x + y

x – y

The sum of a two digit number and the number obtained by interchanging the digits is 99.

If the digits differ by 3. Find the number

Dividing throughout by 11

Let the digit in ten’s place be x and digit in unit’s place be y

∴ The number is 10x + y

The number obtained by interchanging

the digit = 10y + x

According to the first condition

∴

= 99

∴ 11x + 11y

= 99

∴ x + y

= 9

......(i)

Case (I) :

x > y

∴ x – y

......(ii)

= 3

Sol.

What do we need to find ?

We need to find a two digit number.

A two digit number has two parts.

Digit in tens place.

digit in units place.

10x + y

+

10y + x

Number it as equation (i)

Here digits differ by 3,

So there are two cases

Number it as equation (ii)

Q.

Substituting x = 6 in (i)

Adding (i) and (ii)

∴ x

= 9

= 3

2x

= 12

= 6

The two digit number

6 + y

= 9

∴ y

= 9 – 6

∴ y

= 3

= 63

= 10x + y

= 10 (6) + 3

= 60 + 3

Substitute the values of x & y in the original two digit number

Case (ii) :

x < y

∴ y – x

= 3

∴ – x + y

= 3

.....(iii)

Adding (i) and (iii), we get

x + y

= 9

– x + y

= 3

2y

= 12

Substituting y = 6 in (i), we get

∴ x + 6

= 9

∴ x

= 9 – 6

∴ x

= 3

∴ The two digit number

= 10x + y

= 10 (3) + 6

= 30 + 6

= 36

∴ The two digit number is either 63 or 36.

Q. The sum of a two digit number and the number obtained by interchanging the digits is 99. If the digits differ by 3. Find the number.

∴ y

= 6

Number it as equation (iii)

Substitute the values of x & y in the original two digit number