CBSE Class 7 Mathematics Number Play Practice Paper Worksheet Set 02 with Answers

Practice Paper 17 for CBSE Class 7 Maths Ganita Prakash Chapter 6 (Number Play) has 20 questions worth 40 marks, spread over Sections A to E, with a clear worked answer under each one. This paper focuses on the parity of sums, differences and products, the formulas 2n and 2n − 1 for the nth even and odd numbers, two-digit cryptarithms, the 12th to 15th Virahāṅka numbers, and magic squares: why 5 sits at the centre, writing a whole square in terms of its top row, and scaling the Lo Shu square to a magic sum of 60.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the parity of the sum of 6 even numbers and 3 odd numbers.
Answer:
The sum of the 6 even numbers is even.
The 3 odd numbers: one pair makes an even number, and one odd number is left, so their sum is odd.
Total = even + odd = odd.
In simple words: There are 3 odd numbers (an odd count), so the total is odd.

Teacher's Note:
1. Even numbers never change the parity of a sum.
2. Example: 2 + 4 + 6 + 8 + 10 + 12 + 1 + 3 + 5 = 51, which is odd.

Question 12. Without multiplying, determine the parity of the number of squares in a 54 × 77 grid.
Answer:
Number of squares = 54 × 77.
54 is even and 77 is odd.
Even × odd = even, so the grid has an even number of squares.
In simple words: One side (54) is even, so the number of squares is even.

Teacher's Note:
1. Think of 54 rows: they pair up as 27 pairs, so the total is even.
2. Only odd × odd gives an odd product.

Question 13. Write the 12th and 13th terms of the Virahāṅka sequence 1, 2, 3, 5, 8, ….
Answer:
Each term is the sum of the two terms before it.
1, 2, 3, 5, 8, 13, 21, 34, 55, 89 (10th), 144 (11th), 233 (12th), 377 (13th).
So the 12th term = 233 and the 13th term = 377 (144 + 233 = 377).
In simple words: Keep adding the last two terms; the 12th term is 233 and the 13th is 377.

Teacher's Note:
1. Number each term as you write it to avoid miscounting.
2. This sequence starts 1, 2 because it counts rhythms of 1 beat and 2 beats.

Question 14. Find the nth even number in expression form.
Answer:
The even numbers are 2, 4, 6, 8, …, that is 2 × 1, 2 × 2, 2 × 3, 2 × 4, …
The 1st is 2 × 1, the 2nd is 2 × 2, so the nth is 2 × n.
nth even number = 2n. Check: the 5th even number is 2 × 5 = 10.
In simple words: Even numbers are the multiples of 2, so the nth one is 2n.

Teacher's Note:
1. Compare each term with its position number to find the rule.
2. The nth odd number is one less: 2n − 1.

Section C: Short Answer Questions (3 Marks Each)

Question 15. If a 3 × 4 grid has 12 small squares (even), predict the parity of 45 × 99. Explain.
Answer:
The 3 × 4 grid is even because one side (4) is even; even × odd = even.
In 45 × 99 both sides are odd. Picture 45 rows of 99 squares: each row is odd.
An odd number of odd rows adds up to an odd number (the extra squares in each row pair up except one).
So 45 × 99 is odd: odd × odd = odd.
In simple words: Here neither side is even, so unlike 3 × 4, the 45 × 99 grid has an odd number of squares.

Teacher's Note:
1. A product is even if at least one factor is even, and odd only if all factors are odd.
2. Explain the rule, do not just state the answer.

Question 16. Show that the expression 6n + 2 never produces an odd number.
Answer:
6n = 2 × 3n, so 6n is a multiple of 2 for every whole number n; it is even.
2 is also even, and even + even = even.
In fact 6n + 2 = 2(3n + 1), a multiple of 2.
So 6n + 2 is always even and never odd. Examples: n = 0, 1, 2 give 2, 8, 14.
In simple words: 6n + 2 is twice the number 3n + 1, so it is always even.

Teacher's Note:
1. Taking 2 out as a common factor proves evenness for every n.
2. Examples support the answer, but the factorisation proves it.

Teacher's Note:
1. Find the centre first; it links every opposite pair.
2. Always test your expressions with a known magic square.
3. Remember a + b + c = 15 for the top row as well.

Section D: Long Answer Question (5 Marks)

Teacher's Note:
1. Multiplying (or adding) the same number to every cell keeps a square magic.
2. Check all three rows, three columns and two diagonals.
3. The centre is 60 ÷ 3 = 20, as expected.

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

Question 19. Case Study 1:
A furniture workshop is designing wooden panels for a school wall. Panels are made in rectangular grids, and the supervisor wants to know which designs can be cut symmetrically into two equal rectangular pieces. One worker proposes a 39 × 51 panel, another suggests 62 × 85, and another recommends 72 × 111. Instead of calculating the actual product, the supervisor teaches them the parity rule: even × odd = even, odd × odd = odd. The printing partner charges extra for odd-numbered squares. Workers must quickly determine which panels are cost-efficient and which ones must be redesigned.
Answer the following questions:
(a) Parity of 39 × 51?
(b) Parity of 62 × 85?
(c) Which panel cannot be cut into equal halves?
(d) Which panels are cost-efficient?
Answer:
(a) 39 and 51 are both odd, and odd × odd = odd, so 39 × 51 is odd.
(b) 62 is even, and even × odd = even, so 62 × 85 is even.
(c) 72 × 111 is also even (72 is even). An odd number of squares cannot be split into two equal parts, so the 39 × 51 panel cannot be cut into equal halves.
(d) The panels with an even number of squares avoid the extra charge: 62 × 85 and 72 × 111.
In simple words: Only 39 × 51 has an odd number of squares; the other two panels are even, so they can be halved and are cost-efficient.

Teacher's Note:
1. Check whether any side is even; that decides everything here.
2. An even side can be cut in the middle to give two equal rectangles.
3. In (d) name both panels.

Question 20. Case Study 2:
A poetry club is creating rhythmic patterns for an inter-school performance. Each pattern is made using short syllables of 1 beat and long syllables of 2 beats. They must create rhythms of 5, 6, 7 and 8 beats to match different poem recitation speeds. Students discuss Virahāṅka’s rule: the number of rhythms for n beats equals rhythms of (n − 1) plus rhythms of (n − 2). They list a few examples: 5 beats → 8 rhythms, 6 beats → 13 rhythms. Now they want to predict the number for 7 beats and 8 beats without listing all possibilities. They must also justify the growth pattern.
Answer the following questions:
(a) How many rhythms are possible for 8 beats?
(b) How many rhythms are possible for 9 beats?
(c) Which beat length gives more rhythmic options: 8 or 9 beats?
(d) State the parity of the number of rhythms for 9 beats.
Answer:
Using the rule with 5 beats → 8 and 6 beats → 13: 7 beats → 8 + 13 = 21.
(a) 8 beats → 13 + 21 = 34 rhythms.
(b) 9 beats → 21 + 34 = 55 rhythms.
(c) 55 > 34, so 9 beats give more options. Each new count adds the previous two, so the counts keep growing.
(d) 55 is odd (21 is odd, 34 is even, and odd + even = odd).
In simple words: The counts go 21, 34, 55, so 9 beats give the most rhythms (55), and 55 is odd.

Teacher's Note:
1. Write the counts in a row: 8, 13, 21, 34, 55.
2. Parity can be found from the parities of the two previous terms.
3. The growth pattern: every rhythm of n beats ends in a 1-beat or a 2-beat syllable, giving the (n − 1) and (n − 2) counts.