Difference between revisions of "1956 AHSME Problems/Problem 2"

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== Solution ==
 
== Solution ==
  
For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or <math>\frac{6}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{6} * 1.20 = '</math>'1.00$
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For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or <math>\frac{6}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{6} * 1.20 = \$1.00</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 17:06, 31 December 2015

Problem #2

Mr. Jones sold two pipes at $\textdollar{ 1.20}$ each. Based on the cost, his profit on one was

$20$% and his loss on the other was $20$%. On the sale of the pipes, he:

$\textbf{(A)}\ \text{broke even}\qquad \textbf{(B)}\ \text{lost }4\text{ cents} \qquad\textbf{(C)}\ \text{gained }4\text{ cents}\qquad \\ \textbf{(D)}\ \text{lost }10\text{ cents}\qquad \textbf{(E)}\ \text{gained }10\text{ cents}$

Solution

For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or $\frac{6}{5}$ of its original value. This tells us that the original price was $\frac{5}{6} * 1.20 = $1.00$

See Also