enigma123 wrote:
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 trip from Townsend to Smallville took 50% longer than the trip from Smallville to Townsend.
(2) The distance from Townsend to Smallville is 165 miles.
How can the answer be A folks?
A couple of basics:
1. Average Speed = Total Distance/ Total Time
2. Average Speed = Weighted Average of different speeds where 'time taken' is the weight
3. If equal distances (say d) are traveled at different speeds, a and b, average speed = 2d/[d/a + d/b] = 2ab/(a+b)
In this case, average speed is independent of 'd', the distance.
4. If speeds a and b are maintained for equal time intervals, say t,
average speed = (at+bt)/2t = (a+b)/2
In this case, average speed is independent of 't', the time interval.
Given: Car travels equal distances. Speed in first case is 40 mph. We need the speed in the second case to get the average speed (given by 2*40*b/(40+b))
(1) Ratio of time in the two cases
T to S: S to T= 3:2
Then ratio of speed in the two cases = 2:3
We know the speed from T to S = 40
Then speed from S to T must be 60
Since we have the value of b (= 60) i.e. the speed while coming back, we can easily find the average speed..
Sufficient.
(2) As we discussed, distance traveled doesn't affect average speed in this case. Not sufficient
Answer (A)
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55% (01:41) correct
