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Difference between revisions of "1953 AHSME Problems/Problem 30"

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==Solution==
 
==Solution==
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When Mr.A sells the house at a <math>10</math>% loss, he sells it for <math>9000(1 - .1) = 8100</math>. When Mr.B sells the house back to Mr. A at a <math>10</math> % gain he sells it for <math>8100(1 + .1) = 8910</math>. Therefore Mr. A has lost <math>8100-8910 = 900</math> dollars, so the answer is <math>\boxed{D}</math>.
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When Mr.A sells the house at a <math>10</math>% loss, he sells it for <math>9000(1 - .1) = 8100</math>. When Mr.B sells the house back to Mr. A at a <math>10</math> % gain he sells it for <math>8100(1 + .1) = 8910</math>. Therefore Mr. A has lost <math>8100-8910 = 810</math> dollars, so the answer is <math>\boxed{D}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 16:33, 22 April 2020

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 $\boxed{D}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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All AHSME Problems and Solutions

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