2012 AMC 12A Problem 13

Problem 13 of 25IntermediateAlgebra

Paula the painter and her two helpers each paint at constant, but different, rates. They always start at

8{:}00

AM and all three always take the same amount of time to eat lunch. On Monday the three of them painted

50\%

of a house, quitting at

4{:}00

PM. On Tuesday, when Paula wasn’t there, the two helpers painted only

24\%

of the house and quit at

2{:}12

PM. On Wednesday Paula worked by herself and finished the house by working until

7{:}12

PM. How long, in minutes, was each day’s lunch break?

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Solution

Let

m

be the lunch length in minutes. The three worked

480 - m

minutes Monday, the helpers

372 - m

minutes Tuesday, and Paula

672 - m

minutes Wednesday. If Paula paints

p\%

per minute and the helpers together paint

h\%

per minute, then

\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,

h = \tfrac{16}{9}p.

Solving the system gives

p = \tfrac{1}{24}

and

m = 48.

Thus, the correct answer is D.