During a trip from City A to City B, a driver averages 25 mph. On his return trip, he drives faster, averaging 50 mph. What percent of the first travel time does the return trip take at the faster speed?
Answer: The return trip takes 50% of the time it took to complete the first trip. As 25 is half of 50, the return trip is “twice as fast and takes half as long.”
During a trip from City A to City B, a driver averages 40 mph. On his return trip, he drives faster, averaging 50 mph. What percent of the first travel time does the return trip take at the faster speed?
Answer: To solve this problem, we use the formula relating Distance, Rate, and Time, which is D = RT.
In this case, let d = 40 (speed of the first trip) and t = time at 40 mph, and D = 50 (speed of the return trip) and T = time at 50 mph.
Since the distance is the same, rt = RT. We rearrange this into the ratio of “rate to rate” and “time to time” (by dividing both sides by R and by t) which is r/R = t/T. Plugging in the values for time given in the question, we know that T/t = 40/50 = 4/5.
The ratio of the new time (T) to the old time (t) is 4/5, which equals 80%. So, the faster return trip takes only 80% of the time it took to make the first trip.
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