2012 AMC 10A Problem 19

Problem 19 of 25HarderAlgebra

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?

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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.