A number consists of two digits whose sum is 9. If 27 is subtracted from the number its digits are reversed. Find the number.

Let the digit in the units place of the number be x.

The sum of the digits of the number = 9.

∴ The digit in the tens place of the number = 9 – x.

∴ The number = 10 (9 – x) + x

The number whose digits are a reverse of the given number

= 10x + (9 – x).

According to the given condition,

10 (9 – x) + x – 27 = 10x + (9 – x)

⇒ 90 – 10x + x – 27 = 10x + 9 – x (Removing brackets)

-9x + 63 = 9x + 9

⇒ -9x – 9x = 9 – 63 (Transposing 9x to LHS and 63 to RHS)

-18x = -54

⇒ x = (Transposing -18 to RHS)

x = 3

The digit in the units place of the number = x = 3

The digit in the tens place of the number = 9 – x = 9 – 3

= 6

∴ The number is 63.