## NCERT Solutions Class 8 Maths Chapter 16 – Free PDF Download

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers** are prepared by the subject-matter experts at BYJU’S, keeping in mind the requirements of CBSE Class 8 students. These solutions are easy-to-understand for students and help them reach a conclusion using the ideal method. Solving NCERT questions will give students the advantage of preparing themselves better before the final CBSE exams. BYJUâ€™S aims at bringing out the best in students through additional skill-building exercises that are tailored to their grade levels, abilities and interests. Class 8 NCERT Solutions are a valuable resource to help students in their exam preparations.

Practising these **NCERT Solutions**Â enables the students to solve the questions given in the textbooks more easily. These solutions are one of the best resources that provide complete data and knowledge of each and every concept that equips students to face all kinds of questions, irrespective of their toughness. Practice is an essential task to learn and score well in Mathematics.

## NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

### Access Answers to NCERT Class 8 Maths Chapter 16 Playing with Numbers

## Exercise 16.1 Page No. 255

**Find the values of the letters in each of the following and give reasons for the steps involved.**

**1. **

**Solution:**

Say, A = 7, and we get

7+5 = 12

In which oneâ€™s place is 2.

Therefore, A = 7

And putting 2 and carrying over 1, we get

B = 6

Hence, **A = 7 and B = 6.**

**2. **

**Solution: **

If A = 5, we get

8+5 = 13, in which one’s 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, we get

AxA = 6×6 = 36, in which one’s place is 6.

Therefore, **A = 6**

**4. **

**Solution:**

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

Putting 2 at one’s place and carrying 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 0x3 = 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 0x5 = 0 and A = 5, then 5×5 =25

A = 5, C = 2

Hence **A = 5, B = 0 and C =2**

**7. **

**Solution**:

Here, products of B and 6 must be the same as one’s place digit is B.

6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 = 24

On putting B = 4, we get the one’s digit 4, and the remaining two Bâ€™s value should be 44.

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 carrying over 1, we get 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 carrying 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

The tens place digit is â€˜0â€™ and carries over 1. Now 1+6+1 = 8 = A

Hence, **A = 8 and B =1.**

## Exercise 16.2 Page No: 260

**1. 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 only 9.

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

**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 to the last problem. Why is this so?**

**Solution:**

Since 31z5 is a multiple of 9,

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 the two possible answers.

**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:**

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 the 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.

**4. 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,

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 the four possible answers for z.

Also AccessÂ |

CBSE Notes for Class 8 Maths Chapter 16 |

### NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

Class 8 NCERT exercise-wise questions and answers are very important for students to develop logical and problem-solving abilities. Some of the important topics introduced in Solutions of Class 8 NCERT Maths are **Numbers** in General Form, Games with Numbers, Letters for Digits and Tests of Divisibility. In previous classes, students have studied various types of numbers, their properties and the relationships among them. In this section, numbers are explored in more detail.

**NCERT Solutions for Class 8 Maths Chapter 16 Exercises are as follows:**

Get a detailed solution for all the questions listed under the below exercises:

Exercise 16.1 Solutions: 10 Questions (Short answer type)

Exercise 16.2 Solutions: 4 Questions (Short answer type)

### NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

NCERT Class 8 Maths Chapter 16 Playing with Numbers, introduce numbers in general form, games with numbers and divisibility test by the numbers: 2, 3, 5, and 10.

**The main topics covered in this chapter include the following:**

Exercise |
Topic |

16.1 | Introduction |

16.2 | Numbers in General Form |

16.3 | Games with Numbers |

16.4 | Letters for Digits |

16.5 | Tests of Divisibility |

### Key Features of NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers

- The NCERT solutions are meticulously solved by subject experts.
- Simple and easy language is used.
- All questions are solved using a step-by-step approach.
- Practising all the solved examples will make students get ready for the exam.
- These NCERT solutions are a valuable resource for students in their assignments and annual exams.

**Disclaimer:**

**Dropped Topics –** 16.1 Introduction, 16.2 Numbers in General Form, 16.3 Games with Numbers, 16.4 Letters for Digits and 16.5 Tests of Divisibility.

## Frequently Asked Questions on NCERT Solutions for Class 8 Maths Chapter 16

### Write the key benefits of NCERT Solutions for Class 8 Maths Chapter 16.

2. It also provides explanatory diagrams and tables for comparative study, which creates an interest in learning.

3. These solutions facilitate the students to build a good knowledge of basics as well as advanced mathematical concepts.

4. It also helps students in retaining and quickly retrieving the concepts.

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