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?Options91116137

Let's denote the digit in the tens place as T and the digit in the units place as U.

From the problem, we know that:

1. T = U + 5 (since the digit in the tens place exceeds that in the units place by 5)
2. The number is 10T + U (since T is in the tens place and U is in the units place)
3. If 5 times the sum of the digits is added to the number, the digits are reversed. This gives us the equation: 10T + U + 5(T + U) = 10U + T

Substituting T = U + 5 into the second equation, we get:

10(U + 5) + U + 5(U + 5 + U) = 10U + U + 5 10U + 50 + U + 5U + 25 + 5U = 11U + 5 21U + 75 = 11U + 5 10U = -70 U = -7

However, a digit cannot be negative, so there seems to be a mistake in the problem. The digits of the number cannot be reversed by adding 5 times the sum of the digits to the number. Therefore, the problem does not have a valid solution.