2012 AMC 10A Problem 19
Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 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 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at PM. On Wednesday Paula worked by herself and finished the house by working until P.M. How long, in minutes, was each day's lunch break?
Show solution without answering
Key insight
Tuesday's 24% plus Wednesday's 26% equals Monday's 50%. Comparing them gives the helpers' speed compared with Paula's, and then only the lunch break is unknown.
Solution
1. Times. Count minutes after 8:00 AM. Monday ends at 4:00 PM, minutes. Tuesday ends at 2:12 PM, minutes. Wednesday ends at 7:12 PM, minutes.
2. Wednesday's share. Paula finished the house, so she painted .
3. Unknowns. Let be the lunch break in minutes, the percent of the house Paula paints per minute, and the percent the two helpers together paint per minute. Each day, the working time is the time until quitting minus :
4. Compare days. Tuesday and Wednesday together painted , the same as Monday. So
The terms cancel on both sides, leaving , so . Dividing by : . The helpers together paint as fast as Paula.
5. Divide Tuesday by Wednesday. The rates are now linked, so divide the two equations:
Replace by and simplify :
6. Cross-multiply: , that is . Dividing by : , so , which gives and .
7. Check. With : Tuesday is working minutes, so the helpers paint percent per minute. Wednesday is minutes, so Paula paints . Monday is minutes at , which gives . It works.
The answer is .
Why this works
There are three unknowns, but only the lunch break is asked for; the painting speeds are just in the way. Look for two days whose work adds up to another day's work, because that comparison links the speeds. Dividing one day's equation by another's then removes the speeds completely.
Alternative approach
Try the choices, using the check in step 7. For each lunch length, work out the helpers' speed from Tuesday and Paula's speed from Wednesday, then see whether Monday comes out to exactly . For it does. For example, gives helpers and Paula per minute, and Monday's minutes would paint about , not exactly .
Concise solution
A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.
Work in minutes after 8:00 AM: Monday ends at , Tuesday at , Wednesday at . Let be the lunch break, Paula's rate and the helpers' combined rate (percent of house per minute). Wednesday's share is .
Tuesday and Wednesday together also total , so adding those two equations and comparing with Monday:
Now divide the Tuesday equation by the Wednesday equation:
Cross-multiplying, , so , giving and .
The answer is .
Review in the AoPS books
- Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
- Introduction to Algebra, Chapter 5: Multi-Variable Linear Equations (§5.2, §5.3, §5.6)
- Introduction to Algebra, Chapter 7: Proportion (§7.4)
- Intermediate Algebra, Chapter 1: Basic Techniques for Solving Equations (§1.4)
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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.