Extra Questions for Class 7 Maths Chapter 6: Number Play – HOTS

Fresh HOTS practice questions for Class 7 Maths Chapter 6 “Number Play”, testing parity, magic squares, and the Virahanka sequence with new numbers. These Class 7 Mathematics Chapter 6 important questions are handy for last-minute exam practice.

Last Updated: September 23, 2026

Q1. Is the sum of 7 odd numbers always odd or even? Verify with an example.

Answer: Always odd (an odd count of odd numbers always sums to odd). Example: 1+3+5+7+9+11+13 = 49 (odd).

Q2. A 3×3 magic square has centre value 12. What is its magic sum, and what would the magic sum become if every number were tripled?

Answer: Magic sum = 3×12 = 36. After tripling every number, the new magic sum = 36×3 = 108.

Q3. Assertion-Reason — Assertion (A): The product of two odd numbers is always odd. Reason (R): An odd number…

Assertion (A): The product of two odd numbers is always odd.
Reason (R): An odd number can be written as 2k+1 for some integer k, and the product of two such numbers is also of the form 2m+1.

Answer: Both A and R are true, and R correctly explains A.

Q4. What are the next two terms after 4181 in the sequence 1, 2, 3, 5, 8, 13, …, 2584, 4181?

Answer: 2584+4181 = 6765; 4181+6765 = 10946

Q5. Spot the error — A student claims: "Since 6j−4 gives 2, 8, 14, 20 for j=1,2,3,4, it must generate all…

A student claims: “Since 6j−4 gives 2, 8, 14, 20 for j=1,2,3,4, it must generate all even numbers.” Explain the error.

Answer: This is incorrect. The expression 6j−4 only produces numbers that increase in steps of 6 (2, 8, 14, 20, …), so it skips many even numbers like 4, 6, 10, 12, etc. It generates only some even numbers, not all of them.

Q6. In how many ways can 5 be written as an ordered sum of 1s and 2s?

Answer: This equals the Virahanka term for n=5, which is 8 ways (matching the sequence 1,2,3,5,8,…).

Q7. A bulb starts OFF. It is toggled 124 times. Is it ON or OFF at the end?

Answer: Since 124 is even, an even number of toggles returns the bulb to its starting state, so it will be OFF.

Q8. Solve the cryptarithm: AA + B = CDD, where each letter is a single digit and AA means a 2-digit number with both digits equal to A.

Answer: Try A=9: AA=99. 99+B=CDD (a 3-digit number starting with C). For B=1: 99+1=100, so C=1, D=0, D=0. This gives A=9, B=1, C=1, D=0 (99+1=100 checks out).

Q9. Is 10×15 odd or even? Explain using the parity rule for products.

Answer: Even. Since at least one of the two numbers (10) is even, the product must be even, regardless of the other factor. (10×15 = 150, which is even.)

Q10. Create a magic square using the numbers 4 to 12 (9 consecutive numbers). What is its magic sum?

Answer: Add 3 to every cell of the classic 1–9 magic square (since 4 = 1+3). New magic sum = 15 + 3(3) = 24.

NCERT Solutions: Class 7 Maths Chapter 6

Revision Notes: Class 7 Maths Chapter 6 Revision Notes

Frequently Asked Questions

How can you quickly find the Lowest Common Multiple of two numbers using their prime factorisations?
The Lowest Common Multiple is found by taking the highest power of every prime factor that appears in either number and multiplying those together, ensuring the result is divisible by both original numbers.

Why is a number that ends in 0 always divisible by both 2 and 5?
A number ending in 0 is always a multiple of 10, and since 10 is the product of 2 and 5, any multiple of 10 is automatically divisible by both 2 and 5 as well.

Written by Satish

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