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

   

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers.

Gujarat Board Textbook Solutions 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:
∵ We have 2 + 1 + y + 5 = 8 + y
21y5 is a multiple of 9,
∴ (8 + y) must be divisible by 9.
∴ (8 + y) should be 0, 9, 18, 27, …, etc.
∴ 8 + y = 0 is not required.
Since, y is a digit
8 + y = 9 ⇒ y = 9 – 8 or y = 1

GSEB Solutions 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:
We have 3 + 1 + z + 5 = 9 + z
∵ 31z5 is divisible by 9.
∴ (9 + z) must be equal to 0, or 9 or 18 or 27, … But z is a digit.
∴ 9 + z = 0 or 9 + z = 18
If 9 + z = 0, then z = -9 (which is not required since z is a digit) and if 9 + z = 18, then z = 9.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
Solution:
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.

GSEB Solutions 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:
We have 3 + 1 + z + 5 = 9 + z
∵ 31z5 is divisible by 3.
∴ (9 + z) must be divisible by 3.
i.e. (9 + z) = 0 or 3 or 6 or 9 or 15 or 18 since, z is a digit.
∴ If 9 + z = 0, then z = -9.
If 9 + z = 3, then z = -6.
If 9 + z = 6, then z = -3.
If 9 + z = 9, then z = 0.
If 9 + z = 12, then z = 3.
If 9 + z = 15, then z = 6.
If 9 + z = 18, then z = 9.
If 9 + z = 21, then z = 12.
∴ Possible values of z are 0, 3, 6 or 9.

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