RBSE Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 16 Playing with Numbers Ex 16.2

Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer:
Given that 21y5 is a multiple of 9. i.e. 21y5 is divisible by 9.
Therefore, 2 + 1 + y + 5 = (8 + y) must be a multiple of 9 i.e., (8 + y) should be divisible by 9.
Hence, 8 + y should be 0, 9, 18, 27, ...........
⇒ 8 + y = 9 or 8 + y = 18
y = 1 or y = 10
[∵ 8 + y cannot be equal to zero.]
But y is a digit, therefore y = 10 is not possible. Thus y = 1.

Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers for the last problem. Why is this so?
Answer:
Given that 31z5 is a multiple of 9. So, the sum of its digits 3 + 1 + 3 + 5 = 9 + z is a multiple of 9.
∴ (9 + z) is either 0, or 9 or 18 or 27
But z is a digit, therefore (9 + z) must be equal to 9 or 18
i.e., 9 + z = 9 ⇒ z = 0 and 9 + z = 18 ⇒ z = 9

Question 3.
If, 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer:
If 24x is a multiple of 3, then the sum of its digits 2 + 4 + x = (6 + x) is a multiple of 3.
∴ (6 + x) is one of the numbers 0, 3, 6, 9, 12, 15, 18, ....
But x is a digit therefore, (6 + x) must be equal to 6 or 9 or 12 or 15.
i.e., 6 + x = 6 or 9 or 12 or 15
⇒ x = 0 or 3 or 6 or 9
Thus x can have any of the four different values, namely 0, 3, 6 or 9.

Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer:
Since 31z5 is a multiple of 3.
∴ The sum of the digits (3 + 1 + z + 5) is a multiple of 3.
So, 9 + z is one of these numbers 0, 3, 6, 9, 12. ......
When, 9 + z = 0 ⇒ z = - 9 (negative)
When, 9 + z = 3 ⇒ z = 3 - 9 = - 6 (negative)
When, 9 + z = 6 ⇒ z = 6 - 9 = -3 (negative)
When, 9 + z = 9 ⇒ z = 0
When, 9 + z = 12 ⇒ z = 3
When, 9 + z = 15 ⇒ z = 6
When, 9 + z = 18 ⇒ z = 9
When, 9 + z = 21 ⇒ z = 12 (not possible)
∴ The values of z are 0; 3, 6 or 9.
[∵ Negative values of z are not possible, and z is a digit.]