Difference between revisions of "2008 AMC 12A Problems/Problem 7"
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<math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10</math> | ||
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| + | ==Solution== | ||
| + | It will take <math>\frac{1}{4}</math> of an hour or <math>15</math> minutes to get to shore. Since only <math>30</math> gallons of water can enter the boat, only <math>\frac{30}{15}=2</math> net gallons can enter the boat per minute. Since <math>10</math> gallons of water enter the boat per min, LeRoy must bail <math>10-2=8</math> gallons per minute <math>\Rightarrow D</math>. | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2008|num-b=6|num-a=8|ab=A}} | ||
Revision as of 22:09, 19 February 2008
Problem
While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?
Solution
It will take 
of an hour or 
minutes to get to shore. Since only 
gallons of water can enter the boat, only 
net gallons can enter the boat per minute. Since 
gallons of water enter the boat per min, LeRoy must bail 
gallons per minute 
.
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
