Difference between revisions of "1953 AHSME Problems/Problem 1"
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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
