In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number, find the number.

Let's consider the digits at ten's place as x and let the digit at unit's place be y.

Then according to first condition,

⇒ x + y = 7

⇒ x = 7 - y ……(i)

The number if digits are reversed = 10 × y + x.

Given, number with the order of the digits reversed is 28 greater than twice the unit's digits of the original number

∴ (10 × y + x) - 2y = 28

Substituting value of x from (i) we get,

⇒ 10y + 7 - y - 2y = 28

⇒ 7y = 21

⇒ y = 3.

⇒ x = 7 - y = 7 - 3 = 4.

Original number = 10 × x + y = 10 × 4 + 3 = 43.

Hence, original number = 43.