1955 AHSME Problems/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:

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

Since we are buying at $3$ for $10$ cents and $5$ for $20$ cents, let's assume that together, we are buying 15 oranges. That means that we are getting a total of $30$ oranges for $(10x5) + (20x3)$ cents. That comes to a total of $30$ oranges for $110$ cents. $110/30$ = $11/3$ . This leads us to $3$ for $11$ cents which is $C$ and we are done. -Brudder

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1955 AHSME Problems/Problem 6