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

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== Solution ==
 
== Solution ==
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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}\cdot1.20 = \$1.00</math>.
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For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or <math>\frac{4}{5}</math> of its original value. This tells us that the original price was <math>\frac{5}{4}\cdot1.20 = \$1.50</math>.
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Thus, his total cost was <math>\$2.50</math> and his total revenue was <math>\$2.40</math>.
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Therefore, he <math>\boxed{\textbf{(D)}\ \text{lost }10\text{ cents}}</math>.
  
 
== See Also ==
 
== See Also ==
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{{AHSME 50p box|year=1956|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:01, 14 March 2023

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}\cdot1.20 = $1.00$.

For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or $\frac{4}{5}$ of its original value. This tells us that the original price was $\frac{5}{4}\cdot1.20 = $1.50$.


Thus, his total cost was $$2.50$ and his total revenue was $$2.40$.


Therefore, he $\boxed{\textbf{(D)}\ \text{lost }10\text{ cents}}$.

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AHSME Problems and Solutions

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