Difference between revisions of "1950 AHSME Problems/Problem 23"

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A man buys a house for &#036;10,000 and rents it. He puts <math>12\frac{1}{2}\%</math> of each month's rent aside for repairs and upkeep; pays  &#036;325 a year taxes and realizes <math>5\frac{1}{2}\%</math> on his investment. The monthly rent (in dollars) is:
 
A man buys a house for &#036;10,000 and rents it. He puts <math>12\frac{1}{2}\%</math> of each month's rent aside for repairs and upkeep; pays  &#036;325 a year taxes and realizes <math>5\frac{1}{2}\%</math> on his investment. The monthly rent (in dollars) is:
  
<math> \textbf{(A)}\ \64.82\qquad\textbf{(B)}\ \83.33\qquad\textbf{(C)}\ \72.08\qquad\textbf{(D)}\ \45.83\qquad\textbf{(E)}\ \177.08 </math>
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<math> \textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08 </math>
  
 
== Solution ==
 
== Solution ==
<math>12\frac{1}{2}\%</math> is the same as <math>\frac{1}{8}</math>, so the man sets one eighth of each month's rent aside, so he only gains <math>\frac{7}{8}</math> of his rent. He also pays &#036;325 each year, and he realizes <math>5.5\%</math>, or &#036;550, on his investment. Therefore he must have collected a total of &#036;325 +&#036;550 = &#036;875 in rent. This was for the whole year, so he collected <math>\frac{875}{12}</math> dollars each month as rent. This is only <math>\frac{7}{8}</math> of the monthly rent, so the monthly rent in dollars is <math>\frac{875}{12}\cdot \frac{8}{7}=\boxed{\textbf{(B)}\ \83.33}</math>
+
<math>12\frac{1}{2}\%</math> is the same as <math>\frac{1}{8}</math>, so the man sets one eighth of each month's rent aside, so he only gains <math>\frac{7}{8}</math> of his rent. He also pays &#036;325 each year, and he realizes <math>5.5\%</math>, or &#036;550, on his investment. Therefore he must have collected a total of &#036;325 +&#036;550 = &#036;875 in rent. This was for the whole year, so he collected <math>\frac{875}{12}</math> dollars each month as rent. This is only <math>\frac{7}{8}</math> of the monthly rent, so the monthly rent in dollars is <math>\frac{875}{12}\cdot \frac{8}{7}=\boxed{\textbf{(B)}\ \ 83.33}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 20:33, 27 January 2015

Problem

A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:

$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08$

Solution

$12\frac{1}{2}\%$ is the same as $\frac{1}{8}$, so the man sets one eighth of each month's rent aside, so he only gains $\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\%$, or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the whole year, so he collected $\frac{875}{12}$ dollars each month as rent. This is only $\frac{7}{8}$ of the monthly rent, so the monthly rent in dollars is $\frac{875}{12}\cdot \frac{8}{7}=\boxed{\textbf{(B)}\ \ 83.33}$

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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