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 $\boxed{D}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 31
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1953 AHSME Problems/Problem 30

Problem 30

Solution

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