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

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:
+
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>
+
<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 </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.   
+
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.   
+
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}}$
+
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 $3$ 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