Loading...
A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)
Let's understand what we're really being asked. We have a 40-mile trip that we want to complete at an average speed of 60 mph for the entire journey. The driver already completed the first 20 miles at 50 mph, and now we need to figure out how fast they must drive the remaining 20 miles to hit that 60 mph overall average.
Think of it like this: the driver has already "spent" some time on the first half of the trip. To achieve the desired overall average, they have a limited amount of time left for the second half. We need to find what speed will use up exactly that remaining time.
Process Skill: TRANSLATE - Converting the problem setup into a clear mathematical question about time and speed relationships
If we want to average 60 mph for the entire 40-mile trip, let's figure out how much total time we have available.
Using everyday reasoning: if you travel 40 miles at 60 miles per hour, how long does that take? Well, 60 miles would take 1 hour, so 40 miles takes \(\frac{40}{60} = \frac{2}{3}\) of an hour.
Converting to minutes to make this concrete: \(\frac{2}{3}\) hour = \(\frac{2}{3} \times 60\) minutes = 40 minutes total.
So the entire 40-mile trip must be completed in exactly 40 minutes to achieve a 60 mph average.
Now let's calculate how much time the driver already spent on the first 20 miles at 50 mph.
Thinking through this step by step: if the driver goes 50 miles in 1 hour, then 20 miles takes \(\frac{20}{50} = \frac{2}{5}\) of an hour.
Converting to minutes: \(\frac{2}{5}\) hour = \(\frac{2}{5} \times 60\) minutes = 24 minutes.
So the driver has already used 24 minutes of their total 40-minute budget.
This is straightforward subtraction. The driver started with 40 minutes total and has already used 24 minutes.
Remaining time = \(40 - 24 = 16\) minutes
So the driver has exactly 16 minutes to complete the final 20 miles.
Now we can find the required speed: 20 miles must be covered in 16 minutes.
Let's convert back to hours first: 16 minutes = \(\frac{16}{60}\) hours = \(\frac{4}{15}\) hours.
Using the relationship that speed = distance ÷ time:
Required speed = \(20 \div \frac{4}{15} = 20 \times \frac{15}{4} = \frac{300}{4} = 75\) mph
Technical summary: Speed = \(20 \div \frac{4}{15} = 20 \times \frac{15}{4} = 75\) mph
The driver must complete the remaining 20 miles at an average speed of 75 mph.
Looking at our answer choices, this matches choice (D) 75 mph.
We can verify this makes sense: 24 minutes + 16 minutes = 40 minutes total, and 40 miles in 40 minutes gives us exactly 60 mph average for the entire trip.
Students often think they can simply average the two speeds (50 mph and the unknown speed) to get 60 mph. This leads them to set up: \(\frac{50 + x}{2} = 60\), solving for x = 70 mph. However, average speed is total distance divided by total time, not the arithmetic mean of individual speeds. This conceptual error stems from confusing average speed with average of speeds.
Some students assume that since both segments are 20 miles each, the driver should spend equal time on each segment. They might think: "If I want 60 mph average, and I went 50 mph for the first half, I need to go 70 mph for the second half to balance it out." This ignores the fact that different speeds result in different time periods, even for equal distances.
Students might incorrectly set up the problem by focusing on individual segment averages rather than the overall trip constraint. They may try to work with ratios of speeds instead of understanding that the total time for 40 miles at 60 mph creates a fixed time budget that constrains the second segment.
When converting between hours and minutes (or staying in fractional hours), students frequently make arithmetic mistakes. For example, converting \(\frac{2}{5}\) hour to minutes might be calculated as \(\frac{2}{5} \times 100 = 40\) minutes instead of \(\frac{2}{5} \times 60 = 24\) minutes. These unit conversion errors propagate through the entire solution.
The problem involves several fraction calculations (\(\frac{2}{3}\) hour, \(\frac{2}{5}\) hour, \(\frac{4}{15}\) hour). Students often struggle with operations like \(20 \div \frac{4}{15}\), forgetting to flip the fraction when dividing, or making errors in multiplication like \(20 \times \frac{15}{4}\). A common error is calculating \(20 \times \frac{15}{4}\) as \(\frac{20 \times 15}{4} = \frac{300}{4}\) incorrectly.
Students may correctly calculate individual time segments but then make subtraction errors when finding remaining time. For instance, if they calculate total time as 40 minutes and first segment time as 24 minutes, they might compute the remaining time incorrectly due to rushed arithmetic.
After making calculation errors, students often end up with values like 70 mph (from the averaging misconception) and see it matches choice (C), giving them false confidence. They may not double-check their work because 70 mph "feels reasonable" as it's higher than 50 mph but not extremely high.
Some students might calculate correctly but then second-guess their units. If they calculated speed in miles per minute or some other unit during intermediate steps, they might accidentally convert incorrectly for the final answer, leading them to select a different multiple choice option.