NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers, are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers.
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers
NCERT Solutions for 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.
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,
That is, A = 7 and B = 6.
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.
3.
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.
∵ Unit’s digit of 3 x A is A. So, it is essential that A = 0 or 5.
Since C cannot be 0, therefore A = 5.
Hence, we get
∴ 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.
9.
10.
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers 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.
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