In checking the petty cash a clerk counts $ q$ quarters, $ d$ dimes, $ n$ nickels, and $ c$ cents.  Later he discovers that $ x$ of the nickels were counted as quarters and $ x$ of the dimes were counted as cents.  To correct the total obtained the clerk must:

$ \textbf{(A)}\ \text{make no correction} \qquad
\textbf{(B)}\ \text{subtract 11 cents} \qquad
\textbf{(C)}\ \text{subtract 11}x\text{ cents} \\
\textbf{(D)}\ \text{add 11}x\text{ cents} \qquad
\textbf{(E)}\ \text{add }x\text{ cents}$
1. **Initial Calculation**:
   - The value of \( q \) quarters is \( 25q \) cents.
   - The value of \( d \) dimes is \( 10d \) cents.
   - The value of \( n \) nickels is \( 5n \) cents.
   - The value of \( c \) cents is \( c \) cents.
   - Therefore, the total value initially counted is:
     \[
     25q + 10d + 5n + c \text{ cents}
     \]

2. **Miscounted Nickels**:
   - \( x \) nickels were counted as quarters.
   - The value of \( x \) nickels is \( 5x \) cents.
   - The value of \( x \) quarters is \( 25x \) cents.
   - The error introduced by this miscount is:
     \[
     25x - 5x = 20x \text{ cents}
     \]
   - Therefore, the total value was overcounted by \( 20x \) cents.

3. **Miscounted Dimes**:
   - \( x \) dimes were counted as cents.
   - The value of \( x \) dimes is \( 10x \) cents.
   - The value of \( x \) cents is \( x \) cents.
   - The error introduced by this miscount is:
     \[
     10x - x = 9x \text{ cents}
     \]
   - Therefore, the total value was undercounted by \( 9x \) cents.

4. **Net Correction**:
   - To correct the total, we need to account for both errors:
     \[
     20x - 9x = 11x \text{ cents}
     \]
   - Therefore, the clerk must add \( 11x \) cents to the total to correct the miscount.

The final answer is \(\boxed{\text{add } 11x \text{ cents}}\)