A two-digit number is such that the ten's digit exceeds twice the unit's digit by 2 and the number obtained by inter-changing the digits is 5 more than three times the sum of the digits. Find the two digit number.

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

Given,

Ten's digit exceeds twice the unit's digit by 2.

∴ x - 2y = 2

⇒ x = 2y + 2 ……..(1)

Given,

Number obtained by inter-changing the digits is 5 more than three times the sum of the digits.

∴ 10y + x = 3(x + y) + 5

⇒ 10y + x = 3x + 3y + 5

⇒ 10y - 3y + x - 3x = 5

⇒ 7y - 2x = 5 ……..(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ 7y - 2(2y + 2) = 5

⇒ 7y - 4y - 4 = 5

⇒ 3y - 4 = 5

⇒ 3y = 9

⇒ y = $\dfrac{9}{3}$ = 3.

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

⇒ x = 2y + 2 = 2(3) + 2 = 6 + 2 = 8.

Original number = 10x + y = 10(8) + 3 = 80 + 3 = 83.

Hence, original number = 83.