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

**NCERT Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1**

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

Ex 16.1 Class 8 Maths Question 1.

Solution:

Ex 16.1 Class 8 Maths Question 2.

Solution:

Ex 16.1 Class 8 Maths Question 3.

Solution:

Ex 16.1 Class 8 Maths Question 4.

Solution:

Ex 16.1 Class 8 Maths Question 5.

Solution:

3 × B = B

⇒ B = D

3 × A = CA

⇒ 3 × 5 = 15

Thus A = 5 and C = 1

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

Ex 16.1 Class 8 Maths Question 6.

Solution:

5 × B = B

⇒ B = 0 or 5

5 × A = CA

5 × 5 = 25

Only possible when B = 0

Thus A = 5 and C = 2

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

Ex 16.1 Class 8 Maths Question 7.

Solution:

B × 6 = B

6 × 4 = 24 → B = 4 and 2 is carried to

6 × A = BB

⇒ 6 × 7 = 42 + 2 (carried on) = 44

Thus B = 7

Hence A = 7 and B = 4

Ex 16.1 Class 8 Maths Question 8.

Solution:

1 + B = 0

1 + 9 = 10 → unit digit is 0 and 1 is carried to A

+ 1 +1 (carried on) = B = 9

A + 2 = 9 ⇒ A = 9 – 2 = 7

Hence A = 7 and B = 9

Ex 16.1 Class 8 Maths Question 9.

Solution:

B + 1 = 8 ⇒ B = 8 – 1 = 7

A + B = a number with unit digit 1

A + B = 11

⇒ A + 7 = 11

⇒ A = 11 – 7 = 4 (1 Carried to)

Now 1 carried on + 2 + A = B

3 + A = 7

⇒ A = 7 – 3 = 4

Hence A = 4, B = 7

Ex 16.1 Class 8 Maths Question 10.

Solution:

9 = A + B

9 = 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5 or 8 + 1 or 7 + 2 or 6 + 3 or 5 + 4 or 0 + 9 or 9 + 0

Now 0 is required at unit place

2 + A = 10

⇒ A = 10 – 2 = 8

B = 9 – 8 = 1

1 + 6 + 1 (carried on) = A = 8

Hence A = 8 and B = 1

## Class 8 Maths Chapter 16 Playing with Numbers Ex 16.2

**NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.2**

Ex 16.2 Class 8 Maths Question 1.

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

Solution:

A number is divisible by 9 if the sum of its digits is also divisible by 9.

Sum of the digits of 21y5 = 2 + 1 +y + 5 = 8 + y

(8 + y) ÷ 9 = 1

⇒ 8 + y = 9

⇒ y = 9 – 8 = 1

Hence, the required value of y = 1.

Ex 16.2 Class 8 Maths Question 2.

If 31z5 is a multiple of 9, where z is a digit, what is the value of z?

Solution:

A number is a multiple of 9 when the sum of its digits is also divisible by 9.

Sum of the digits of 31z5 = 3 + 1 + z + 5

3 + 1 + 2 + 5 = 9k where k is an integer.

For k = 1,

3 + 1 + z + 5 = 9

⇒ z = 9 – 9 = 0

For k = 2,

3 + 1 + z + 5 = 18

⇒ z = 18 – 9 = 9

k = 3 is not possible because 3 + 1 + z + 5 = 27

⇒ z = 27 – 9 = 18 which is not a digit.

Hence the required value of z is 0 or 9

Ex 16.2 Class 8 Maths Question 3.

If 24x is a multiple of 3, where x is a digit, what is the value of x?

Solution:

Since 24x is a multiple of 3, the sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers; 0, 3, 6, 12, 15, 18, ……..

6 + x = 3k where k is any integer.

For k = 0,

6 + x = 3 × 0

⇒ 6 + x = 0

x = -6. Not possible

For k = 1,

6 + x = 3 × 1

⇒ 6 + x = 3

⇒ x = 3 – 6 = -3. Not possible

For k = 2,

6 + x = 3 × 2

⇒ 6 + x = 6

⇒ x = 6 – 6 = 0

2 + 4 + 0 = 6 multiple of 3

For k = 3,

6 + x = 3 × 3

⇒ x = 9 – 6 = 3

2 + 4 + 3 = 9 multiple of 3

For k = 4,

6 + x = 3 × 4

⇒ 6 + x = 12

⇒ x = 12 – 6 = 6

2 + 4 + 6 = 12 which is multiple of 3

For k = 5,

6 + x = 3 × 5

⇒ x = 15 – 6 = 9

2 + 4 + 9 = 15 which is multiple of 3

For k = 6,

6 + x = 3 × 6

⇒ x = 18 – 6 = 12 not possible as x is digit

Hence the required values of x are 0, 3, 6 or 9.

Ex 16.2 Class 8 Maths Question 4.

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

Solution:

A number is a multiple of 3 if the sum of its digits is divisible by 3.

3 + 1 + z + 5 = 3k where k is an integer

⇒ 9 + z = 3k

⇒ z = 3k – 9

Here, k = 0, 1, 2 is not possible as z is a digit of the number.

For k = 3,

z = 3 × 3 – 9 = 9 – 9 = 0

9 + 0 = 9 multiple of 3

For k = 4,

z = 3 × 4 – 9 = 12 – 9 = 3

9 + 3 = 12 multiple of 3

For k = 5,

z = 3 × 5 – 9 = 15 – 9 = 6

9 + 6 = 15 multiple of 3

For k = 6,

z = 3 × 6 – 9 = 18 – 9 = 9

9 + 9 = 18 multiple of 3

For k = 7,

z = 3 × 7 – 9 = 21 – 9 = 12 not possible as z is a digit

Hence, the required values of 2 are 0, 3, 6 and 9.

## NCERT Solutions for Class 8 Maths

**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 Indirect proportions****Chapter 14: Factorisation****Chapter 15: Introduction to Graphs****Chapter 16: Playing with Numbers**