A water pump for a dredging project can remove 180 gallons of water per minute, but it can only work for $2.5$ consecutive hours, at which time it requires 20 minutes of maintenance before it can be brought back online. While it is offline, a smaller pump is used in its place, which can pump 80 gallons per minute. Using this system, what is the least amount of time it would take to pump 35,800 gallons of water?

If we want to pump out the water as quickly as possible, we want to use the faster pump for the maximum $2.5$ hours. To find the total amount of water pumped in that time, we do the conversion: $2.5\text{ hours }\times \frac{60\text{ minutes }}{1\text{ hour }}\times \frac{180\text{ gallons }}{1\text{ minute }}=27,000\text{ gallons }$ So after $2.5$ hours, there are still $35,800-27,000=8,800$ gallons left to pump. At that point, the smaller pump must be used for a minimum of 20 minutes, which can pump 20 minute $\times \frac{80\text{ gallons }}{1\text{ minute }}=1,600$ gallons which still leaves 8,800-1,600 $=7,200$ gallons. Note that we have already taken 2 hours 50 minutes, and yet have not finished the job. So how long will it take to pump the remaining 7,200 gallons? Now that we can bring the stronger pump online, it will only take 7,200 gallons $\times \frac{1\text{ minute }}{180\text{ gallons }}=40$ more minutes for a total of 3 hours 30 minutes; therefore, the correct answer is D. Although you don't need to use a graph to solve this problem, a graph can help to show the overall picture: Note that the line has a slope of 180 gallons per minute for the first 150 minutes, 80 gallons per minute for the next 20 minutes, and 180 again for the final 40 minutes and crosses the line $y=35,800$ at the 210 -minute mark.