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# GMAT Diagnostic Test Question 25

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Re: GMAT Diagnostic Test Question 26 [#permalink]

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08 Nov 2010, 11:02
Fijisurf wrote:
I set a little bit different equation:

v-speed
T-time in either direction

v(T-3) - for the way there
2v(T-5) - on the way back

v(T-3)=2v(T-5)
vT-3v=2vT-10v
7v-vT=0
v(7-T)=0
T=7

The total time is (do not forget to add extra hour of waiting before return trip) 7+7+1=15

I am able to understand by your method.
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Argument : If you love long trips, you love the GMAT.
Conclusion : GMAT is long journey.

What does the author assume ?
Assumption : A long journey is a long trip.

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Re: GMAT Diagnostic Test Question 26 [#permalink]

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04 Jan 2011, 13:22
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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Re: GMAT Diagnostic Test Question 26 [#permalink]

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09 Feb 2011, 16:33
can someone explain the flaw in my logic?

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

so d/(t-3) = 2d/(t-5)
solving I get t = 1

Obviously this is not correct. Any ideas?
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Re: GMAT Diagnostic Test Question 26 [#permalink]

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13 Oct 2011, 09:19
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
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Re: GMAT Diagnostic Test Question 26 [#permalink]

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01 Jan 2012, 00:11
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?
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Re: GMAT Diagnostic Test Question 26 [#permalink]

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07 Apr 2012, 13:49
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Expert's post
dzyubam wrote:
Explanation:
 Rating:

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

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

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Re: GMAT Diagnostic Test Question 26 [#permalink]

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14 Aug 2013, 15:10
Ikowill wrote:
Another way of solving this problem:

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?
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Re: GMAT Diagnostic Test Question 26 [#permalink]

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15 Aug 2013, 02:07
Ikowill wrote:
Another way of solving this problem:

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?

Can you please elaborate a bit? Thank you.

Meanwhile check this: gmat-diagnostic-test-question-79355-20.html#p1071311 Hope it helps.
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Re: GMAT Diagnostic Test Question 26 [#permalink]

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05 Jan 2014, 22:27
Ikowill wrote:
Another way of solving this problem:

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"...
Re: GMAT Diagnostic Test Question 26   [#permalink] 05 Jan 2014, 22:27

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# GMAT Diagnostic Test Question 25

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