# 1953 AHSME Problems/Problem 30

## Problem 30

A house worth \$ $9000$ is sold by Mr. A to Mr. B at a $10$ % loss. Mr. B sells the house back to Mr. A at a $10$ % gain. The result of the two transactions is:

$\textbf{(A)}\ \text{Mr. A breaks even} \qquad \textbf{(B)}\ \text{Mr. B gains }900 \qquad \textbf{(C)}\ \text{Mr. A loses }900\\ \textbf{(D)}\ \text{Mr. A loses }810\qquad \textbf{(E)}\ \text{Mr. B gains }1710$

## Solution

When Mr.A sells the house at a $10$% loss, he sells it for $9000(1 - .1) = 8100$. When Mr.B sells the house back to Mr. A at a $10$ % gain he sells it for $8100(1 + .1) = 8910$. Therefore Mr. A has lost $8100-8910 = 810$ dollars, so the answer is $\boxed{D}$.

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