# NCERT Solutions for Class 8 Maths

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1.    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.

∴ A = 7

And  putting 2 and carry over 1, we get

B = 6

Hence A = 7 and B = 6

2.   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.

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

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

∴ A = 6

Hence A = 6

4.   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.   Here on putting B = 0, we get 0 × 3 = 0.

And A = 5, then 5 × 3 = 15

A = 5 and C = 1

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

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

⇒ A = 5, C = 2

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

7.    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 be 44.

For 6 × 7 = 42 + 2 = 44

Hence A = 7 and B = 4

8.   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.   On putting B = 7,

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.  Putting A = 8 and B = 1, we get

8 + 1 = 9

Now again we add           2 + 8 = 10

Tens place digit is ‘0’ and carry over 1.

Now 1 + 6 + 1 = 8 = A

Hence A = 8 and B = 1

11.  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.

∴ 2 + 1 + y + 5 = 8 + y

8 + y = 9            ⇒          y = 1

12.  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.

∴ 3 + 1 + z  + 5 = 9 + z

⇒ 9 + z = 9          ⇒           z = 0

If          3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 18         ⇒          z = 9

Hence 0 and 9 are two possible answers.

13.  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.

∴    2 + 4 + x = 6 + x

Since x is a digit.

⇒ 6 + x = 6             ⇒      x = 0

⇒ 6 + x = 9             ⇒      x = 3

⇒ 6 + x = 12            ⇒      x = 6

⇒ 6 + x = 15            ⇒      x = 9

Thus, x can have any of four different values.

14.  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

Since z is a digit.

∴ 3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 9          ⇒        z = 0

If          3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 12         ⇒       z = 3

If          3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 15         ⇒       z = 6

If          3 + 1 + z + 5 = 9 + z

⇒ 9 + z = 18         ⇒       z = 9

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

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