Let the original two digit number 'd' be 10x+y

According to the question, half of 'd' exceeds one third of 'd' by the sum of the digits in 'd'

In other words,

$$\begin{gathered} \frac{10x+y}{2}-\frac{10x+y}{3}=x+y \\ \Rightarrow 3(10x+y)-2(10x+y)=6(x+y) \\ \Rightarrow 30x+3y-20x-2y=6(x+y) \\ \Rightarrow 10x+y=6(x+y) \\ \Rightarrow 4x=5y \\ \Rightarrow \frac{x}{y}=\frac{5}{4} \end{gathered}$$

∴ Sum of digits (x+y) is 9.

Hence, option C is correct.