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Bunuel

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14 Jun 2025, 01:05

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CapnSal

I did not quite understand the solution. Can we not equate the distance for the initial journey and the return journey (since they are the same) and then solve?

The solution does exactly that: it equates the distances for the initial and return trips.

Here’s the key step:

Distance from A to B = speed * time = r * (t - 3)
Distance from B to A = speed * time = 2r * (t - 5)

Since both distances are the same:

r * (t - 3) = 2r * (t - 5)

Then solve for t.

So yes, we’re equating distances, the solution just already includes that step.

Gmat232323

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05 Jul 2025, 05:03

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I like the solution - it’s helpful.

ishneetkaur28

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19 Jul 2025, 03:30

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AswinSundar

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15 Aug 2025, 07:51

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22 Aug 2025, 09:42

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Hi,

How are both these equations right and what do they signify/mean individually?

r∗(t−3)=2r∗(t−5)

3+(D/X)=5+(D/2x)

Bunuel

Official Solution:

A train travels at a constant speed from city A to city B. During the trip, the train makes three one-hour stops before arriving at city B, where it stops for another hour. The train then returns to city A at a constant speed, which is twice the constant speed of the initial trip. However, on the way back, the train makes ten thirty-minute stops before finally arriving at city A. If both the trips took the same amount of time, how many hours did the round trip take?

A. 14
B. 15
C. 16
D. 17
E. 18

Assume that the time taken for each trip is \(t\) hours, so the roundtrip duration is \(t + t + 1 = 2t + 1\) hours (as both trips have taken the same amount of time, plus an extra hour waiting time at city B after the first trip).

During the first trip, the train made three one-hour stops, and for the second trip, it made ten thirty-minute stops, which totals to three hours and five hours, respectively. Therefore, during the first trip, the train moved at a constant speed (let's call it \(r\)) for \(t-3\) hours, and for the second trip, it moved at a constant speed of \(2r\) for \(t-5\) hours.

Since the distance covered during both trips is the same, we can write the equation \(r*(t-3) = 2r*(t-5)\), which simplifies to \(t = 7\).

Thus, the roundtrip duration is \(2t+1 = 15\) hours.

Answer: B

Bunuel

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22 Aug 2025, 10:37

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harshitasinghal

Hi,

How are both these equations right and what do they signify/mean individually?

r∗(t−3)=2r∗(t−5)

3+(D/X)=5+(D/2x)

They’re just two ways of writing the same condition:

The first frames it as “distance = distance.”
The second frames it as “time = time.”

Equation 1:

r*(t − 3) = 2r*(t − 5)

Here, t is the total time of one trip.
On the first leg, the train moves at speed r for (t − 3) hours (since 3 hours were spent stopped).
On the return leg, it moves at 2r for (t − 5) hours (since 5 hours were spent stopped).
The distances must be equal, so r*(t − 3) = 2r*(t − 5).

Equation 2:

3 + (D/r) = 5 + (D/2r)

Here, the approach is to explicitly write travel time plus stop time.
First leg: travel time D/r plus 3 hours of stops.
Return leg: travel time D/(2r) plus 5 hours of stops.
Since both trips took the same time, 3 + (D/r) = 5 + (D/2r).

Hope it helps.

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10 Sep 2025, 07:49

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Hi,

I dont understand why are we adding +1 at the end as part of total time taken for roundtrip. Since the question states that train makes 3 hr stop before arriving at city 5, where it stops for another hour.

so similar to t - 3, shouldnt we subtract 1 from 2t?

Bunuel

Bunuel

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10 Sep 2025, 08:20

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MegB07

t is only the travel time of each leg. The extra 1 hour is a layover at city B between the two trips, so it must be added, not subtracted.

Total time = t + 1 + t = 2t + 1.

If you still have troubles understanding the official solution, you can check the previous discussion for alternate solutions. Hope it helps.

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25 Oct 2025, 06:32

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08 Nov 2025, 08:02

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10 Dec 2025, 08:50

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Updated on: 02 Jan 2026, 12:15

Originally posted by Vivek1707 on 02 Jan 2026, 12:14.
Last edited by Bunuel on 02 Jan 2026, 12:15, edited 1 time in total.

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Question can be done using the speed and time inverse relation concept.
If the speed gets multiplied by 2 then the travel time gets divided by 2 (since speed is inversely proportional to time when distance is constant)

So let time taken by Train to go from A to B (excluding stops) be t hours
So total time including stops = t + 3 + 1 hours

Time taken to travel from B to A = t/2 hours (excluding stops)
Hence total time including stops would be (t/2 + 5 ) hours

Since the question says both trips took the same amount of time
t+3 = t/2 + 5 | Mind you here on the LHS take t + 3 and not t+4 cause the extra 1 hour is not part of travel time (this was the mistake I made the first time

So t = 4

So total time for round trip = t + 3 + 1 + t/2 + 5
i.e. 4 + 3 + 1 + 2 + 5 i.e. 15

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01 Mar 2026, 04:28

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I like the solution - it’s helpful. Good

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25 Mar 2026, 19:55

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I like the solution - it’s helpful. By taking distance as 100km, calculation become a lot easier, initially i misinterpret it:
like i have considered that 1 hr halt at station B while calculating T(A-B) later i found what a silly mistake i was doing