PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers Ex 16.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1

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

Question 1.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 1
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 2
By observing the ones column, we have A + 5 and we get 2 from this.
1. e., a number whose ones digit is 2, for this A has to be 7.
∴ A + 5 = 7 + 5 = 12

Now, for the sum in tens column, we have 1 + 3 + 2 = B
∴ B = 6
Thus,
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 3
Thus, A = 7 and B = 6

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 2.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 4
Solution:
Here, we have three letters A, B and C, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 5
By observing the ones column, we have A + 8 and we get 3 from this.
i. e., a number whose ones digit is 3, for this A has to be 5.
∴ A + 8 = 5 + 8 = 13
Now, for the sum in tens column, we have 1 + 4 + 9 = CB
∴ CB = 14
Here, B = 4 and C = 1
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 6
Thus, A = 5, B = 4 and C = 1

Question 3.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 7
Solution:
Here, we have A, whose value is to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 8
Since, product of ones digit
A × A = A, so it must be 1, 5 or 6.
When A = 1, then 1 1 But, the product is 9 A, so A = 1 is not possible.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 9
When A = 5, then
But, the product is 9 A, so A = 5 is not possible.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 10
When A = 6, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 11
Thus, A = 6

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 4.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 12
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 13
By observing the ones column, we have B + 7 and we get A from this, i. e., a number whose ones digit is A.
Now, for the sum in tens column, we have A + 3 and we get 6 from this. Therefore, the value of A must be 2. (Keeping in mind that carry over 1 is to be considered.)
If A = 2, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 14
Then B + 7 gives 2, so B + 37 must be 5 and sum in tens 6 2 column is 1 + 2 + 3 = 6, so it is correct.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 15

Question 5.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 16
Solution:
Here, we have three letters A, B and C, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 17
Units digit of 3 × B is B, so B must be either 0 or 5.

When B = 0, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 18

When B = 5, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 19

Now, units digit of 3 × A is A. So A must be either 0 or 5, but A cannot be 0, because if A = 0, then AB becomes one-digit number. So A must be 5 and multiplication is either 55 × 3 or 50 × 3.
55 × 3 = 165, here A = 6, so this is not possible.
∴ 50 × 3 = 150
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 20
Thus, A = 5, B = 0 and C = 6

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 6.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 21
Solution:
Here, we have three letters A, B and C, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 22
Units digit B × 5 = B, so B must be either 0 or 5.
When B = 0, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 23

When B = 5, then
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 24

Now, units digit of 5 × A = A, so A must be either 0 or 5.
There are three letters as a product. So A ≠ 0, but A = 5.
So multiplication is either 50 × 5 or 55 × 5.
50 × 5 = 250 and 55 × 5 = 275, so 55 × 5 is not correct.
So 50 × 5 = 250
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 25
Thus, A = 5, B = 0 and C = 2.

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 7.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 26
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 27
Units digit B × 6 = B, so B must be 2, 4, 6 or 8.
∴ Possible values of product BBB are 222, 444, 666 or 888.
If we divide these numbers by 6, then quotient should be A2, A4, A6 or A8.
Now, 222 ÷ 6 = 37, remainder = 0
But, the quotient is not as A2, so B = 2 is not possible.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 28
Thus, A = 7 and B = 4

Question 8.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 28
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 30
By observing the sum in unit column, we have 1 + B = 0. So here is a number whose unit digit is 0, so B must be 9.
Now, for sum in tens column, we have 1 + A + 1 = 9.
So A must be 7.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 31
Thus, A = 7 and B = 9

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 9.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 32
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 33
By observing the sum in units column, we have B + 1 = 8.
∴ B must be 7.
Now,
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 34
By observing the sum in tens digit column, we have A + 7 = 1, i.e., whose unit’s digit is 1, so A must be 4.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 35
Thus, A = 4 and B = 7

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1

Question 10.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 36
Solution:
Here, we have two letters A and B, whose values are to be found.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 37
By observing the sum in tens column, we have 2 + A = 0, so A must be 8.
Then,
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 38
Now, by observing the sum in units column, we have 8 + B = 9, so B must be 1.
PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers Ex 16.1 39
Thus, A = 8 and B = 1.

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