On a examination of $n$ questions a student answers correctly $15$ of the first $20$.  Of the remaining questions he answers one third correctly.  All the questions have the same credit.  If the student's mark is $50\%$, how many different values of $n$ can there be?
$ \textbf{(A)}\ 4 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ \text{the problem cannot be solved} $
1. Let \( n \) be the total number of questions on the examination.
2. The student answers 15 out of the first 20 questions correctly.
3. For the remaining \( n - 20 \) questions, the student answers one third correctly. Therefore, the number of correctly answered questions in the remaining part is \( \frac{1}{3}(n - 20) \).
4. The total number of correctly answered questions is:
   \[
   15 + \frac{1}{3}(n - 20)
   \]
5. The student's mark is 50%, which means the student answered half of the total questions correctly. Therefore, we set up the equation:
   \[
   \frac{15 + \frac{1}{3}(n - 20)}{n} = \frac{1}{2}
   \]
6. To solve for \( n \), we first clear the fraction by multiplying both sides by \( n \):
   \[
   15 + \frac{1}{3}(n - 20) = \frac{n}{2}
   \]
7. Simplify the equation:
   \[
   15 + \frac{n - 20}{3} = \frac{n}{2}
   \]
8. Multiply through by 6 to clear the denominators:
   \[
   6 \cdot 15 + 6 \cdot \frac{n - 20}{3} = 6 \cdot \frac{n}{2}
   \]
   \[
   90 + 2(n - 20) = 3n
   \]
9. Distribute and simplify:
   \[
   90 + 2n - 40 = 3n
   \]
   \[
   50 + 2n = 3n
   \]
10. Subtract \( 2n \) from both sides:
    \[
    50 = n
    \]
11. Therefore, the total number of questions \( n \) is 50.

The final answer is \( \boxed{1} \)