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

These NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2

Question 1.
It 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Answer:
21Y5 is a multiple of 9
∴ 2 + 1 + y + 5 = 8 + y must be divisible by 9
(8 + y) should be 9, 18, 27,… etc.
Since y is a digit.
8 + y = 9
y = 9 – 8 = 1
∴ The value of y = 1.

NCERT Solutions for Class 7 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?
Answer.
31z5 = 3 + l + z + 5 = 9 + z
31z5 is divisible by 9
∴ 9 + z must be equal to 9, 18, 27,… etc.
Z is a single digit number.
9 + z is one of these numbers.
9 + z = 9 then z = 0
9 + z=18 thenz= 18-9 = 9
∴ The value of z = 0 or 9.

Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
Answer:
Since 24x is a multiple of 3, so sum of digits i.e. 6 + x is a multiple of 3.
6 + x is one of these numbers: 0, 3,6, 9 12, 15, 18,… But since x is a digit,
∴ 6 + x = 6 or 9 or 12 or 15
∴ x = 0, or 3 or 6 or 9
The value of x = 0 or 3 or 6 or 9

NCERT Solutions for Class 7 Maths Chapter 16 Playing with Numbers Ex 16.2

Question 4.
31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Answer:
Sum of the digits =3+l+z+5=9+z 9 + z is a multiple of 3
So, 9 + z is one of the numbers: 0, 3, 6, 9, 12, 15,…
But since z is a digit, 9 + z = 9 or 12 or 15 or 18 or 21…
∴ The Value of z = 0 or 3 or 6 or 9 or 15 or 18.

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