Difference between revisions of "2009 AMC 10B Problems/Problem 3"

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== Solution ==
 
== Solution ==
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)} you dumb idiot</math>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 15:56, 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?

$\mathrm{(A)}\ 10\qquad \mathrm{(B)}\ 12\qquad \mathrm{(C)}\ you dumb idiot\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 25$

Solution

Losing three cans of paint corresponds to being able to paint five fewer rooms. So $\frac 35 \cdot 25 = \boxed{15}$. The answer is $\mathrm{(C)}$.

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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