1955 AHSME Problem 6

A merchant buys a number of oranges at 3 for 10 cents and an equal number at 5 for 20 cents. To "break even" he must sell all at:

\begin{array}{l} \textbf{(A)}\ \text{8 for 30 cents}\qquad\textbf{(B)}\ \text{3 for 11 cents}\qquad\textbf{(C)}\ \text{5 for 18 cents}\\ \textbf{(D)}\ \text{11 for 40 cents}\qquad\textbf{(E)}\ \text{13 for 50 cents} \end{array}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.