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1953 AHSME Problems/Problem 1

Revision as of 13:10, 31 July 2015 by Mathsolver101 (talk | contribs) (Created page with "The boy buys <math>3</math> oranges for <math>10</math> cents or <math>1</math> orange for <math>\frac{10}{3}</math> cents. He sells them at <math>\frac{20}{5}=4</math> cents...")
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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

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