Question

The sum of a two-digit number and the number obtained by reversing the order of its digits is 99. If ten’s digit is 3 more than the unit’s digit, then find the number.

Answer 1

Let the digit in the unit's place be x and the digit at the ten's place be y.

Then,

Number = x + 10y

The number obtained by reversing the order of the digits is 10x + y.

According to the question,

(x + 10y) + (10x + y) = 99

11x + 11y = 99

11(x + y) = 99

x + y = `99/11`

x + y = 9  .....(i)

According to the question,

y = 3 + x

x − y = −3  ......(ii)

On adding equations (i) and (ii), we get

2x = 6

x = `6/2`

∴ x = 3

Putting the value x = 3 in equation (i), we get

x + y = 9

3 + y = 9

y = 9 − 3

∴ y = 6

∴ Number = x + 10y

= 3 + 10(6)

= 3 + 60

= 63

Therefore, the required number is 63.