NCERT Solutions Class 8 Maths Chapter 16 Playing with Numbers

Question

Chapter 16 NCERT Solutions for Class 8 Math Playing with Numbers explains all the facts about numbers and how they can be divided by other numbers. Numbers play an important role in math education. They serve as the foundation for almost every math topic and arithmetic calculation. As a result, developing a thorough understanding of numbers and their properties is critical. Students will become familiar with numbers, their properties, and divisibility tests as they practice NCERT Solutions Class 8 Maths Chapter 16.

Topics in the chapter

• Numbers in General Form
• Tests of Divisibility
• Two-digit numbers and three-digit numbers

NCERT Solutions for Class 8 Maths Chapter 16 are valuable resources that guide students in exam preparation. Our NCERT Solutions Class 8 Math Chapter 16 Playing with Numbers will assist you in understanding and resolving all NCERT question-answer concerns. We've covered every topic and answered every in-text and exercise question. For one-stop solutions for all concepts of Playing with Numbers, please see our solution. Our NCERT solutions are created by a subject matter specialist who understands all of the concepts in this field. Experts have used various methods to make these solutions easier to understand for users, such as diagrams, different equations, graphs, shortcuts, tips, and tricks.

Answer 1

NCERT Solutions Class 8 Maths Chapter 16 Playing with Numbers

Find the values of the letters in each of the following and give reasons for the steps involved (1 to 10).

1. 

Solution:

Say, A = 7 and we get,

7 + 5 = 12

In which one’s place is 2.

Therefore, A = 7

And putting 2 and carry over 1, we get

B = 6

Hence A = 7 and B = 6

2.

Solution:

If A = 5 and we get,

8 + 5 = 13 in which ones place is 3.

Therefore, A = 5 and carry over 1 then

B = 4 and C = 1

Hence, A = 5, B = 4 and C = 1

3.

Solution:

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

A x A = 6 × 6 = 36 in which ones place is 6.

Therefore, A = 6

4.

Solution:

Here, we observe that B = 5 so that 7 + 5 =12

Putting 2 at ones place and carry over 1 and A = 2, we get

2 + 3 + 1 = 6

Hence A = 2 and B =5

5.

Solution:

Here on putting B = 0, we get 0 x 3 = 0.

And A = 5, then 5 × 3 =15

A = 5 and C = 1

Hence A = 5, B = 0 and C = 1

6.

Solution:

On putting B = 0, we get 0 x 5 = 0 and A = 5, then 5 × 5 =25

A = 5, C = 2

Hence A = 5, B = 0 and C =2

7.

Solution:

Here product of B and 6 must be same as ones place digit as B.

6 × 1 = 6, 6 × 2 = 12, 6 × 3 = 18, 6 × 4 =24

On putting B = 4, we get the ones digit 4 and remaining two B’s value should be44.

Therefore, for 6 × 7 = 42 + 2 =44

Hence A = 7 and B = 4

8.

Solution:

On putting B = 9, we get 9 + 1 = 10

Putting 0 at ones place and carry over 1, we get for A = 7

7 + 1 + 1 =9

Hence, A = 7 and B = 9

9.

Solution:

On putting B = 7, we get 7 + 1 =8

Now A = 4, then 4 + 7 =11

Putting 1 at tens place and carry over 1, we get

2 + 4 + 1 =7

Hence, A = 4 and B = 7

10.

Solution:

Putting A = 8 and B = 1, we get

8 + 1 = 9

Now, again we add 2 + 8 = 10

Tens place digit is ‘0’ and carry over 1.

Now 1 + 6 + 1 = 8 = A

Hence A = 8 and B = 1.

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

Solution:

Suppose 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.

That is, 2 + 1 + y + 5 = 8 + y

Therefore, 8+y is a factor of 9.

This is possible when 8+y is any one of these numbers 0, 9, 18, 27, and so on

However, since y is a single digit number, this sum can be 9 only.

Therefore, the value of y should be 1 only i.e. 8 + y = 8 + 1 = 9.

12. 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 a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

3 + 1 + z + 5 = 9 + z

Therefore, 9 + z is a multiple of 9

This is only possible when 9+z is any one of these numbers: 0, 9, 18, 27, and so on.

This implies, 9 + 0 = 9 and 9 + 9 = 18

Hence 0 and 9 are two possible answers.

13. 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:

Let’s say, 24x is a multiple of 3.

Then, according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

2 + 4 + x = 6 + x

So, 6+x is a multiple of 3, and 6+x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 and so on.

Since, x is a digit, the value of x will be either 0 or 3 or 6 or 9, and the sum of the digits can be 6 or 9 or 12 or 15 respectively.

Thus, x can have any of the four different values: 0 or 3 or 6 or 9.

14. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

Solution:

Since 31z5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

That is, 3 + 1  + z + 5 = 9 + z

Therefore, 9 + z is a multiple of 3.

This is possible when the value of 9+z is any of the values: 0, 3, 6, 9, 12, 15, and so on.

At z = 0, 9 + z = 9 + 0 = 9

At z = 3, 9 + z = 9 + 3 = 12

At z = 6, 9 + z = 9 + 6 = 15

At z = 9, 9 + z = 9 + 9 = 18

The value of 9 + z can be 9 or 12 or 15 or 18.

Hence 0, 3, 6 or 9 are four possible answers for z.