (b) Every rhythm begins with either a 1-beat sound, after which the rest is an (n − 1)-beat rhythm, or a 3-beat sound, after which the rest is an (n − 3)-beat rhythm, so the two counts add, in the same way the chapter counts rhythms of 1-beat and 2-beat syllables. (c) With 1, 1 and 2 rhythms of 1, 2 and 3 beats: 4 beats: 2 + 1 = 3; 5 beats: 3 + 1 = 4; 6 beats: 4 + 2 = 6; 7 beats: 6 + 3 = 9. (Counting rhythms by their first syllable, pages 140 and 141.)

Question 8. In the multiplication A × A = BA, A and B stand for digits, and BA is a 2-digit number whose units digit is A. Find every pair of digits that works.

Answer: A = 5, B = 2 (5 × 5 = 25) and A = 6, B = 3 (6 × 6 = 36). A × A must end in the digit A itself and must have two digits, so A is at least 4. Check the squares: 4 × 4 = 16, 5 × 5 = 25, 6 × 6 = 36, 7 × 7 = 49, 8 × 8 = 64, 9 × 9 = 81. Only 25 ends in 5 and 36 ends in 6, so the answers are 25 and 36. (Letters standing for digits, page 143; this puzzle uses multiplication, while the chapter's puzzles use addition.)

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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 + 126 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.