The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 12 less than twice the original number. Find the original number.

Let the ten's and unit's digits of required number be x and y.

Number = 10x + y

Given,

Sum of digits = 12.

⇒ x + y = 12

⇒ x = 12 - y     …..(1)

Given,

Number obtained by reversing the digits = (10y + x)

If the digits are reversed, the new number is 12 less than twice the original number.

⇒ (10y + x) = 2(10x + y) - 12

⇒ 10y + x = 20x + 2y - 12

⇒ 10y - 2y + x - 20x = -12

⇒ 8y - 19x = - 12     …..(2)

Substituting the value of x from equation (1) in equation (2),

⇒ 8y - 19(12 - y) = -12

⇒ 8y - 228 + 19y = -12

⇒ 27y = -12 + 228

⇒ 27y = 216

⇒ y = $\dfrac{216}{27}$

⇒ y = 8.

Substituting value of y in equation (1), we get :

⇒ x = 12 - 8

⇒ x = 4.

The number is,

⇒ (10x + y) = 10 × 4 + 8 = 48.

Hence, the number is 48.