A salesman drove from his home to a nearby city at an average speed of 40 mi/. He returned home at an average speed of 50 mi/h. What was his average speed for the entire trip?
I tried this question, and apparently the answer isn't 45 mph. It's something of a trick question. Any help/hints would be much appreciated!
Average speed = total distance / total time
time for first trip = d/40
time for return trip = d/50
total time = d/40 + d/50 = 9d/200
total distance = 2d
avg speed = 2d/(9d/200) = 400/9 = 44.444.. mph
Thank you!
To find the average speed for the entire trip, we need to calculate the total distance traveled and the total time taken.
Let's assume the distance from the salesman's home to the nearby city is "d" miles. Since he traveled to the city and back, the total distance traveled is 2d miles.
Now, let's find the total time taken.
Time taken for the outbound journey = distance / speed = d / 40 hours.
Time taken for the return journey = distance / speed = d / 50 hours.
To find the total time taken for the entire trip, we add the time taken for the outbound and return journeys:
Total time = (d / 40) + (d / 50) = (5d + 4d) / (40 * 50) = 9d / 200 hours.
Finally, to find the average speed, we divide the total distance by the total time:
Average speed = total distance / total time = 2d / (9d / 200) = 2 * 200 / 9 = 400 / 9 ≈ 44.44 mph.
So, the average speed for the entire trip is approximately 44.44 mph, not 45 mph.