Class 8 | NCERT Solution Maths Chapter 16 | Playing with Numbers | Exercise 16.1

Exercise 16.1 Page No: 255

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

1.

Solution:

Say, A = 7 and we get,

7+5 = 12

In which one’s place is 2.

Therefore, A = 7

And putting 2 and carry over 1, we get

B = 6

Hence A = 7 and B = 6

2.

Solution:

If A = 5 and we get,

8+5 = 13 in which ones place is 3.

Therefore, A = 5 and carry over 1 then

B = 4 and C = 1

Hence, A = 5, B = 4 and C = 1

3.

Solution:

On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get,

AxA = 6×6 = 36 in which ones place is 6.

Therefore, A = 6

4.

Solution:

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.

Solution:

Here on putting B = 0, we get 0x3 = 0.

And A = 5, then 5×3 =15

A = 5 and C=1

Hence A = 5, B = 0 and C = 1

6.

Solution:

On putting B = 0, we get 0x5 = 0 and A = 5, then 5×5 =25

A = 5, C = 2

Hence A = 5, B = 0 and C =2

7.

Solution:

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

6×1 = 6, 6×2 = 12, 6×3 = 18, 6×4 =24

On putting B = 4, we get the ones digit 4 and remaining two B’s value should be44.

Therefore, for 6×7 = 42+2 =44

Hence A = 7 and B = 4

8.

Solution:

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

Putting 0 at ones place and carry over 1, we get for A = 7

7+1+1 =9

Hence, A = 7 and B = 9

9.

Solution:

On putting B = 7, we get 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.

Solution:

Putting A = 8 and B = 1, we get

8+1 =9

Now, again we add2 + 8 =10

Tens place digit is ‘0’ and carry over 1. Now 1+6+1 = 8 =A

Hence A = 8 and B =1