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

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

• Class 8 Maths Playing with Numbers Exercise 16.1
• Class 8 Maths Playing with Numbers Exercise 16.2

NCERT Solutions for 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

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