Difference between revisions of "2009 AMC 10B Problems/Problem 3"
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Losing three cans of paint corresponds to being able to paint five fewer rooms. So <math>\frac 35 \cdot 25 = \boxed{15}</math>. The answer is <math>\mathrm{(C)}</math>. | Losing three cans of paint corresponds to being able to paint five fewer rooms. So <math>\frac 35 \cdot 25 = \boxed{15}</math>. The answer is <math>\mathrm{(C)}</math>. | ||
| + | == See also == | ||
{{AMC10 box|year=2009|ab=B|num-b=2|num-a=4}} | {{AMC10 box|year=2009|ab=B|num-b=2|num-a=4}} | ||
{{AMC12 box|year=2009|ab=B|num-b=1|num-a=3}} | {{AMC12 box|year=2009|ab=B|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 14:58, 9 June 2021
- The following problem is from both the 2009 AMC 10B #3 and 2009 AMC 12B #2, so both problems redirect to this page.
Problem
Paula the painter had just enough paint for 30 identically sized rooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for 25 rooms. How many cans of paint did she use for the 25 rooms?
Solution
Losing three cans of paint corresponds to being able to paint five fewer rooms. So 
. The answer is 
.
See also
| 2009 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2009 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
