A number consists of two digits and the digit in the ten’s place exceeds that in the unit’s place by 5. if 5 times the sum of the digits be added from the numbers, the digits of the number are reversed. Then the sum of digits of the number is?

Let's break down the problem step by step:

1. Let's denote the digit in the tens place as T and the digit in the units place as U. According to the problem, T = U + 5.

2. The number itself can be expressed as 10T + U.

3. The problem also states that if 5 times the sum of the digits is added to the number, the digits are reversed. This can be expressed as 10T + U + 5(T + U) = 10U + T.

4. Substituting T = U + 5 into the equation from step 3, we get 10(U + 5) + U + 5((U + 5) + U) = 10U + (U + 5).

5. Simplifying the equation, we get 10U + 50 + U + 5U + 25 = 10U + U + 5.

6. Simplifying further, we get 16U + 75 = 11U + 5.

7. Solving for U, we get U = -70/5 = -14. However, this is not possible as U is a digit and cannot be negative.

8. Therefore, there seems to be a mistake in the problem as it does not have a valid solution.