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

You can Download KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Questions and Answers helps you to revise the complete syllabus.

KSEEB 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.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q1
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q1.1
In column one A + 5, we get 2 i.e., the number whose one’s digit is 2. Thus A should be 7.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q1.2
That is B = 6

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

Question 2.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q2
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q2.1
Here, A + 8 = 3 (a number whose ones digit is 3)
∴ ‘A’ should be 5
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q2.2
Thus A = 5, B = 4, and C = 1

Question 3.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q3
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q3.1
Since the ones digit of A × A = A
So, it must be 6, because the product of a number and A is 9.
i.e., 1 × A = 9 ⇒ A = 9
but A = 9 does not satisfy
So, A = 6
Now
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q3.2

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

Question 4.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q4
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q4.1
Here, B + 7 is ‘A’ that is a number whose one digit is ‘A’.
Let us put the values of B starting from 0.
If B = 0, then A = 7
∴ 7 + 6 ≠ 6
If B = 1, then A = 8
∴ 8 + 3 ≠ 6
If B = 2, then A = 9
∴ 9 + 3 ≠ 6
If B – 5, then A = 2
∴ (2 + 3) + 1 = 6
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q4.2
So, B = 5 and A = 2 satisfy the given condition.

Question 5.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5.1
Since the one digit of B × 3 is B.
It must be B = 0 or B = 5.
Now, look at A.
If A = 1, then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5.2
(Do not satisfy the condition)
If A = 2 then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5.3
(Do not satisfy the condition)
If A = 3 then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5.4
(Do not satisfy the condition)
If A = 5 then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q5.5
So B = 0, A = 5 and C = 1

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

Question 6.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q6
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q6.1
Since the one digit of B × 5 is B, B must be 0 or 5.
Now, look at A.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q6.2
∴ A = 7, B = 5 and C = 3

Question 7.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q7
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q7.1
Since the one’s digit of B × 6 is 6
∴ Possible values of B are 0, 4, 6, and 8.
Now, look at A.
So, the value of B B B maybe 444, 666, or 888. [000 is not posible]
666 ÷ 6 = 111 which is a 3 digits number
888 ÷ 6 = 148 which is a 3 digits number
444 ÷ 6 = 74 which is a 2 digits number
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q7.2
Thus B = 4 and A = 7

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

Question 8.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q8
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q8.1
If you study the addition in the one’s column
i.e., (1 + B), you get ‘0’
∴ B should be 9.
Thus, A must be 7.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q8.2
Thus, B = 9 and A = 7

Question 9.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q9
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q9.1
If you study the addition in the one’s column i.e., (B + 1) then you get ‘8’
∴ ‘B’ should be 7 and A should be 4.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q9.2
So, the given condition is satisfied.
Hence B = 7 and A = 4

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

Question 10.
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q10
Solution:
We have
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q10.1
Since the addition of ones, the column gives the value 9.
∴ A + B should (1, 8) or (8, 1)
Now, let us supply hit and trial method
If A = 1 and B = 8 then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q10.2
If A = 8 and B = 1 then
KSEEB Solutions for Class 8 Maths Chapter 16 Playing with Numbers Ex 16.1 Q10.3
Thus A = 8 and B = 1 satisfied the given condition.