Difference between revisions of "1950 AHSME Problems/Problem 23"
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== 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 $325 each year, and he realizes <math>5.5\%</math>, 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 <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 == | ||
− | {{AHSME box|year=1950|num-b=22|num-a=24}} | + | {{AHSME 50p box|year=1950|num-b=22|num-a=24}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 08:10, 23 April 2012
Problem
A man buys a house for $10,000 and rents it. He puts of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes on his investment. The monthly rent (in dollars) is:
$\textbf{(A)}\ \64.82\qquad\textbf{(B)}\ \83.33\qquad\textbf{(C)}\ \72.08\qquad\textbf{(D)}\ \45.83\qquad\textbf{(E)}\ \177.08$ (Error compiling LaTeX. Unknown error_msg)
Solution
is the same as , so the man sets one eighth of each month's rent aside, so he only gains of his rent. He also pays $325 each year, and he realizes , 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 dollars each month as rent. This is only of the monthly rent, so the monthly rent in dollars is $\frac{875}{12}\cdot \frac{8}{7}=\boxed{\textbf{(B)}\ \83.33}$ (Error compiling LaTeX. Unknown error_msg)
See Also
1950 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AHSME Problems and Solutions |