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

CBSE Class 7 Maths Ganita Prakash Chapter 6 Number Play Practice Paper 16 is a 40-mark test of 20 questions arranged in Sections A to E, and every question below comes with a step-by-step answer. It checks the parity of sums and of grid products, algebraic expressions for odd and even numbers, the Virahāṅka–Fibonacci sequence and rhythm counting, a simple cryptarithm, and 3 × 3 magic squares, including building one from the numbers 2 to 10.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the parity of the sum: 2 even numbers + 3 odd numbers.
Answer:
The sum of the 2 even numbers is even.
The 3 odd numbers: odd + odd = even, and even + odd = odd. So their sum is odd.
Total = even + odd = odd.
Example: 2 + 4 + 1 + 3 + 5 = 15, which is odd.
In simple words: Three odd numbers make the total odd; the even numbers do not change that.

Teacher's Note:
1. Look only at how many odd numbers there are.
2. Give a quick example to check your rule.

Question 12. A grid is 41 × 39. Without multiplication, decide the parity of the number of small squares.
Answer:
Number of small squares = 41 × 39.
41 is odd and 39 is odd.
Odd × odd = odd, so the number of small squares is odd.
In simple words: Both sides are odd, so the number of squares is odd.

Teacher's Note:
1. Odd × odd is the only way to get an odd product.
2. If even one side were even, the count would be even.

Question 13. Write two expressions that always produce odd numbers.
Answer:
For any whole number n, 2n is even, so 2n + 1 is always odd (1, 3, 5, …).
Similarly 4n is even, so 4n + 1 is always odd (1, 5, 9, …).
Reason: even + odd = odd. (Other answers such as 2n + 3 or 6n + 5 are also correct.)
In simple words: Add an odd number to a multiple of 2: 2n + 1 and 4n + 1 are always odd.

Teacher's Note:
1. Any multiple of 2 plus an odd number is odd.
2. Check by putting n = 0, 1, 2.

Question 14. Write the next Virahāṅka number after 89.
Answer:
Each Virahāṅka number is the sum of the two numbers before it.
The numbers before 89 are …, 34, 55, 89.
Next number = 55 + 89 = 144.
In simple words: Add the last two numbers: 55 + 89 = 144.

Teacher's Note:
1. You need the number just before 89, which is 55.
2. Write a few terms of the sequence to be sure.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Lakpa has an odd number of ₹1 coins, an odd number of ₹5 coins, and an even number of ₹10 coins. Can the total be ₹205?
Answer:
Value of the ₹1 coins = odd × 1 = odd.
Value of the ₹5 coins = odd × 5 = odd (odd × odd = odd).
Value of the ₹10 coins = even × 10 = even.
Total = odd + odd + even = even + even = even.
But 205 is odd. An even total can never be 205.
So No, the total cannot be ₹205.
In simple words: The coins always add up to an even amount, and 205 is odd, so it is impossible.

Teacher's Note:
1. Find the parity of each part first, then add the parities.
2. A parity argument proves 'impossible' without trying every case.

Question 16. Construct an expression that generates all even numbers. Explain.
Answer:
Every even number is a multiple of 2, that is, 2 × (some whole number).
So the expression is 2n, where n is a whole number.
n = 0, 1, 2, 3, 4, … gives 0, 2, 4, 6, 8, …: every even number appears, and no odd number appears.
In fact 2n is the nth even number (taking n = 1, 2, 3, …).
In simple words: Doubling any whole number gives an even number, so 2n produces all even numbers.

Teacher's Note:
1. Explain both parts: 2n is always even, and every even number can be written as 2n.
2. Expressions like 2n + 4 also give only even numbers, but they miss 0 and 2.

Question 17. A magic square row contains the numbers a, b, c. If the magic sum is 15, express c in terms of a and b.
Answer:
In a magic square every row adds up to the magic sum.
So a + b + c = 15.
Subtracting a + b from both sides: c = 15 − a − b, that is, c = 15 − (a + b).
Check with the row 4, 9, 2: c = 15 − 4 − 9 = 2.
In simple words: The three numbers make 15, so c is 15 minus the other two.

Teacher's Note:
1. Write the rule as an equation first, then rearrange it.
2. Check your expression with a real row of the Lo Shu square.

Section D: Long Answer Question (5 Marks)

Teacher's Note:
1. Adding the same number to every cell keeps a magic square magic.
2. The centre is always the middle number of the list (here 6).
3. Do not forget to check both diagonals.

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

Question 19. Case Study 1:
A school science club is preparing display boards for an exhibition. They design rectangular grids made of small squares to arrange facts, pictures and diagrams. Meena is assigned a board of 27 × 13 squares, Prakash gets a 40 × 41 board, and Riya gets a 58 × 79 board. They want to know whether their total number of squares is even or odd because the printing press charges differently for even and odd totals. They try to calculate fast without multiplication. They also want to check which boards can be divided exactly into equal halves for symmetrical printing.
Answer the following questions:
(a) What is the parity of Meena’s board (27 × 13)?
(b) What is the parity of the 40 × 41 board?
(c) Which among the three boards cannot be divided into two equal rectangular halves?
(d) If they want all boards with even squares, which one must be replaced?
Answer:
(a) 27 and 13 are both odd. Odd × odd = odd, so Meena’s board has an odd number of squares.
(b) 40 is even. Even × odd = even, so the 40 × 41 board has an even number of squares.
(c) Riya’s board 58 × 79 is even (58 is even). An even total can be split equally, e.g. 40 × 41 into two 20 × 41 halves and 58 × 79 into two 29 × 79 halves. An odd total cannot be split into two equal whole numbers, so Meena’s 27 × 13 board cannot be halved.
(d) Only Meena’s board has an odd number of squares, so the 27 × 13 board must be replaced.
In simple words: Only the board with both sides odd (27 × 13) has an odd number of squares, so only it cannot be halved and must be replaced.

Teacher's Note:
1. Look at the sides of each grid, not at the product.
2. To cut a grid into two equal rectangles, one side must be even.
3. Odd totals can never be shared equally into two whole parts.

Question 20. Case Study 2:
A music teacher is planning a rhythmic performance using Indian classical beats. Each rhythm is made using short beats (1 beat) and long beats (2 beats). Students must create patterns that add up to 6, 7, or 8 beats. The teacher explains the Virahāṅka–Fibonacci rule: the number of possible rhythms for n beats equals the sum of the rhythms for (n − 1) and (n − 2) beats. Students test patterns, listing possibilities on chart papers. They want to compare difficulty levels and predict the number of rhythms for 9 beats without actually listing them.
Answer the following questions:
(a) Number of rhythms of 6 beats?
(b) Number of rhythms of 7 beats?
(c) Number of rhythms of 8 beats?
(d) Predict the number of rhythms for 9 beats.
Answer:
1 beat has 1 rhythm (S) and 2 beats have 2 rhythms (SS, L). Using the rule, the counts for 1, 2, 3, 4, 5 beats are 1, 2, 3, 5, 8.
(a) 6 beats: 5 + 8 = 13 rhythms.
(b) 7 beats: 8 + 13 = 21 rhythms.
(c) 8 beats: 13 + 21 = 34 rhythms.
(d) 9 beats: 21 + 34 = 55 rhythms.
In simple words: Keep adding the last two counts: 13, 21, 34 and then 55 rhythms for 9 beats.

Teacher's Note:
1. A rhythm of n beats ends either in a short beat (leaving n − 1 beats) or a long beat (leaving n − 2 beats); that is why the rule works.
2. Start the list from 1 beat = 1 and 2 beats = 2.