Difference between revisions of "2008 AMC 12A Problems/Problem 7"
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{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #7]] and [[2008 AMC 10A Problems/Problem 11|2008 AMC 10A #11]]}} | {{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #7]] and [[2008 AMC 10A Problems/Problem 11|2008 AMC 10A #11]]}} | ||
==Problem== | ==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 | + | 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 towards 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? |
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10</math> | <math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10</math> | ||
| − | ==Solution== | + | ==Solution 1== |
It will take <math>\frac{1}{4}</math> of an hour or <math>15</math> minutes to get to shore. | 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 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 each minute, LeRoy must bail <math>10-2=8</math> gallons per minute <math>\Rightarrow\mathrm{(D)}</math>. | + | Since <math>10</math> gallons of water enter the boat each minute, LeRoy must bail <math>10-2=8</math> gallons per minute <math>\Rightarrow\mathrm{(D)}</math>. |
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| + | ==Solution 2== | ||
| + | We set up the following equation, where <math>x</math> is the answer: | ||
| + | |||
| + | <math>\frac{30}{10-x} = 15 \Rightarrow\mathrm{x=8}</math>, so the answer is <math>(D)</math>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | Let <math>r</math> be the slowest rate LeRoy can bail out water. It will take them <math>15</math> minutes to get to shore. As stated in solution 1, the number of gallons that enter the boat can be at most <math>2</math> for the boat to stay afloat. Water enters at a rate of <math>10</math> gallons per minute. | ||
| + | |||
| + | Therefore: | ||
| + | |||
| + | <math>2 \geq 10-r</math> | ||
| + | |||
| + | <math>r \geq 8</math> | ||
| + | |||
| + | The minimum possible value for <math>r</math> is <math>8\ \mathrm{(D)}</math>. | ||
| + | |||
| + | ~mobius247 | ||
==See Also== | ==See Also== | ||
Latest revision as of 17:08, 4 June 2021
- The following problem is from both the 2008 AMC 12A #7 and 2008 AMC 10A #11, so both problems redirect to this page.
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 towards 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 1
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 each minute, LeRoy must bail 
gallons per minute 
.
Solution 2
We set up the following equation, where 
is the answer:
, so the answer is 
.
Solution 3
Let 
be the slowest rate LeRoy can bail out water. It will take them minutes to get to shore. As stated in solution 1, the number of gallons that enter the boat can be at most 
for the boat to stay afloat. Water enters at a rate of gallons per minute.
Therefore:
The minimum possible value for is 
.
~mobius247
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 | |
| 2008 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
