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

You can Download KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers helps you to revise the complete syllabus.

KSEEB Solutions for Class 8 Maths 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?
Solution:
Since 21y5 is multiple of 9
∴ 21y5 is divisible by 9.
So 2 + 1 + y + 5 = (8 + y) must be multiple of 9
8 + y = 9 or 18
when 8 + y = 9
⇒ y = 9 – 8
⇒ y = 1

When 8 + y = 18
⇒ y = 18 – 8
⇒ y = 10
Since y is a digit.
∴ y ≠ 10 hence y = 1

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

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?
Solution:
Since 31z5 is divisible by 9
∴ 3 + 1 + z + 5 = (9 + z) must be multiple of 9.
9 + z = 9, 18 or 27
when 9 + z = 9
z = 9 – 9
z = 0

when 9 + z = 18
z = 18 – 9
z = 9

when 9 + z = 27
z = 27 – 9
z = 18
So values of z = 0, 9 not 18 [Since z is a digit]

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

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, …. But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values).
Solution:
We have 24x is a multiple of 3
∴ 2 + 4 + x = (6 + x), is multiple of 3.
if (6 + x) = 0, 3, 6, 9, 12, 15, ……….
6 + x = 0 ⇒ x = -6
6 + x = 3 ⇒ x = 3 – 6 = -3
6 + x = 6 ⇒ x = 6 – 6 = 0
6 + x = 9 ⇒ x = 9 – 6 = 3
6 + x = 12 ⇒ x = 12 – 6 = 6
6 + x = 15 ⇒ x = 15 – 6 = 9
6 + x = 18 ⇒ x = 18 – 6 = 12
Since x is a digit
∴ x = 0, 3, 6, 9

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

Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution:
31z5 is a multiple of 3
If (3 + 1 + z + 5) is multiple of 3
9 + z = 0, 3, 6, 9, 12
9 + z = 0 ⇒ z = 0 – 9 = -9 (Rejected gives -ve value of z)
9 + z = 3 ⇒ z = 3 – 9 = -6 (Rejected gives -ve value of z)
9 + z = 6 ⇒ z = 6 – 9 = -3 (Rejected gives -ve value of z)
9 + z = 9 ⇒ z = 9 – 9 = 0
9 + z = 12 ⇒ z = 12 – 9 = 3
9 + z = 15 ⇒ z = 15 – 9 = 6
9 + z = 18 ⇒ z = 18 – 9 = 9
9 + z = 21 ⇒ z = 21 – 9 = 12
∴ The values of z = 0, 3, 6, 9.