# 1956 AHSME Problems/Problem 2

## 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 $\frac{5}{6} * 1.20 = 1.00$.

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 $\frac{5}{4} * 1.20 = 1.50$.

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

Therefore, he $\fbox{(D) lost 10 cents}$.

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