Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$. The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$
First, put the six numbers we have in order, since we are concerned with the median:  $3, 5, 5, 7, 8, 9$.
We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list.  If we top-load the list by making all three of the numbers greater than $9$, the median will be the highest it can possibly be.  Thus, the maximum median is the fifth piece of data in the list, which is $8$, giving an answer of $\boxed{D}$.
(In fact, as long as the three new integers are greater than $8$, the median will be $8$.)
This illustrates one important fact about medians:  no matter how high the three "outlier" numbers are, the median will never be greater than $8$.  The arithmetic mean is, generally speaking, more sensitive to such outliers, while the median is resistant to a small number of data that are either too high or too low.