# NCERT Solutions for Class 8 Maths Chapter 16

NCERT Solutions for Class 8 Maths Chapter 16 PLAYING WITH NUMBERS Exercise 16.1 and Exercise 16.2 in English as well as Hindi Medium in PDF format to free download updated for new academic session 2021-2022.

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## NCERT Solutions for Class 8 Maths Chapter 16

### Class 8 Maths Chapter 16 all Exercises Solution

 Class: 8 Subject: Maths – गणित Chapter 16: Playing with Numbers

### Class 8 Maths Chapter 16 Solutions

Class 8 Maths Chapter 16 Playing with Numbers all exercises in English Medium and Hindi Medium are given here to use online or download in PDF format. For any inconvenience regarding website, please call us, we will immediately solve the problem. Offline Apps 2021-22 are in updated format to use offline without internet.

• ### 8 Maths Chapter 16 Solutions in Hindi Medium

#### Class 8 Maths Exercise 16.1 Solution in Videos

Class 8 Maths Exercise 16.1 Solution
Class 8 Maths Exercise 16.1 Explanation

#### Class 8 Maths Exercise 16.2 Solution in Videos

Class 8 Maths Exercise 16.2 Solution
Class 8 Maths Exercise 16.2 Explanation

### Important Terms about Class 8 Maths Chapter 16

##### If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

Since 21y5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
So, 2 + 1 + y + 5 = 8 + y
Now 8 + y = 9
Hence, y = 1

##### If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

Since 31z5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.
So, 3 + 1 + z + 5 = 9 + z should be divisible by 9.
If 9 + z = 9, then z = 0.
If 9 + z = 18, then z = 9.
Hence, 0 and 9 are two possible answers.

##### If 24x, is multiple of 3, where x is a digit. Find x?

Since, 24x, is a multiple of 3,
Therefore, it should be divisible by 3.
Sum of digits = 2 + 4 + x = 6 + x
If, 6 + x = 6 then x = 0.
If, 6 + x = 9 then x = 3.
If, 6 + x = 12 then x = 6.
If, 6 + x = 15 then x = 9.
Hence, the possible values of x are 0, 3, 6 and 9.

##### If 31z5, is multiple of 3, where z is a digit. Find z?

Since, 31z5, is multiple of 3.
Therefore, the sum of digits should be divisible by 3.
Sum of digits = 3 + 1 + z + 5 = 9 + z
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.
Then, the possible values of z are 0, 3, 6 and 9.

In Chapter 16 Playing with Numbers, we will study about the Numbers in usual form as well as general form or expanded form. This chapter also tells us the divisibility tests of 2, 3, 5, 9 and 11. A number is divisible by 2 or 3 or any other number, that can be determined only by oral calculation not by actual division. In the last exercise, few a questions are given as tricky one on the basis of divisibility test. Divisibility tests are important tools in Vedic Maths also to simplify the questions.

##### Divisibility Check

When checking for divisibility by 2, we only check the digits of the unit whether it is divisible by 2 or not. Similarly, if a number is zero at the end, it will be divisible by 10 and 5. If it is 5 at the end, it is divisible by 5. To check for divisibility from 3 and 9, one has to find the sum of all the digits and check whether the sum is divisible by 3 or 9.      