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 $7:12$ P.M.

How long, in minutes, was each day's lunch break?

$30$

$36$

$42$

$48$

$60$

Solution:

Let the lunch break be $m$ minutes, Paula's rate be $p$ percent per minute, and the helpers' combined rate be $h$ percent per minute.

Monday gives $(p+h)(480-m)=50$ . Tuesday gives $h(372-m)=24$ . Since the remaining work on Wednesday was $26$ percent, Wednesday gives $p(672-m)=26$ .

Adding the Tuesday and Wednesday equations and subtracting the Monday equation gives $108h-192p=0$ , so $h=\dfrac{16}{9}p$ .

Substituting into Monday gives $\dfrac{25}{9}p(480-m)=50$ , while Wednesday gives $p(672-m)=26$ . Solving these two equations gives $m=48$ .

Thus, D is the correct answer.

2012 AMC 10A Problem 19

Solution:

Problem 19 in Other Years