NCERT Solutions For Class 8 Maths Chapter 16

Ncert Solutions For Class 8 Maths Chapter 16 PDF Free Download

NCERT Solutions for class 8 Maths Chapter 16, Playing with Numbers is available for students to understand the concepts included in this chapter in an easy way. Students can download the pdf of these materials to excel in their class 8th examination. These solutions help students to frame a better understanding of the topic with the help of solved questions. These NCERT solutions are designed by our subject experts in accordance with CBSE syllabus (2018-2019), prescribed by the board for class 8 maths subject.

While solving the exercise questions available in the NCERT book, students often face some difficulties to solve some of the typical problems and eventually pile them up. To help the students clear their doubts instantly, the NCERT class 8 maths solutions for chapter 16 (Playing with Numbers) are provided here, to refer them as per they need.

Class 8 Maths NCERT Solutions Playing with Numbers

Students should practice sample papers and previous year question paper, as it will give an idea about the question pattern and marks contained by each question of the particular chapter. Also, check with other learning materials such as notes, books, question papers, exemplar problems, tips and tricks for solving maths problems, to practice for class 8 final exams and score well in Maths with good marks.

NCERT solutions for chapter 16 of class 8 contains topics such as;

  • Writing numbers in general form
  • Reversing the digits – two digit number
  • Reversing the digits – three digit number
  • Divisibility by 10
  • Divisibility by 2
  • Divisibility by 9 and 3

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.

 

Students should prepare for exams with the help of online learning materials available for class 8, to score well. The NCERT Solutions for class 8 for all the chapters are available with us. Visit BYJU’S to get updated materials and prepare well for your final exams. Download and try BYJU’s – The Learning app for a more interactive and seamless learning experience with the help of videos which explains the concepts of numbers and other math-related topics.