1950 AMC 12 Problem 23

Attempt Problem 23 of the 1950 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1950 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

A man buys a house for $10,000\$10{,}000 and rents it. He puts 1212%12\dfrac12\% of each month’s rent aside for repairs and upkeep; pays $325\$325 a year taxes and realizes 512%5\dfrac12\% on his investment. The monthly rent is:
$64.82\$64.82
$83.33\$83.33
$72.08\$72.08
$45.83\$45.83
$177.08\$177.08
Answer: B
Concepts:moneypercentagelinear equation
Small hint:
Let rr be the monthly rent and express the annual rent as 12r12r
Big hint:
After upkeep and taxes, the annual return must be 5.5%5.5\% of $10,000\$10,000
Solution:
Let the monthly rent be rr dollars. The yearly rent is 12r,12r, of which 1−0.125=781-0.125=\tfrac78 remains after setting aside the upkeep money. The desired yearly return is 0.055(10,000)=5500.055(10,000)=550 dollars, so 78(12r)−325=550. \frac78(12r)-325=550. Hence 10.5r=875,10.5r=875, giving r=83.3‾,r=83.\overline3, which is $83.33\$83.33 to the nearest cent.
Thus, the correct answer is B.
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