1. Use the information about the median:

The median of the 5 numbers is 15 .
For the median to be 15 , when the numbers are arranged in ascending order, 15 must be the middle (3rd) term.

Let's arrange the known numbers: $5,15,40,55$.
If 15 is the median, then $X$ must satisfy certain conditions:
- If $X<15$, the sorted list would be $X, 5,15,40,55$ (if $X<5$ ), or $5, X, 15,40,55$ (if $5<$ $X<15$ ). In both cases, 15 is the 3rd term.
- If $X=15$, the sorted list would be $5,15,15,40,55$. The median is 15 .
- If $X>15$, the sorted list would be 5,15 , something, 40,55 . For 15 to remain the median, $X$ cannot be greater than 40 . If $X$ is between 15 and 40 (exclusive), the sorted list would be $5,15, X, 40,55$. In this case, $X$ would be the median, not 15 .
- If $X>15$, for 15 to be the median, $X$ must be one of the larger numbers. This means 15 must be the 3rd number.
The sorted list must look like: $A, B, 15, C, D$.
Given numbers are $5,15,40,55$.
So, for 15 to be the median, $X$ must be less than or equal to 15 .
- If $\(X \leq 5\)$, the sorted list is $X, 5,15,40,55$. Median is 15 .
- If $\(5<X \leq 15\)$, the sorted list is $5, X, 15,40,55$. Median is 15 .
- If $X>15$, then $X$ would be the 3 rd or 4 th or 5 th number.
- If $15<X<40: 5,15, X, 40,55$. Median is $X$. (Not 15)
- If $X=40: 5,15,40,40,55$. Median is 40. (Not 15)
- If $X>40: 5,15,40,55, X$. Median is 40. (Not 15)

Therefore, for the median to be $15, X$ must be less than or equal to $\(15(X \leq 15)\)$.

2. Maximize the mean:

The mean of the five numbers is $\(\frac{5+15+40+55+X}{5}=\frac{115+X}{5}\)$.
To maximize the mean, we need to maximize the value of $X$.
3. Combine the conditions:

We need to maximize $X$ such that $X \leq 15$.
The largest possible integer value for $X$ that satisfies $X \leq 15$ is 15 .
Let's check the options:
(A) 5 : If $X=5$, median is 15 . Mean $=(115+5) / 5=120 / 5=24$.
(B) 14: If $X=14$, median is 15 . Mean $=(115+14) / 5=129 / 5=25.8$.
(C) 15: If $X=15$, median is 15 . Mean $=(115+15) / 5=130 / 5=26$.
(D) 40: If $X=40$, the list is $5,15,40,40,55$. The median is 40 , not 15 . So $X=40$ is not possible.
(E) 55 : If $X=55$, the list is $5,15,40,55,55$. The median is 40 , not 15 . So $X=55$ is not possible.

Among the valid options (A, B, C), the largest value of $X$ is 15, which yields the maximum mean.
The final answer is 15 .