If a car traveled from Townsend to Smallville at an average

[Guest]

If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?
(1) The return trip took 50% longer than the trip there.

(2) The distance from Townsend to Smallville is 165 miles.


I think the answer should be E. Average speed is total distance traveled divided by total time taken. We are not given the amount of time spent relaxing in Smallville, so we do not know the total time for the trip. Am I being too paranoid in solving this problem or can I assume that the car turned around and returned immediately? I thought the whole point of DS was not to assume anything, which leads me to fear traps like this.

Fawad

[Guest]

I think A is sufficient.

I setup an RTD equation. RT=D

In the problem we are given the rate for the trip there as 40 mph, so 40t=d or 40 = d/t.

We know that the distance on the way back will be the same, and the time will be 50% longer.
Therefore the equation for the return trip is R2 * 1.5t = d.

To calculate the average rate we need to divide the total distance traveled by the total time travelling.

The total distance traveled is 2d. The total time travelling is t + 1.5t or 2.5t.

From this we can setup the equation Average Rate of Trip = 2d / 2.5t or (2/2.5) * d/t

At the beginning of the problem we said that d/t = 40 so we can substitute

Average Rate of Trip = (2/2.5) * 40 = 32mph

[StaceyKoprince]

Even if the person sat in Smallville for the entire day, you wouldn't include non-travel time. Average speed for a trip is based on the time spent traveling in the vehicle (car, train, or whatever).