1953 AHSME Problems/Problem 1: Difference between revisions
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..." |
No edit summary |
||
| Line 1: | Line 1: | ||
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> </math>1.00<math>, he must sell: | |||
To make a profit of <math>100< | </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== | |||
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 </math>4-\frac{10}{3}=\frac{2}{3}<math> cents. | |||
To make a profit of </math>100<math> cents, he needs to sell </math>\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$ | |||
~mathsolver101 | ~mathsolver101 | ||
Revision as of 13:11, 31 July 2015
A boy buys oranges at 
for 
cents. He will sell them at 
for 
cents. In order to make a profit of $$ (Error compiling LaTeX. Unknown error_msg)1.00
\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}$==Solution==
The boy buys$ (Error compiling LaTeX. Unknown error_msg)3
10
1
\frac{10}{3}
\frac{20}{5}=4
4-\frac{10}{3}=\frac{2}{3}$cents.
To make a profit of$ (Error compiling LaTeX. Unknown error_msg)100
\frac{100}{\frac{2}{3}}=\boxed{150}=\boxed{\text{B}}$
~mathsolver101
