chapter 16 maths class 8 deals with Playing with Numbers. Find ncert solutions for class 8 maths chapter 16 here. In Class 8 maths chapter 16, we explored finding factors and multiples and the relationships among natural numbers, whole numbers, integers and rational numbers.

**NCERT solutions for Class 8 Chapter 16**

Exercises and solutions of chapter 16 maths class 8 available here so students can read online view.

mensuration.in provides all exercises with answers for chapter 8 in class 8 maths.

Find solutions to exercise 16.1 and exercise 16.2 in chapter 8 in class 8 maths.

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Find solutions to ncert maths class 8 chapter 16.

## class 8 maths chapter 16 exercise 16.1

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

class 8 maths chapter 16 exercise 16.1 (1)

1. 3 A

+ 2 5

B 2

Solution to class 8 maths chapter 16 exercise 16.1(1)

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

7+5 = 12 in which ones place is 2.

∴ A=7

And putting 2 and carry over 1, we get

B = 6

Hence, A = 7 and B = 6

class 8 maths chapter 16 exercise 16.1 (2)

2. 4 A

+ 9 8

C B 3

Solution to class 8 maths chapter 16 exercise 16.2 (2)

On putting A = 1, 2, 3, 4, 5, and so on and we get, 8 + 5 = 13 in which ones place is 3.

A = 5

And putting 3 and carry over 1, we get B = 4 and C = 1

Hence, A= 5, B = 4 and C=1

class 8 maths chapter 16 exercise 16.1 (3)

3. 1 A

× A

9 A

Solution to class 8 maths chapter 16 exercise 16.2 (3)

On putting A = 1, 2, 3, 4, 5, 6.. and so on and we get, A×A = 6 ×6 = 36 in which ones place is 6.

A = 6

And putting 6 and carry over 3, we get A×1=9.

Hence, A= 6

class 8 maths chapter 16 exercise 16.2 (4)

4. A B

+ 3 7

6 A

Solution to class 8 maths chapter 16 exercise 16.2 (4)

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

class 8 maths chapter 16 exercise 16.2 (5)

5. A B

× 3

C A B

Solution to class 8 maths chapter 16 exercise 16.2 (5)

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

And A = 5, then 5 ×3 = 15

A = 5 and C= 1

Hence, A= 5, B = 0 and C= 1

class 8 maths chapter 16 exercise 16.2 (6)

6. A B

× 5

C A B

Solution to class 8 maths chapter 16 exercise 16.2 (6)

On putting B = 0, we get

0 × 5 = 0 and A = 5, then 5 ×5 = 25

A = 5, C = 2

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

class 8 maths chapter 16 exercise 16.2 (7)

7. A B

× 6

B B B

Solution to class 8 maths chapter 16 exercise 16.2 (7)

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 be 44. For 6×7 = 42+2 = 44

Hence, A=7 and B = 4

class 8 maths chapter 16 exercise 16.2 (8)

8. A 1

+ 1 B

B 0

Solution to class 8 maths chapter 16 exercise 16.2 (8)

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

class 8 maths chapter 16 exercise 16.2 (9)

9. 2 A B

+ A B 1

B 1 8

Solution to class 8 maths chapter 16 exercise 16.2 (9)

On putting B = 7

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

class 8 maths chapter 16 exercise 16.2 (10)

10. 1 2 A

+ 6 A B

A 0 9

Solution to class 8 maths chapter 16 exercise 16.2 (10)

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

class 8 maths chapter 16 exercise 16.2

class 8 maths chapter 16 exercise 16.2 (1)

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

Solution to class 8 maths chapter 16 exercise 16.2 (1)

Since 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

∴ 2+1 + y + 5 = 8+y

Given y is a digit, so 0≤y ≤9,

For y = 1, 8+y is divisible by 9.

∴ y=1

class 8 maths chapter 16 exercise 16.2 (2)

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 to class 8 maths chapter 16 exercise 16.2 (2)

Since 31z5 is a multiple of 9.

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

3+1+z+5=9+z

‘z’ is a digit. 0≤z≤9

For z = 9 and z=0, 9+z is divisible by 9.

∴ z=0 or z=9

Hence 0 and 9 are two possible answers.

class 8 maths chapter 16 exercise 16.2 (3)

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 to class 8 maths chapter 16 exercise 16.2 (3)

Since 24x 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.

2+4+x=6+x

Since x is a digit.

For x=0 ⇒6+x=6+0=6

For x=3 ⇒6+ x=6+3=9

For x=6⇒6+ x=6+6=12

And for x=9 ⇒6+x=6+9=15

For x=0, x=3, x=6 and x=9

9+z is multiple of 3.

Thus, ‘x’ can have any of four different values.

class 8 maths chapter 16 exercise 16.2 (4)

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

Solution to class 8 maths chapter 16 exercise 16.2 (4)

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.

Since ‘z’ is a digit.

3+1+z+5=9+z

For z=0 ⇒z+9=0+9=9

For z=3 ⇒z+9=3+9=12

For z=6 ⇒z+9=6+9=15

And For z=9 ⇒z+9=9+9=18

For z=0, z=3, z=6 and z=9

9+z is multiple of 3.

Hence, 0,3, 6 and 9 are four possible answers

By

**Sudheer Kumar B**