AMC 10 · 2008 · #11

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

rateunit-conversionoptimization dimensional-analysisextremal-construction ↑ Prerequisites: rate

📏 Short solution 💡 2 insights

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?

Pick an answer.

(A)

$\ 2$

(B)

$\ 4$

(C)

$\ 6$

(D)

$\ 8$

(E)

$\ 10$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A boat is $1$ mile from shore. Water leaks in at $10$ gallons per minute. The boat sinks if it ever takes in more than $30$ gallons. Steve rows toward shore at $4$ miles per hour while LeRoy bails water out. Find the slowest steady rate, in gallons per minute, at which LeRoy can bail so the boat reaches shore without sinking.

Givens: Distance to shore is $1$ mile; Water enters at $10$ gallons per minute; The boat sinks if it takes in more than $30$ gallons total; Rowing speed toward shore is $4$ miles per hour; Answer choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $10$ gallons per minute

Unknowns: The slowest bailing rate, in gallons per minute, that keeps the boat afloat

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #14 Extreme Principle, #4 Introduce a Variable

The numbers come in different units: miles, miles per hour, gallons, and gallons per minute. Tool #8 (Analyze the Units) says line the units up first — turn the trip into a number of minutes, then everything can be measured in gallons per minute. Because the question asks for the slowest safe rate, Tool #14 (Extreme Principle) points to the boundary case where the water builds up to exactly the $30$-gallon limit and not a drop more. Tool #4 (Introduce a Variable) names the bailing rate so the balance between water coming in and going out becomes one clean inequality.

Execute — Answer: D

#8 Analyze the Units 6.RP.A.3 Step 1

Find how long the trip takes

Time equals distance divided by speed.
Rowing $1$ mile at $4$ miles per hour takes $\tfrac{1}{4}$ of an hour.
To compare with the leak, change that into minutes: $\tfrac{1}{4}\times 60 = 15$ minutes.

$$t = \dfrac{1\ \text{mi}}{4\ \text{mi/hr}} = \tfrac{1}{4}\ \text{hr} = 15\ \text{min}$$

💡 Going $1$ mile at $4$ miles each hour uses a quarter of an hour, and a quarter of $60$ minutes is $15$.

#14 Extreme Principle 6.RP.A.3 Step 2

Set the largest net gain that stays safe

The boat can hold at most $30$ extra gallons, and it must survive all $15$ minutes.
Spreading that limit evenly over the trip, the water in the boat may grow by at most $30 \div 15 = 2$ gallons per minute on net.
The slowest safe bailing happens exactly at this boundary, where the water reaches $30$ gallons right as they land.

$$\text{max net rate} = \dfrac{30\ \text{gal}}{15\ \text{min}} = 2\ \text{gal/min}$$

💡 If $30$ gallons is the whole budget for $15$ minutes, then $2$ gallons a minute is all the boat can afford to gain.

#4 Introduce a Variable 6.EE.B.8 Step 3

Balance water in against water out

Let $b$ be LeRoy's bailing rate in gallons per minute.
Water comes in at $10$ and leaves at $b$, so the net gain each minute is $10 - b$.
To stay safe this net gain must be at most $2$: $10 - b \le 2$, which rearranges to $b \ge 8$.
The smallest bailing rate that works is $b = 8$ gallons per minute, choice (D).

$$10 - b \le 2 \;\Rightarrow\; b \ge 8 \;\Rightarrow\; b_{\min} = 8\ \textbf{(D)}$$

💡 Bailing has to erase all but $2$ of the $10$ gallons pouring in each minute, so it must clear at least $8$.

[1] #8 6.RP.A.3 Time equals distance divided by speed. Rowing $1$ mile at $4$ miles per hour tak

[2] #14 6.RP.A.3 The boat can hold at most $30$ extra gallons, and it must survive all $15$ minut

[3] #4 6.EE.B.8 Let $b$ be LeRoy's bailing rate in gallons per minute. Water comes in at $10$ an

Review

Reasonableness: Check the boundary directly: bailing $8$ gal/min against a $10$ gal/min leak leaves a net gain of $2$ gal/min, and over $15$ minutes that is $2 \times 15 = 30$ gallons — exactly the limit, so $8$ is safe and nothing slower is. A slower rate like $6$ would let $4$ gal/min build up, reaching $60$ gallons and sinking the boat, so smaller choices (A), (B), (C) fail; $10$ would keep the boat bone dry but is more bailing than needed, so (E) is not the slowest.

Alternative: Work in total gallons instead of rates. Over $15$ minutes the leak brings in $10 \times 15 = 150$ gallons. LeRoy must bail all but $30$ of them, so he removes at least $150 - 30 = 120$ gallons in $15$ minutes, which is $120 \div 15 = 8$ gallons per minute — the same choice (D).

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Turning the $1$ mile at $4$ mph trip into $15$ minutes and finding the largest safe net inflow $30 \div 15 = 2$ gallons per minute.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Writing the safety condition $10 - b \le 2$ as $b \ge 8$ and reading off the slowest bailing rate $b = 8$.)

⭐ Turn everything into gallons-per-minute over the $15$-minute trip: the boat can only gain $2$ of the $10$ gallons leaking in each minute, so LeRoy must bail at least $8$.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Find how long the trip takes

Set the largest net gain that stays safe

Balance water in against water out

Review

CCSS standards used (min grade 6)

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