1953 AHSME Problems/Problem 1

A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:

$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$

Solution

The boy buys $3$ oranges for $10$ cents or $1$ orange for $\frac{10}{3}$ cents. He sells them at $\frac{20}{5}=4$ cents each. That means for every orange he sells, he makes a profit of $4-\frac{10}{3}=\frac{2}{3}$ cents.

To make a profit of $100$ cents, he needs to sell $\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$

~mathsolver101

Solution 2

The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$ . So, he buys $15$ for $50$ cents and sells them a $15$ for $60$ cents, so he makes $10$ cents of profit on every $5$ oranges. To make $100$ cents of profit, he needs to sell $10\cdot 5$ , or $50$ oranges.

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1953 AHSME Problems/Problem 1

Solution

Solution 2