NCERT Class 8 Maths Chapter 16 – Playing with Numbers Solutions

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.

Ncert solutions class 8 chapter 16-1

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.

Ncert solutions class 8 chapter 16-2

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.

Ncert solutions class 8 chapter 16-3

Solution:

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

AxA = 6×6 = 36 in which ones place is 6.

Therefore, A = 6

4.

Ncert solutions class 8 chapter 16-4

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.

Ncert solutions class 8 chapter 16-5

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.

Ncert solutions class 8 chapter 16-6

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.

Ncert solutions class 8 chapter 16-7

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.

Ncert solutions class 8 chapter 16-8

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.

Ncert solutions class 8 chapter 16-9

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.

Ncert solutions class 8 chapter 16-10

Solution:

Putting A = 8 and B = 1, we get

8+1 =9

Now, again we add2 + 8 =10

Tens place digit is ‘0’ and carry 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 9 only.

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 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.

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 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.

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.

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.

Access Class-8 Chapter-wise Solutions:

NCERT Solutions of Maths Class 8 Chapter-wise.
Chapter 1 Rational Numbers
Chapter 2 Linear Equations in One Variable
Chapter 3 Understanding Quadrilaterals
Chapter 4 Practical Geometry
Chapter 5 Data Handling
Chapter 6 Squares and Square Roots
Chapter 7 Cubes and Cube Roots
Chapter 8 Comparing Quantities
Chapter 9 Algebraic Expressions and Identities
Chapter 10 Visualising Solid Shapes
Chapter 11 Mensuration
Chapter 12 Exponents and Powers
Chapter 13 Direct and Inverse Proportions
Chapter 14 Factorisation
Chapter 15 Introduction to Graphs
Chapter 16 Playing with Numbers