Vincen wrote:Greg travels from city A to city B along route K and returns along route L. Greg's average speed during the onward journey is 40kmph.
What is his average speed for the entire trip?
1) Greg's average speed for the return journey is 3/4ths that for the onward journey.
2) Route K is 2/5ths longer than route L.
The OA is C
Experts I need your help. I get stuck with such kind of DS questions.
Why statement 1 alone is not sufficient?
Say the route AKB is x km and the route ALB is y km.
Average speed for the entire trip = Total distance / Total time = (x + y) / Total time
Thus, time taken in onward journey = x/40 hours
1) Greg's average speed for the return journey is 3/4ths that for the onward journey.
=> The average speed for the return journey = 3/4 of 40 = 30 kmph
=> Time taken in return journey = y/30 hours
Average speed for the entire trip = (x + y) / Total time = (x + y) / (x/40 + y/30) = 120(x+y) / (3x+4y) kmph
Can't get the answer. Insufficient.
Since the length of the routes is not the same, we cannot get the answer with the help of Statement 1 alone.
2) Route K is 2/5ths longer than route L.
=> y = (1 + 2/5) of x = 7x/5
Average speed for the entire trip = (x + y) / Total time = (x + 7x/5) / Total time. We do not have the value of Total time. Insufficient.
(1) and (2) together
From Statement 1, we have Average speed for the entire trip = 120(x+y)/(3x+4y) kmph and from (2), we have y = 7x/5
Thus, Average speed for the entire trip = 120(x + 7x/5) / (3x + 28x/5) = 120[(1 + 7/5) / (3 + 28/5)] = 120[(12/5) / (43/5)] = 120*(12/43) = a finite value. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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