Hi,

I did it this way. the way I set up my eqns were based on the assumption that the two journeys have equal distance and are completed in equal time T.

now for Journey 1: We travel distance d, at speed s, and time=d/s

for Journey 2: We travel distance d, at speed 2s and time =d/2s

Now let us account for the times the train was stopped

We know that on Journey 1 it spent 3 hrs stopped and journey 2 it spent 10*0.5=5hrs stopped. Now we know that both journeys were of equal overall duration.
d/s+3=d/2s + 5

solving these eqns yields d=4s

This way we can now compute the time for journey 1 was t=d/s = 4s/s=4 hrs
the second journey was d/s=4s/2s=2 hrs

total=> 4+3+1+2+5=15
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Hey guys, had a bit of trouble with this one. Maybe I missed something but I think its because the definition of when one trip begins and the other one ends is a bit ambiguous here (that or I'm just dumb

I created an RTD chart for this. Since we havent been given any conrete info on distance, I assumed the distances are the same and set d=300 (picked off top of my head). I represented the first trip as Rxt= D and also put an entry for the three hour long stops. I then assumed this is the end of trip 1.

For trip 2, I entered in the break periods (5 hours during the trip and the one hour before the train sets off), and the equation for the train's movement. Rate is 2R since we are told the rate is two times faster, and initially I represented the time as "M". Since we know the times are equal, I set T+3 (total time from the first trip plus breaks) equal to M+6 (total time from second trip plus break, which becomes M= T-3. Now we have two equations; 600-6R=300 and RT=300, which when solved equals R=50, therefore, because RT=300, T=6. This means the length of each total trip including breaks is 9 hours, and the chart appears to add up. This would make the answer choice E.

Any and all help greatly appreciate,

-Ken

R x T = D
Trip 1
R T 300
0 3 0
Trip 2
0 1 0
0 5 0
2R (T-3) 300

Let the time for each trip be t hours, then the roundtrip took t+t+1=2t+1 hours (each trip took the same amount of time plus 1 hour of waiting after the first half of the trip).

Next,, during the first trip the train was stopped for 3*1=3 hours and for the second trip it was stopped for 10*0.5=5 hours, so for the first trip it was moving at some constant rate, say r, for t-3 hours and for the second trip it was moving for the constant rate of 2r, for t-5 hours.

Now, since the distance for the first and the second trip was the same then r*(t-3)=2r*(t-5) --> t=7.

So, the roundtrip took 2t+1=15 hours.

Answer: B.
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