# Difference between revisions of "2008 AMC 12A Problems/Problem 7"

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?

$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10$

## Solution 1

It will take $\frac{1}{4}$ of an hour or $15$ minutes to get to shore.

Since only $30$ gallons of water can enter the boat, only $\frac{30}{15}=2$ net gallons can enter the boat per minute.

Since $10$ gallons of water enter the boat each minute, LeRoy must bail $10-2=8$ gallons per minute $\Rightarrow\mathrm{(D)}$.

## Solution 2

We set up the following equation, where $x$ is the answer:

$\frac{30}{10-x} = 15 \Rightarrow\mathrm{x=8}$, so the answer is $(D)$.

## Solution 3

Let $r$ be the slowest rate LeRoy can bail out water. It will take them $15$ minutes to get to shore. As stated in solution 1, the number of gallons that enter the boat can be at most $2$ for the boat to stay afloat. Water enters at a rate of $10$ gallons per minute.

Therefore:

$2 \geq 10-r$

$r \geq 8$

The minimum possible value for $r$ is $8 \mathrm{(D)}$