CBSE Class 7 Mathematics Number Play Practice Paper Assignment Set 03 with Answers

Here is CBSE Class 7 Maths Ganita Prakash Chapter 6 Number Play Practice Paper 18: 20 questions for 40 marks in Sections A to E, each followed by a detailed, step-by-step solution. The questions test parity rules for sums, differences and products such as 75 × n, expressions for multiples of 4 and for odd numbers, the smallest and largest possible row sums of a 1–9 grid, the cryptarithm KP + KP = PRR, the Virahāṅka rule for rhythms of up to 11 beats, and a magic square made from the numbers 12 to 20.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the parity of 2 odd + 6 even numbers.
Answer:
The 2 odd numbers: odd + odd = even.
The 6 even numbers add up to an even number.
Total = even + even = even.
Example: 1 + 3 + 2 + 4 + 6 + 8 + 10 + 12 = 46.
In simple words: There are 2 odd numbers (an even count), so the total is even.

Teacher's Note:
1. Count only the odd numbers.
2. An even count of odd numbers gives an even sum.

Question 12. Without multiplying, find the parity of 129 × 131.
Answer:
129 is odd (last digit 9) and 131 is odd (last digit 1).
Odd × odd = odd.
So 129 × 131 is odd.
In simple words: Both numbers are odd, so their product is odd.

Teacher's Note:
1. The parity of a number depends only on its last digit.
2. Only a product of odd numbers is odd.

Question 13. Give any two expressions always giving odd numbers.
Answer:
For a whole number n, 2n is even, so 2n + 1 is always odd.
6n is even (6n = 2 × 3n), so 6n + 5 is always odd.
Reason: even + odd = odd. Check: n = 0, 1, 2 gives 1, 3, 5 and 5, 11, 17.
In simple words: A multiple of 2 plus an odd number is always odd, like 2n + 1 and 6n + 5.

Teacher's Note:
1. Expressions like 6n − 5 also work, but only for n ≥ 1.
2. Show a quick check with small values of n.

Question 14. Write the rule for the nth term of the Fibonacci/Virahāṅka sequence.
Answer:
Each term after the first two is the sum of the two terms just before it.
If V_n is the nth term, then V_n = V_{n − 1} + V_{n − 2}, starting with V₁ = 1 and V₂ = 2.
Example: V₃ = 2 + 1 = 3, V₄ = 3 + 2 = 5, V₅ = 5 + 3 = 8.
In simple words: To get any term, add the two terms before it.

Teacher's Note:
1. The rule needs two starting terms.
2. In rhythms, V_n is the number of rhythms of n beats.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Explain why the sum of an even number of odd numbers is always even.
Answer:
Odd + odd = even, because the single extra unit of each odd number joins up to make one more pair.
With an even count of odd numbers, they can all be grouped into pairs, and each pair has an even sum.
Adding even numbers always gives an even number.
So the sum is always even. Example: 3 + 5 + 7 + 9 = 24.
In simple words: Group the odd numbers in twos; each pair is even, so the whole sum is even.

Teacher's Note:
1. Use pairing in your explanation.
2. With an odd count, one odd number is left unpaired, so the sum would be odd.

Question 16. Show that the parity of 75 × n depends only on n.
Answer:
75 is odd.
If n is even: odd × even = even, so 75 × n is even.
If n is odd: odd × odd = odd, so 75 × n is odd.
So 75 × n has the same parity as n; it depends only on n. Examples: 75 × 2 = 150 (even), 75 × 3 = 225 (odd).
In simple words: 75 is odd, so multiplying by it keeps the parity of n unchanged.

Teacher's Note:
1. Check both cases, n even and n odd.
2. If the fixed number were even, the product would always be even.

Question 17. In a 3 × 3 grid filled with the numbers 1 to 9 (each used once), can the three row sums be 10, 22 and 5? Explain.
Answer:
The smallest possible row sum is 1 + 2 + 3 = 6, so a row sum of 5 is impossible.
Also, the three rows together use all the numbers, so the row sums must add up to 1 + 2 + … + 9 = 45. But 10 + 22 + 5 = 37, not 45.
So No, the row sums cannot be 10, 22 and 5.
In simple words: A row cannot add up to less than 6, and the three row sums must total 45, so 10, 22, 5 is impossible.

Teacher's Note:
1. Possible row sums lie between 6 and 24.
2. The three row sums of a 1–9 grid always add up to 45.

Section D: Long Answer Question (5 Marks)

Teacher's Note:
1. Adding the same number to every cell keeps the square magic.
2. Check: 12 + 13 + … + 20 = 144 and 144 ÷ 3 = 48.
3. The centre is the middle number, 16.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1:
A mathematics club prepares giant floor grids for a “Walk on the Numbers” activity. Each floor mat uses rectangular grids. Three mats proposed are 63 × 97, 84 × 111, and 92 × 115. The school wants only mats with an even number of squares because they need to be folded exactly into two identical halves for transporting. The club does not use multiplication; they apply parity rules: odd × odd = odd; even × odd = even. Students must decide which mats qualify and which must be redesigned for symmetrical folding and cost-effective printing.
Answer the following questions:
(a) Parity of 63 × 97?
(b) Parity of 92 × 115?
(c) Which mats are unusable for equal folding?
(d) Which mats should be chosen?
Answer:
(a) 63 and 97 are both odd, so 63 × 97 is odd.
(b) 92 is even, so 92 × 115 is even.
(c) 84 × 111 is even too, because 84 is even. Only an odd total cannot be folded into two identical halves, so only the 63 × 97 mat is unusable.
(d) Both even mats qualify: 84 × 111 and 92 × 115.
In simple words: Only 63 × 97 has an odd number of squares; the other two mats are even and can be chosen.

Teacher's Note:
1. Check every mat; 84 × 111 is even even though 111 is odd.
2. Folding along the even side gives two identical halves.

Question 20. Case Study 2:
A theatre group is composing rhythmic background music using 1-beat and 2-beat syllables. They require rhythms of 7, 8, 9, and 11 beats. The music director teaches Virahāṅka’s principle: rhythms(n) = rhythms(n − 1) + rhythms(n − 2). Students know rhythms(5) = 8, rhythms(6) = 13. They compute rhythms(7) = 21, rhythms(8) = 34. They now attempt 9 and 11-beat compositions without listing combinations. The group must decide which beat length allows the most rhythmic variety and which lengths produce odd or even counts.
Answer the following questions:
(a) Rhythms for 9 beats?
(b) Rhythms for 11 beats?
(c) Which beat length gives maximum patterns among 7, 8, 9, 11?
(d) Parity of 144?
Answer:
(a) rhythms(9) = rhythms(8) + rhythms(7) = 34 + 21 = 55.
(b) rhythms(10) = 55 + 34 = 89, so rhythms(11) = 89 + 55 = 144.
(c) The counts are 21, 34, 55 and 144 for 7, 8, 9 and 11 beats, so 11 beats give the most patterns (144).
(d) 144 ends in 4, so it is even.
In simple words: Adding the last two counts gives 55 for 9 beats and 144 for 11 beats; 11 beats give the most, and 144 is even.

Teacher's Note:
1. You must find rhythms(10) before rhythms(11).
2. The counts always grow, so the longest length gives the most patterns.
3. Parity check: 89 (odd) + 55 (odd) = even.