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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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let distance be d; let the overall time be t t-3 is the time it takes for the train to reach point B from A t-5 is the time it takes for the train to return

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

There are a lot of great explanations out there, but I found someone posted one earlier that seemed to be simpler, if worded a little differently. To me this approach shows a simplicity of thought, and there is the added *minor* incentive of solving the problem in about 30 seconds -

first journey travel time = t, waiting time = 3 second journey travel time = t/2 (since double the speed), waiting time = 5

since the second journey saved 2 hours t-t/2 = 2 hours so t=4

So first journey = 4+3 = 7 hours = second journey total time = 7+7+1

Probably simpler if we can think this straight in the actual exam, eh?

First, we have to calculate the amount of time the train spent for the stops. \(3*1=3\) hours for the first trip and \(10*0.5=5\) for the return trip. Now, we can write an equation with \(S\) for one-way distance and \(V\) for train's speed:

\(\frac{S}{V} + 3 = \frac{S}{2V} + 5\)

\(\frac{S}{V} - \frac{S}{2V} = 2\)

\(\frac{2S-S}{2V} = 2\)

\(\frac{S}{2V} = 2\)

\(\frac{S}{V} = 4\)

So, the roundtrip lasted for \(7+7+1=15\) hours (we should count the 1 hour stop in the destination point as well).

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 \(r*(t-3)=2r*(t-5)\) --> \(t=7\).

We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.

t = total time. This is important because t is the total time for each way of the round trip including stops Total time of stop first part of the trip = 1 hour * 3 = 3 hours Total time of stop second part of the trip = 0.5 hour *10 = 5 hours

d = s (t -3) First part of the trip d= 2s (t-5) Second part of the trip

Both distance are the same s (t – 3) = 2s (t – 5) (t – 3) = 2 (t – 5) t – 3 = 2t – 10 t = 7 (7 * 2) both ways + 1 hour stop when reached destination 14 + 1 = 15 hours Total Trip

How do you know to do (t-3) and (t-5)? Someone explain that part in plain English?

We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.

t = total time. This is important because t is the total time for each way of the round trip including stops Total time of stop first part of the trip = 1 hour * 3 = 3 hours Total time of stop second part of the trip = 0.5 hour *10 = 5 hours

d = s (t -3) First part of the trip d= 2s (t-5) Second part of the trip

Both distance are the same s (t – 3) = 2s (t – 5) (t – 3) = 2 (t – 5) t – 3 = 2t – 10 t = 7 (7 * 2) both ways + 1 hour stop when reached destination 14 + 1 = 15 hours Total Trip

How do you know to do (t-3) and (t-5)? Someone explain that part in plain English?

We know that each trip took the same amount of time and has the same distance.Then we can equal either distance or time. I chose distance, dzyubam chose time.

t = total time. This is important because t is the total time for each way of the round trip including stops Total time of stop first part of the trip = 1 hour * 3 = 3 hours Total time of stop second part of the trip = 0.5 hour *10 = 5 hours

d = s (t -3) First part of the trip d= 2s (t-5) Second part of the trip

Both distance are the same s (t – 3) = 2s (t – 5) (t – 3) = 2 (t – 5) t – 3 = 2t – 10 t = 7 (7 * 2) both ways + 1 hour stop when reached destination 14 + 1 = 15 hours Total Trip

.

WHY YOU SAY "This is important because t is the total time for each way of the round trip including stops"...

gmatclubot

Re: GMAT Diagnostic Test Question 26
[#permalink]
05 Jan 2014, 23:27