Class 7 Maths Chapter 6 Short and Long Question Answers - Number Play

Q1: Priya is experimenting with sums of odd numbers. What happens to the parity of the sum when she adds 
(a) 3 odd numbers
(b) 4 odd numbers?

Sol: For 3 odd numbers (e.g., 1 + 3 + 5 = 9), the sum is odd because each odd number contributes one unpaired unit, and 3 unpaired units remain unpaired.

For 4 odd numbers (e.g., 1 + 3 + 5 + 7 = 16), the sum is even because the 4 unpaired units can be paired.
(a) Odd, (b) Even

Q2: Anjali has 5 boxes and number cards with odd numbers (1, 3, 5, 7, 9, ...). She needs to place one card in each box so they sum to 20. Is this possible?

Sol: The sum of 5 odd numbers is always odd (since each odd number adds an unpaired unit, and 5 unpaired units remain odd). Since 20 is even, it’s impossible.
 No, it’s not possible.

Q3: Rohan observes the Virahãnka-Fibonacci sequence: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... What is the parity of the 7th term (21)?

Sol: The 7th term is 21. Since 21 ÷ 2 = 10 remainder 1,
it’s odd.

Q4: In the cryptarithm P + P + P = QP, where P and Q are digits, and QP is a two-digit number, find P and Q.

Sol: Let P be a digit (0–9). 
Then, 3P = QP (a two-digit number with units digit P). 
This implies 3P = 10Q + P. Subtract P: 2P = 10Q. 
Thus, P = 5Q. Since P and Q are digits, Q = 1, P = 5 (as 5 × 1 = 5). 
Check: 5 + 5 + 5 = 15, and QP = 15 (Q = 1, P = 5).

Q5: Meena wants to write 6 as a sum of 1s and 2s in all possible ways. How many ways can she do this?

Sol: This follows the Virahãnka-Fibonacci sequence, where the number of ways to write n as a sum of 1s and 2s is the (n+1)th term. 
For n = 6, the 7th term is 13 (sequence: 1, 2, 3, 5, 8, 13, 21, ...).
 Hence there are 13 ways

Q6: Siblings Vikram and Nisha, born one year apart, claim the sum of their ages is 45. Is this possible?

Sol: Their ages are consecutive (e.g., n and n+1). 
The sum n + (n+1) = 2n + 1 is always odd (since 2n is even, adding 1 makes it odd). 
Since 45 is odd, it’s possible.
Example: 22 + 23 = 45.
 Yes, it’s possible.

Q7: What is the 50th odd number in the sequence 1, 3, 5, 7, ...?

Sol: The nth odd number is given by the formula 2n - 1. For n = 50, 2 × 50 - 1 = 100 - 1 = 99.

Q8: Arjun has a 15 × 22 grid. What is the parity of the number of small squares in this grid?

Sol: The number of squares is 15 × 22. 
Parity depends on the factors: 15 is odd, 22 is even. 
Odd × Even = Even (e.g., 15 × 22 = 330, which is even).
 Hence,  Even