## NCERT Solutions for Chapter 16 Playing with Numbers Class 8 Mathematics

Page No: 220**Exercise 16.1**

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

1.

**Answer**

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, 7 + 5 = 12 in which ones place is 2.

âˆ´ A = 7

And putting 2 and carry over 1, we get

B = 6

Hence A = 7 and B = 6

2.

**Answer**

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,8 + 5 = 13 in which ones place is 3.

âˆ´ A = 5

And putting 3 and carry over 1, we get

B = 4 and C = 1

Hence A = 5, B = 4 and C = 1

3.

**Answer**

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, A x A = 6 x 6 = 36 in which ones place is 6.

âˆ´ A = 6

Hence A = 6

4.

**Answer**

**Here, we observe that B = 5**

So that 7 + 5 = 12.

Putting 2 at ones place and carry over 1 and A = 2, we get

2 + 3 + 1 = 6

Hence A = 2 and B = 5

5.

**Answer**

**Here on putting B = 0,**

we get 0Ã—3 = 0.

And A = 5, then 5Ã—3 = 15

â‡’ A = 5 and C = 1

Hence A = 5, B = 0 and C = 1

6.

**Answer**

**On putting B = 0, we get 0 , and A = 5, then 5Ã— 5 = 25**

â‡’ A = 5, C = 2

Hence A = 5, B = 0 and C = 2

7.

**Answer**

**Here product of B and 6 must be same as ones place digit as B.**

6 Ã— 1 = 6, 6 Ã— 2 = 12, 6 Ã— 3 = 18,

6 Ã— 4 = 24

On putting B = 4, we get the ones digit 4 and remaining two Bâ€™s value should be 44.

âˆ´ For 6 Ã— 7 = 42 and carry over 2 = 44

Hence A = 7 and B = 4

8.

**Answer**

**On putting B = 9, we get 9 + 1 = 10**

Putting 0 at ones place and carry over 1, we get

For A = 7 â‡’ 7 + 1 + 1 = 9

Hence A = 7 and B = 9

9.

**Answer**

**On putting B = 7,**

â‡’ 7 + 1 = 8

Now A = 4, then 4 + 7 = 11

Putting 1 at tens place and carry over 1, we get

2 + 4 + 1 = 7

Hence A = 4 and B = 7

10.

**Answer**

**Putting A = 8 and B = 1, we get**

8 + 1 = 9

Now again we add 2 + 8 = 10

Tens place digit is â€˜0â€™ and carry over 1.

Now 1 + 6 + 1 = 8 = A

Hence A = 8 and B = 1

Page No. 260

**Exercise 16.2**

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

**Answer**

Since 21y5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

âˆ´ 2+2+y+5= 8+y

â‡’ 8+y = 9

â‡’ y = 1

Since 21y5 is a multiple of 9.

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 for the last problem. Why is this so?

**Answer**

Since 31z5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

âˆ´ 3+1+z+5 = 9+z

â‡’ 9+z = 9

â‡’ z= 0

If 3 +1+z+5 = 9+z

â‡’ 9+a = 18

â‡’ z= 9

Hence, 0 and 9 are two possible answers.

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 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.)

**Answer**

Since, 24x is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

âˆ´ 2+4+x = 6 +x

Since x is a digit.

â‡’6+x= 6 â‡’ x= 0

â‡’ 6+x = 9 â‡’ x= 3

â‡’ 6+ x = 12 â‡’x= 6

â‡’ 6+x = 15 â‡’ x = 9

Thus, x can have any of four different values.

4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?

**Answer**

Since 31z5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Since z is a digit.

âˆ´ 3+1+z+5 = 9+z

â‡’ 9+z=9 â‡’ z= 0

If 3+1+z+5 = 9+z

â‡’ 9 +z = 12 â‡’ z= 3

If 3+1+z+5 = 9+z

â‡’ 9 + z= 15 â‡’ z= 6

If 3+1+z+5 = 9+z

â‡’ 9 +z = 18 â‡’ z= 9

Hence 0, 3, 6 and 9 are four possible answers.

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