Students can refer and download RD Sharma Class 8 Maths Chapter 5 Playing with Numbers from the links provided below. Subject experts have solved theÂ RD Sharma Class 8 Solutions to help students grasp the concepts easily, and also helps in boosting their confidence level, which plays an essential role in their examinations. Hence, students whose intention is to score high in the exams are advised to go through RD Sharma Class 8 Solutions.

In Exercise 5.3 we shall discuss problems based on addition and multiplication of cryptarithms, (cryptarithms are puzzles, which are operated on various operations and numbers, in which letters take the place of digits and one has to find out which letter represents which digit).

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### Access answers to Maths RD Sharma Solutions For Class 8 Exercise 5.3 Chapter 5 – Playing with Numbers

** Solve each of the following Cryptarithms:
1.
**

**Solution:**

Firstly let us solve for unitâ€™s place,

7 + B = A

And for tenâ€™s place,

3 + A = 9

Which means that A = 6 and B = -1 which is not possible.

So, there should be one carry in tenâ€™s place which means 7 + B >9

Now solving for tenâ€™s place with one carry,

3 +A +1 = 9

A = 9-1-3 = 5

For unitâ€™s place subtracting 10 as one carry is given to tenâ€™s place,

7 + B – 10= 5

B = 5+10-7 = 8

âˆ´A = 5 and B = 8

**2.**

**Solution:**

Firstly let us solve for unitâ€™s place,

B + 7 = A

And for tenâ€™s place,

A + 3 = 9

Which means that A = 6 and B = -1 which is not possible.

So, there should be one carry in tenâ€™s place, which means B + 7 >9

Now solving for tenâ€™s place with one carry,

A +3 +1 = 9

A = 9-4 = 5

For unitâ€™s place subtracting 10 as one carry is given to tenâ€™s place,

B + 7 – 10= 5

B = 5+10-7 = 8

âˆ´ A = 5 and B = 8

**3.
**

**Solution:**

Firstly let us solve for unitâ€™s place,

1 + B = 0

Which means that B = -1 which is not possible.

So, there should be one carry in tenâ€™s place,

A + 1 +1 = B **—-** (1)

For unitâ€™s place, we need to subtract 10 as one carry is given in tenâ€™s place,

1 + BÂ â€“Â 10 = 0

B = 10-1 = 9

Substituting B = 9 in (1),

A + 1 + 1 = 9

A = 9-1-1 = 7

âˆ´ A = 7 and B = 9

**4.**

**Solution:**

Firstly let us solve for unitâ€™s place,

B + 1 = 8

B = 7

Now let us solve for tenâ€™s place,

A + B = 1

A + 7 = 1

A = -6 which is not possible.

Hence, A + B > 9

We know that now there should be one carry in hundredâ€™s place and so we need to subtract 10 from tenâ€™s place,

i.e., A + B â€“ 10 = 1

A + 7 = 11

A = 11-7 = 4

Now to check whether our values of A and B are correct, we should solve for hundredâ€™s place.

2 + A + 1 = B

2 + 4 + 1 = 7

7 = 7

i.e., RHS = LHS

âˆ´ A = 4 and B = 7

**5.**

**Solution:**

Firstly let us solve for unitâ€™s place,

A + B = 9 **—-** (1)

With this condition we know that sum of 2 digits can be greater than 18.

So, there is no need to carry one from tenâ€™s place.

Now let us solve for tenâ€™s place,

2 + A = 0

Which means A = -2 which is never possible

Hence, 2 + A > 9

Now, there should be one carry in hundredâ€™s place and hence we need to subtract 10 from tenâ€™s place,

i.e., 2 + A â€“ 10 = 0

A = 10-2 = 8

Now, substituting A=8 in 1,

A + B = 9

8 + B = 9

B = 9 â€“ 8

B = 1

âˆ´ A = 8 and B = 1

**6.**

**Solution:**

Firstly let us solve for unitâ€™s place,

We have two conditions here, 7 + B â‰¤ 9 and 7 + B > 9

For 7 + B â‰¤ 9

7 + B = A

AÂ â€“Â B = 7 **—-** (1)

Now let us solve for tenâ€™s place,

B + A = 8 **—-** (2)

Solving 1 and 2 simultaneously,

2A = 15 which means A = 7.5 which is not possible

So, our condition 7 + B â‰¤ 9 is wrong.

âˆ´Â 7 + B > 9 is correct condition

Hence, there should be one carry in tenâ€™s place and subtracting 10 from unitâ€™s place,

7 + B â€“ 10 = A

BÂ â€“Â A = 3Â **—-** (3)

For tenâ€™s place,

B + A + 1 = 8

B + A = 8-1

B + A = 7Â **—-** (4)

Solving (3) and (4) simultaneously,

2B = 10

B = 10/2 = 5

Substituting the value of B in equation 4

B + A = 7

5 + A = 7

A = 7-5 = 2

âˆ´Â B = 5 and A = 2

**7.** **Show that the CryptarithmÂ does not have any solution.**

**Solution:**

If B is multiplied by 4 then only 0 satisfies the above condition.

So, for unit place to satisfy the above condition, we should have B = 0.

Similarly for tenâ€™s place, only 0 satisfies the above condition.

But, AB cannot be 00 as 00 is not a two digit number.

So, A and B cannot be equal to 0

âˆ´ there is no solution satisfying the condition

.