1953 AHSME Problems/Problem 1: Difference between revisions
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A boy buys oranges at <math>3</math> for <math>10</math> cents. He will sell them at <math>5</math> for <math>20</math> cents. In order to make a profit of <math> | A boy buys oranges at <math>3</math> for <math>10</math> cents. He will sell them at <math>5</math> for <math>20</math> cents. In order to make a profit of <math> \$1.00</math>, he must sell: | ||
< | <math>\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}</math> | ||
Solution | ==Solution== | ||
The boy buys < | 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 each. | ||
That means for every orange he sells, he makes a profit of < | That means for every orange he sells, he makes a profit of <math>4-\frac{10}{3}=\frac{2}{3}</math> cents. | ||
To make a profit of < | To make a profit of <math>100</math> cents, he needs to sell <math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}</math> | ||
~mathsolver101 | ~mathsolver101 | ||
Revision as of 13:12, 31 July 2015
A boy buys oranges at 
for 
cents. He will sell them at 
for 
cents. In order to make a profit of 
, he must sell:
Solution
The boy buys oranges for cents or 
orange for 
cents. He sells them at 
cents each.
That means for every orange he sells, he makes a profit of 
cents.
To make a profit of 
cents, he needs to sell 
~mathsolver101
