Class 8 Maths NCERT Solutions 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.
Solution.
1.
Here, there are two letters A and B whose values are to be found out.
Let us see the sum in unit’s column. It is A + 5 and we get 2 from this. So,
2.
Here, there are three letters A, B and C whose values are to be found out.
Let us see the sum in unit’s column. It is A + 8 and we get 3 from this. So A has to be 5
That is, A = 5, B = 4 and C = 1.
4.
Here, there are two letters A and B whose values are to be found out.
5.
Here, there are three letters A, B and C whose values are to be found out.
∴ A = 5, B = 0 and C = 1
6.
Here, there are three letters A, B and C, whose values are to be found out.
7.
Here, there are two letters A and B whose values are to be found out. We have
8.
Here, there are two letters A and B whose values are to be found out.
10.
We are to find out values of A and B
Ex – 16.2
Question 1.
If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution.
Since 21y5 is a multiple of 9, its sum of digits 2 + 1+ y + 5 = 8+ y isa multiple of 9; so 8 + y is one of these numbers: 0, 9, 18, 27, 36, 45,… .
But since y is a digit, it can only be possible that 8 + y = 9. Therefore, y = 1.
Question 2.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z?
You will find that there are two answers to the last problem. Why is this so?
Solution.
Since 31z5 is a multiple of 9, its sum of digits 3 + 1 + z + 5 = 9 + z isa multiple of 9; so 9 + z is one of these numbers: 0, 9, 18, 27, 36, 45, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 18. Therefore, z = 0 or 9.
Question 3.
If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .
But since xis a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution.
The solution is given with question.
Question 4.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution.
Since 31z5 is a multiple of 3, its sum of digits 3 + 1 + z + 5 = 9 + z is a multiple of 3; so 9 + z is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … .
But since z is a digit, it can only be possible that 9 + z = 9 or 12 or 15 or 18. Therefore, z = 0 or 3 or 6 or 9.