Joseph traveled from Boston to Framingham at 50 mph and then back to Boston at 40 mph. What was Joseph’s average speed on the round trip?
The correct answer and explanation is:
Correct Answer: 44.44 mph (or 400/9 mph)
Explanation:
To calculate the average speed for a round trip where the distance is the same both ways but the speeds are different, we do not take the simple average of the two speeds (which would be (50 + 40)/2 = 45 mph — this is incorrect). Instead, we use the harmonic mean formula for average speed: Average speed=2aba+b\text{Average speed} = \frac{2ab}{a + b}
Where:
- a=a = speed going (50 mph)
- b=b = speed returning (40 mph)
Plug in the values: Average speed=2⋅50⋅4050+40=400090=4009≈44.44 mph\text{Average speed} = \frac{2 \cdot 50 \cdot 40}{50 + 40} = \frac{4000}{90} = \frac{400}{9} \approx 44.44 \text{ mph}
Why This Works:
Let’s break it down using actual distances and times.
Assume the distance between Boston and Framingham is D miles.
- Time to go from Boston to Framingham at 50 mph: t1=D50t_1 = \frac{D}{50}
- Time to return at 40 mph: t2=D40t_2 = \frac{D}{40}
- Total distance of the round trip: 2D2D
- Total time: t=D50+D40t = \frac{D}{50} + \frac{D}{40}
Find a common denominator: t=D(150+140)=D(4+5200)=D(9200)t = D\left(\frac{1}{50} + \frac{1}{40}\right) = D\left(\frac{4 + 5}{200}\right) = D\left(\frac{9}{200}\right)
Now, average speed = total distance / total time: 2DD⋅9200=29200=4009≈44.44 mph\frac{2D}{D \cdot \frac{9}{200}} = \frac{2}{\frac{9}{200}} = \frac{400}{9} \approx 44.44 \text{ mph}
Conclusion:
Joseph’s average speed on the round trip is 400/9 mph, or approximately 44.44 mph, not the simple average of 50 and 40.