# NCERT Solutions For Class 8 Maths Chapter 16

## NCERT Solutions Class 8 Maths Playing with Numbers

### Ncert Solutions For Class 8 Maths Chapter 16 PDF Free Download

NCERT Solutions for class 8 Maths chapter 16 Playing with Numbers is crucial for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 16 Playing with Numbers. Student can download the NCERT Solution for class 8 Maths Chapter 16 pdf or can view it online by following the link. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 8 maths chapter 16. It is important to have a great understanding of the chapter 16 and a student can download and learn more about these chapters by simply downloading the PDF below. It is important to have a greater understanding of the variety of different topics in the subject of mathematics for class 8.

### NCERT Solutions Class 8 Maths Chapter 16 Exercises

Exercise 16.1

1. In the given addition, find the value of the letters and give reason.

4    A

+ 3    4

———-

B    3

Answer:

When we add A and 4, we are getting number 1 that is it gives number 1 in onesplace. This is possible only when A is number 9. Therefore, adding A that is 9 to 4, it gives 13. 1 will be carry for the next step.

In next step, the addition is 1 + 4 + 3 = 8. So the addition is as given below.

4   9

+ 3   4

————

8   3

Therefore, B is 8.

So, A and B are 4 and 8 respectively.

1. In the given addition, find the value of the letters and also give reason.

5   X

+ 8   7

———-

Y  Z  2

Answer:

The addition of X and 7 gives 2 that is a no. whose ones place is 2. It is possible only when digit X is 5. So, the addition of 5 and 7 gives 12. 1 will be carry for the next step.

1 + 5 + 8 = 14

Hence, the addition is as given below.

5   5

+ 8   7

—————

1   4   2

The value of X, Y and Z is 5, 1 and 4 respectively.

1. In the given multiplication, find the value of the letters and give reason.

2   Y

×   Y

—————–

1    2   Y

Answer:

The multiplication of a number with itself gives a number whose ones placeis that number itself. This occurs when that number is 1, 5 or 6.

Y = 1

The multiplication will be 21 × 1 = 21. Here, there is hundreds place as well. So, Y = 1 is not possible.

Y = 5

The multiplication will be 25 × 5 = 125. Here, tens as well as hundreds place match. So, Y = 5 is the correct answer.

1. In the given addition, find the value of the letters and also give reason.

X   Y

+ 4   6

————–

7   X

Answer:

The addition of X and 3 is giving 7. There can be two cases.

(i)No carry

The value of X will be 3 so we get 7 when we add 3 and 4 that is 3 + 4 = 7. Now consider the first step, Y + 6 = 3 so the value of Y has to be 7. Then we get 3 in ones place. But the value X is single digit so it is not possible.

(ii) With carry

The value of X will be 2 as 1 + 2 + 4 = 7. Now consider the first step where Y is added to 6 to give 2 in ones place. For that the value of Y will be 6 + Y = 12. Therefore, the value of Y is 6.

2   6

+ 4   6

——————

7    2

Hence, the value of X and Y is 2 and 6 respectively.

1. In the given multiplication, find the value of the letters and give reason.

X   Y

×   3

————-

Z  X   Y

Answer:

When 3is multiplied with Y it gives a number whose onesplace is Y again. So, Y must be 5 or 0.

Let Y = 5

First step: 5 × 3 = 15

1 will be carried forward. Therefore, (X × 3) + 1 = ZX. This is not possible for any number.

Therefore, value of Y has to be 0 only.

If Y = 0, then there will be no carry. So we get X × 3 = ZX.

When a number is multiplied with 3, its ones placeshould be the number itself. That is possible only for X = 0 or 5. But X cannot be 0 as it has to be two digit numbers. Therefore, the value of X is 5. Thus we get the following

5   0

×   3

—————

15 0

The value of X, Y and Z is 5, 0 and 1 respectively.

1. In the given multiplication, find the value of the letters and also give reason.

X   Y

×   5

————

Z X   Y

Answer:

When 5 is multiplied with Y it gives a number whose ones placeis Y again. So, Y must be 5 or 0.

Let Y = 5

First step: 5 × Y = 5 × 5 = 25

2 will be carried forward. Therefore, (X × 5) + 2 = ZX. This is possible for number X = 2 or 7.

The multiplication is as given below.

2   5                                                     7   5

×   5                                                   ×   5

———-                                                ———-

1  2  5                                                   3  7  5

Let Y = 0

First step: 5 × Y = 5

5 × 0 = 0

There will not be any carry in this case.

In the next step, 5 × X = ZX

This can happen only when the value of X is 5 or 0.

However, X cannot be 0 as XY is two digit numbers. Therefore, the value of X is 5.

5   0

×   5

———–

2  5  0

Therefore, there three possible values of X, Y and Z.

(i) 2, 5 and 1 respectively

(ii) 5, 0 and 2 respectively

(iii) 7, 5 and 3 respectively

1. In the given multiplication, find the value of the letters and also give reason.

X   Y

×   6

———-

Y  YY

Answer:

When 6 is multiplied with Y, it gives a number whose ones placeis Y. It is possible only if Y = 0, 2, 4, 6 or 8.

Y = 0;

The product will be 0 in this case so it is not possible.

Y = 2;

Y × 6 = 12 and 1 will be carried forward for the next step.

6X + 1 = YY = 22. Then integer value of X is not possible.

Y = 6;

Y × 6 = 36 and 3 will be carried forward for the next step.

6X + 3 = YY = 66. Then integer value of X is not possible.

Y = 8;

Y × 6 = 48 and 4 will be carried forward for the next step.

6X + 4 = YY = 88.

6X = 84.

X = 14

But X is single digit number.

Then value of X is not possible.

Y = 4;

Y × 6 = 24 and 2 will be carried forward for the next step.

6X + 2 = YY = 44.

6X = 42.

X = 7

The multiplication is given below

7   4

×   6

———-

4  4  4

Thus integer value of X and Y is 7 and 4 respectively.

1. In the given addition, find the value of the letters and also give reason.

X   1

+ 1   Y

————

Y    0

Answer:

When 1 is added to Y, it gives 0 that is a number whose ones place is 0. This is possible when digit Y is 9.

So the addition of 1 and Y will be 10 so 1 will be carried forward for the next step.

In the next step,

1+X+1=9

Therefore, X is 7.

1 + 7 + 1 = 9 = Y

Hence, the addition is as given below.

7   1

+ 1   9

———–

9   0

Thus value of X and Y is 7 and 9 respectively.

1. In the given addition, find the value of the letters and also give reason.

2   X   Y

+ X   Y   1

—————-

Y   1   8

Answer:

When 1 is added to Y, it gives 8 that is a number whose ones place is 8. This is possible when digit Y is 7.

In the next step, X + Y = 1. Therefore, the value of X is 4.

4 + 7 = 11 and 1 will be carried forward for the next step.

In the next step,

1+ 2 + X = Y

1+ 2 + 4 = 7

Hence, the addition is as given below.

2   4   7

+ 4   7   1

————–

7   1   8

Thus value of X and Y is 4 and 7 respectively.

1. In the given addition, find the value of the letters and also give reason.

1   2   X

+ 6X   Y

—————-

X   0   9

Answer:

When X is added to Y, it gives 9 that is a number whose ones place is 9.Sum can be 9 only as summation of two single digits cannot be 19. So no carry generated.

In the next step, X + 2 = 0

It is possible if X = 8.

Therefore, 2 + 8 = 10 and 1 will be carried forward for the next step.

1 + 1 + 6 = 8. Therefore, value of X = 8.

When X is added to Y, it gives 9.

X + Y = 9

8 + Y = 9

Therefore, value of Y = 1

1   2   8

+ 6   8   1

—————–

8   0   9

Thus value of X and Y is 8 and 1 respectively.

Exercise 16.2

1. What would be the value of m to make 23m3 to be a multiple of 9?

Answer:

As, 23m3 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.

∴ 2 +3 +m + 3 = 8 + m

m + 8 = 9

m = 1

1. If 31p5 is a multiple of 9, where p is a digit, find the value of p?

In this problem you will get two results. Explain why??

Answer:

Since 31p5 is a multiple of 9

So now, if a number is a multiple of 9, then the sum of the digits will be divisible by 9.

∴ $3+1+p+5=9+p$

9+p = 9      [Since, 9+p should be multiple of 9]

p = 0

If $3+1+p+5=9+p$

9 +p = 18

p = 9

However, since p is a single digit number, the sum can be either 9 or 18.

But in this case, 0 and 9 are two possible answers.

1. If 24p is a multiple of 3, where p is a digit, find the value of p? (Since 24p is a multiple of 3, its sum of digits 6 + p is a multiple of 3; so 6 + p is one of these numbers: 0, 3, 6, 9, 12, 15, 18 … But since p is a digit, it can only be that 6 + p = 6 or 9 or 12 or 15. Therefore, p = 0 or 3 or 6 or 9. Thus, p can have any of four different values.)

Answer:

Since 24p 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.

Here, the sum of digits of 24p is = 2 + 4+ p

∴  2 + 4+ p=6 + p

Since ‘p’ is a digit.

6 +p = 6

p = 0

6 + p = 9

p = 3

6 + p = 12

p = 6

6 + p = 15

p = 9

Since p is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of p comes to 0 or 3 0r 6 or 9 respectively.

Therefore, p can have any of the four different values.

1. If 31p5 is a multiple of 3, where p is a digit, find out the values of p.

Answer:

Since 31p5 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.

Here p is a digit.

∴ 3+1+p+5=9+p

9 + p = 9

p = 0

∴ 3+1+p+5=9+p

9 + p = 12

p = 3

∴ 3+1+p+5=9+p

9 + p = 15

p = 6

∴ 3+1+p+5=9+p

9 + p = 18

p = 9

Since p is a single digit number, the sum of the digits can be 9 or 12 or 15 or 18 and thus, the value of p comes to 0 or 3 0r 6 or 9 respectively.

Therefore, p can have any of the four different values.

Thus, the above topic deals with the concept of the topics of NCERT Solutions, Chapter 16 deals with the topics of mathematics. The NCERT Solutions for class 8 maths chapter 16 contains the problems for the topics dealing with the addition of letters. One can know more about the various types of mathematical problems by solving with multiplication, addition and division formulas. One needs to find the value of the various letters in the process of addition, subtraction and multiplication. Figuring out the missing values from the equations is another important aspect of solving the exercises to achieve the required result. One can know more about the different topics such as Rational Numbers, Cubes and Cube Roots, Exponents and Powers. There are many other topics that can be learnt through the concepts.