2008 AMC 10A Problem 11

P11-15~2 minPrint

Linear Equations & Word Problems Set up the equation/formula and compute; no special trick needed

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?

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Key insight

The trip to shore takes 15 minutes, so the boat may gain at most 30 ÷ 15 = 2 gallons each minute. LeRoy must bail the other 8.

Solution

1. First find how long they are on the water. One mile at $4$ miles per hour takes $\frac{1}{4}$ of an hour, which is $15$ minutes.
2. The water rates are in gallons per minute, so work in minutes. Over $15$ minutes the boat can take in at most $30$ gallons in total.
3. Spread that over the trip: the water level may rise by at most
$\frac{30}{15} = 2 \text{ gallons each minute}.$
4. Water leaks in at $10$ gallons each minute. For only $2$ of them to stay, LeRoy must throw out the rest:
$10 - 2 = 8 \text{ gallons each minute}.$
5. Check: at $8$ gallons per minute the boat gains $2$ gallons per minute, so after $15$ minutes it holds $30$ gallons. That is exactly the limit, not more, so they make it. Any slower and they sink.

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

Water comes in and water goes out at the same time, and only the difference matters. The limit is on the total water over the whole trip, so divide that total by the time to get the largest allowed gain per minute. The key is to put the travel time into the same units as the rates.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

First find how long they are in the water. One mile at $4$ miles per hour takes $\frac{1}{4}$ hour, which is $15$ minutes.

During those $15$ minutes the boat may accumulate at most $30$ gallons, so the net rate of water coming in can be at most
$\frac{30}{15} = 2\ \text{gallons per minute}.$

Water enters at $10$ gallons per minute, so LeRoy must remove at least $10 - 2 = 8$ gallons per minute.

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)