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2012 AMC 12A Problem 13

Problem 13 of 25IntermediateAlgebraArithmetic

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:008{:}00 AM and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50%50\% of a house, quitting at 4:004{:}00 PM. On Tuesday, when Paula wasn’t there, the two helpers painted only 24%24\% of the house and quit at 2:122{:}12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:127{:}12 PM. How long, in minutes, was each day’s lunch break?

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Solution

Let mm be the lunch length in minutes. The three worked 480−m480 - m minutes Monday, the helpers 372−m372 - m minutes Tuesday, and Paula 672−m672 - m minutes Wednesday. If Paula paints p%p\% per minute and the helpers together paint h%h\% per minute, then (p+h)(480−m)=50,h(372−m)=24,p(672−m)=26. \begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26. \end{aligned} Adding the last two equations and subtracting from the first gives 108h−192p=0,108h - 192p = 0, so h=169p.h = \tfrac{16}{9}p. Solving the system gives p=124p = \tfrac{1}{24} and m=48.m = 48. Thus, the correct answer is D.
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