Home » class 8 Maths » NCERT Solutions for Class 8 Maths – Chapter 16 – Playing with NumbersClass 8

# NCERT Solutions for Class 8 Maths – Chapter 16 – Playing with NumbersClass 8

Exercise 16.1 : Solutions of Questions on Page Number : 255

Q1 : Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 5 is giving 2 i.e., a number whose ones digit is 2. This is possible only when digit A is 7. In that case, the addition of A (7) and 5 will give 12 and thus, 1 will be the carry for the next step. In the next step,
1 + 3 + 2 = 6
Therefore, the addition is as follows. Clearly, B is 6.
Hence, A and B are 7 and 6 respectively.

Q2 :Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 8 is giving 3 i.e., a number whose ones digit is 3. This is possible only when digit A is 5. In that case, the addition of A and 8 will give 13 and thus, 1 will be the carry for the next step. In the next step,
1 + 4 + 9 = 14
Therefore, the addition is as follows. Clearly, B and C are 4 and 1 respectively.
Hence, A, B, and C are 5, 4, and 1 respectively.

Q3 : Find the value of the letter in the following and give reasons for the steps involved. The multiplication of A with A itself gives a number whose ones digit is A again. This happens only when A = 1, 5, or 6.
If A = 1, then the multiplication will be 11 × 1 = 11. However, here the tens digit is given as 9. Therefore, A = 1 is not possible. Similarly, if A = 5, then the multiplication will be 15 × 5 = 75. Thus, A = 5 is also not possible.
If we take A = 6, then 16 × 6 = 96. Therefore, A should be 6.
The multiplication is as follows. Hence, the value of A is 6.

Q4 :Find the values of the letters in the following and give reasons for the steps involved. The addition of A and 3 is giving 6. There can be two cases.
(1) First step is not producing a carry
In that case, A comes to be 3 as 3 + 3 = 6. Considering the first step in which the addition of B and 7 is giving A (i.e., 3), B should be a number such that the units digit of this addition comes to be 3. It is possible only when B = 6. In this case, A = 6 + 7 = 13. However, A is a single digit number. Hence, it is not possible.
(2) First step is producing a carry
In that case, A comes to be 2 as 1 + 2 + 3 = 6. Considering the first step in which the addition of B and 7 is giving A (i.e., 2), B should be a number such that the units digit of this addition comes to be 2. It is possible only when B = 5 and 5 + 7 = 12. Hence, the values of A and B are 2 and 5 respectively.

Q5 : Find the values of the letters in the following and give reasons for the steps involved. The multiplication of 3 and B gives a number whose ones digit is B again.
Hence, B must be 0 or 5.
Let B is 5.
Multiplication of first step = 3 × 5 = 15
1 will be a carry for the next step.
We have, 3 × A + 1 = CA
This is not possible for any value of A.
Hence, B must be 0 only. If B = 0, then there will be no carry for the next step.
We should obtain, 3 × A = CA
That is, the one’s digit of 3 × A should be A. This is possible when A = 5 or 0.
However, A cannot be 0 as AB is a two-digit number.
Therefore, A must be 5 only. The multiplication is as follows. Hence, the values of A, B, and C are 5, 0, and 1 respectively.

Q6 : Find the values of the letters in the following and give reasons for the steps involved. The multiplication of B and 5 is giving a number whose ones digit is B again. This is possible when B = 5 or B = 0 only.
In case of B = 5, the product, B × 5 = 5 × 5 = 25
2 will be a carry for the next step.
We have, 5 × A + 2 = CA, which is possible for A = 2 or 7
The multiplication is as follows.  If B = 0,
B × 5 = B ⇒ 0 × 5 = 0
There will not be any carry in this step.
In the next step, 5 × A = CA
It can happen only when A = 5 or A = 0
However, A cannot be 0 as AB is a two-digit number.
Hence, A can be 5 only. The multiplication is as follows. Hence, there are 3 possible values of A, B, and C.
(i) 5, 0, and 2 respectively
(ii) 2, 5, and 1 respectively
(iii) 7, 5, and 3 respectively

Q7 : Find the values of the letters in the following and give reasons for the steps involved. The multiplication of 6 and B gives a number whose one’s digit is B again.
It is possible only when B = 0, 2, 4, 6, or 8
If B = 0, then the product will be 0. Therefore, this value of B is not possible.
If B = 2, then B × 6 = 12 and 1 will be a carry for the next step.
6A + 1 = BB = 22 ⇒ 6A = 21 and hence, any integer value of A is not possible.
If B = 6, then B × 6 = 36 and 3 will be a carry for the next step.
6A + 3 = BB = 66 ⇒ 6A = 63 and hence, any integer value of A is not possible.
If B = 8, then B × 6 = 48 and 4 will be a carry for the next step.
6A + 4 = BB = 88 ⇒ 6A = 84 and hence, A = 14. However, A is a single digit number. Therefore, this value of A is not possible.
If B = 4, then B × 6 = 24 and 2 will be a carry for the next step.
6A + 2 = BB = 44 ⇒ 6A = 42 and hence, A = 7
The multiplication is as follows. Hence, the values of A and B are 7 and 4 respectively.

Q8 : Find the values of the letters in the following and give reasons for the steps involved. The addition of 1 and B is giving 0 i.e., a number whose ones digits is 0. This is possible only when digit B is 9. In that case, the addition of 1 and B will give 10 and thus, 1 will be the carry for the next step. In the next step,
1 + A + 1 = B
Clearly, A is 7 as 1 + 7 + 1 = 9 = B
Therefore, the addition is as follows. Hence, the values of A and B are 7 and 9 respectively.

Q9 : Find the values of the letters in the following and give reasons for the steps involved. The addition of B and 1 is giving 8 i.e., a number whose ones digits is 8. This is possible only when digit B is 7. In that case, the addition of B and 1 will give 8. In the next step,
A + B = 1
Clearly, A is 4.
4 + 7 = 11 and 1 will be a carry for the next step. In the next step,
1 + 2 + A = B
1 + 2 + 4 = 7
Therefore, the addition is as follows. Hence, the values of A and B are 4 and 7 respectively.

Q10 : Find the values of the letters in the following and give reasons for the steps involved. The addition of A and B is giving 9 i.e., a number whose ones digits is 9. The sum can be 9 only as the sum of two single digit numbers cannot be 19. Therefore, there will not be any carry in this step.
In the next step, 2 + A = 0
It is possible only when A = 8
2 + 8 = 10 and 1 will be the carry for the next step.
1 + 1 + 6 = A
Clearly, A is 8. We know that the addition of A and B is giving 9. As A is 8, therefore, B is 1.
Therefore, the addition is as follows. Hence, the values of A and B are 8 and 1 respectively.

Exercise 16.2 : Solutions of Questions on Page Number : 260

Q1 :  If 21y5 is a multiple of 9, where y is a digit, what is the value of y?

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.
Sum of digits of 21y5 = 2 + 1 + y + 5 = 8 + y
Hence, 8 + y should be a multiple of 9.
This is possible when 8 + y is any one of these numbers 0, 9, 18, 27, and so on …
However, since y is a single digit number, this sum can be 9 only. Therefore, y should be 1 only.

Q2 : 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 for the last problem. Why is this so?

If a number is a multiple of 9, then the sum of its digits will be divisible by 9.
Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z
Hence, 9 + z should be a multiple of 9.
This is possible when 9 + z is any one of these numbers 0, 9, 18, 27, and so on …
However, since z is a single digit number, this sum can be either 9 or 18. Therefore, z should be either 0 or 9.

Q3 :  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 x is 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)

Since 24x is a multiple of 3, the sum of its digits is a multiple of 3.
Sum of digits of 24x = 2 + 4 + x = 6 + x
Hence, 6 + x is a multiple of 3.
This is possible when 6 + x is any one of these numbers 0, 3, 6, 9, and so on …
Since x is a single digit number, the sum of the digits can be 6 or 9 or 12 or 15 and thus, the value of x comes to 0 or 3 or 6 or 9 respectively.
Thus, x can have its value as any of the four different values 0, 3, 6, or 9.

Question 4:
if 31z5 is a multiple of 3,where z is a digit,what might be the values of z?