GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

Gujarat Board GSEB Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Questions and Answers.

Gujarat Board Textbook Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

Question 1.
Find the values of the letters in each of the following and give reasons for the steps involved?
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 1
Solution:
1. ∵ A + 5 = GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 2 2 or 22 or 32, etc.
∴ A = 12 – 5
= 7 or 22 – 5 = 17
Since A = 17 is not possible
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 3
∴ A = 7
Since GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 4 + 3 + 2 = 6
∴ B = 6
Thus, A = 7, B = 6

GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

2. If A + 8 = GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 5 3
Then A = 13 – 8 = 5
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 6 + 4 + 9 = B
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 7
or B = GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 8 4
Clearly C = 1
Thus A = 5, B = 4 and C = 1

3. ∵ A × A = 1
∴ A = 1, or A = 5
or A = 6
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 9
If A = 1, then
∵ 11 ≠ 9A
∴ A = 1, or A = 5
75 ≠ 9A
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 10
∴ A = 5
If A = 6, then
∵ 96 = 9A
∴ A = 6 is the required values of A.
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 11

GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

4. We need B + 7 = A
∴ Possible value of B can be 5,
[∵ 5 + 7 = 12]
∵ 6 – 3 – 1 = 2
∴ A can be equal to 2
Now
∴ A = 2 and B = 5
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 12

5. We need B × 3 = GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 13 B
Since 5 × 3 = 15
∴ Possible value of B can be 5.
Also 0 × 3 = 0, i.e; B = 0 can be another possible value.
∵ A × 3 = A + 0 = A
∴ Possible value of A = 5 or A = 0
∴ Since C ≠ 0
∴ Possible value of A = 5
30, B must be
Thus, B = 0.
∴ A = 5, B = 0 and C = 1
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 14

GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

6. B can be either 5 or 0.
∵ A × 5 = A
∴ B must be o
Again A can either be 5 0r 0.
∴ C ≠ 0 ∴ A ≠ 0
∴ A must be equal to 5
Thus, we have
Therefore, A = 5, B = 0, C = 2
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 15

7. Be can be 2, 4, 6 or 8
We need product 111 or 222, or 333 or 444
or 888 out of them 111 and 333 are rejected.
Possible products are 222, 444 or 888
To obtain
The possible value is B = 4
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 16
∴ A can be either 2 or 7
A × 6 means 2 × 6 = 12 or 7 × 6 = 42
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 17
∴ A × 6 = 7 × 6 is the accepted value
Now,
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 18
Thus, A = 7 and B = 4

GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

8. ∵ 10 – 1 = 9
∴ B = 9
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 19
Thus, A = 7 and B = 9

9. ∵ 8 – 1 = 7
∴ B = 7
∴ 7 + 4 = 11 ∴ A = 4
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 20
Now, we have
Thus, A = 4 and B = 7

GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

10. ∵ 10 – 2 = 8
∴ A = 8
Also, 9 – 8 = 1
∴ B = 1
Now, we have
Thus, A = 8 and B = 1
GSEB Solutions Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 img 21

Leave a Comment

Your email address will not be published. Required fields are marked *