The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

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

Number = 10 × y + x = 10y + x

Given,

Sum of digits = 7

∴ x + y = 7

⇒ x = 7 - y ……..(1)

On reversing digits,

New number = 10 × x + y = 10x + y

Given,

The new number decreased by 2, equals twice the original number.

⇒ (10x + y) - 2 = 2(10y + x)

⇒ 10x + y - 2 = 20y + 2x

⇒ 10x - 2x = 20y - y + 2

⇒ 8x = 19y + 2

⇒ 8(7 - y) = 19y + 2 ……..[From (1)]

⇒ 56 - 8y = 19y + 2

⇒ 56 - 2 = 19y + 8y

⇒ 54 = 27y

⇒ y = $\dfrac{54}{27}$ = 2.

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

⇒ x = 7 - 2 = 5.

Number = 10y + x = 10 × 2 + 5 = 25.

Hence, number = 25.