Question

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?

$30$

$36$

$42$

$48$

$60$

Answer: D

Concepts:rate system of equations

Difficulty rating: 1810

Small Hint:

Let the lunch break be $m$ minutes and let $p$ and $h$ be the painting rates (percent per minute)

Big Hint:

Working minutes are $480 - m,$ $372 - m,$ and $672 - m$ ; set up $(p+h)(480-m) = 50,$ $h(372-m) = 24,$ $p(672-m) = 26$

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 $108h - 192p = 0,$ so $h = \tfrac{16}{9}p.$ Solving the system gives $p = \tfrac{1}{24}$ and $m = 48.$

Thus, the correct answer is D.

2012 AMC 12A Problem 13

Answer: D

Difficulty rating: 1810

Small Hint:

Big Hint:

Solution:

Problem 13 in Other Years