Practice Questions: Number Play | Mathematics (Ganita Prakash) Class 7 - New NCERT PDF Download

Q1. A door is initially unlocked. Maya toggles the lock 60 times, where each toggle switches the state (unlocked to locked or locked to unlocked). Will the door be locked or unlocked at the end?

Sol: Maya toggles the lock 60 times.
Each toggle changes the door’s state (locked to unlocked or unlocked to locked).
Starting from unlocked:
An odd number of toggles will leave the door locked, and an even number of toggles will leave it unlocked.
Since 50 is even, after 60 toggles, the door will be unlocked.

Q2. In a cryptarithm where T + T = ST, where T and S are digits and ST is a two-digit number, what is the value of S?

Sol:
2T = ST means 2T = 10S + T. 
Subtract T: 
T = 10S. 
Since T and S are digits, S = 1 gives T = 10 (not a digit). 
Testing T = 5: 
2 × 5 = 10, 
ST = 10 (S = 1). 
Thus, S = 1

Q3. What is the 15th odd number in the sequence 1, 3, 5, 7, …?

Sol:
The nth odd number is given by 2n - 1. 
For n = 15: 
2 × 15 - 1 
= 30 - 1 
= 29

Q4. In the Virahanka-Fibonacci sequence (1, 2, 3, 5, 8, 13, 21, …), what is the parity of the 7th term?

Sol:
The sequence is 1, 2, 3, 5, 8, 13, 21. 
The 7th term is 21. 
Since 21 ÷ 2 leaves a remainder of 1, 
it is odd

Q5. How many ways can 6 be expressed as a sum of 1s and 2s?

Sol:
The number of ways to write n as a sum of 1s and 2s corresponds to the (n+1)th Virahanka-Fibonacci number. 
For n = 6, the sequence is 1, 2, 3, 5, 8, 13, 21, …, 
so the 7th term is 13.

Q6.In a 3 × 3 magic square with center number 10, what is the magic sum?

Sol: In a 3 × 3 magic square with numbers 1 to 9, the center is 5, and the magic sum is 15 (sum of 1 to 9 = 45, divided by 3). 
If the center is scaled to 10 (multiplied by 2), 
all numbers are scaled by 2, 
so the magic sum is 15 × 2 = 30.

Q7. What is the parity of the number of small squares in a 3 × 6 grid?

Sol:
The number of squares is 3 × 6 = 18. Since 3 is odd and 6 is even, odd × even = even. Thus, the parity is even.

 Q8. Priya has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?

Sol: 
​Suppose there are 50 random sheets from a book. 
Each sheet has two page numbers, one odd page number on the front in the form 2n – 1, one even page number on the back, in the form 2n.
So, the total of both page numbers on one sheet is: (2n – 1) + (2n) = 4n – 1
For 50 sheets, if n₁, n₂, ….., n₅₀ are the respective sheet numbers (not necessarily consecutive), then the total sum of all 50 sheets is:
(4n₁ – 1) + (4n₂ – 1) + ….. + (4n₅₀ – 1)
This can be written as: 4(n₁ + n₂ + …… + n₅₀) – 50
Now, 4(n₁ + n₂ + …… + n₅₀) is always divisible by 4 because it is a multiple of 4. But when we subtract 50 from it, the result is not divisible by 4 (because 50 is not a multiple of 4). So, the total sum of the 50 page numbers will not be divisible by 4. 
Hence, the sum of the page numbers of the loose sheets can never be 6000, because 6000 is divisible by 4 (since 6000 ÷ 4 = 1500)