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No hours stopped from A to B = 3 hrs
No hours stopped from B to A = 5 hrs (10*0.5)
No hours stopped at B = 1 hour

Let Distance be D
Speed traveled by train from A to B = x
Speed traveled by train from B to A = 2x
3+(D/X)=5+(D/2x) (given)
D = 4x ------------------(1)

Total time = Time taken from AtoB + Time taken from BtoA + hours stopped in B
Total time = 3+(D/x) + 5+(D/2x) + 1
Sub (1) in above

total time = 15hrs Option (B)
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I think this is a high-quality question and I agree with explanation.
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ALTERNATIVE SOLUTION for those who are not able to fully understand Bunuel's approach.


Make a grid as per the given data

Rate..................Time........Distance.......... Stops(hrs)
A to B r..............1/r.................1..............3
B to A 2r............1/2r ..............1.............. 5

It is given that the time it took for both trips is the same.
Use algebra to set up the equation

TOTAL TIME for A to B = TOTAL Time for B to A
1/r + 3 = 1/2r + 5
r=1/4

Substitute r=1/4 in equations for total time for round trip.

Time for ROUNDTRIP = Total time for A to B + Total time for B to A + 1 hour stop at B
1/r + 3 + 1/2r + 5 + 1
7 + 7 + 1

Total time for roundtrip = 15hours.

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Bunuel
A train is traveling at a constant speed from city A to city B. Along this trip the train makes three one-hour stops and reaches city B. At city B the train is stopped again for 1 hour. After that the train makes the return trip from city B to city A at a constant speed which is twice the speed of the first trip. Along this return trip the train makes ten thirty minutes stops and reaches city A. If both trips took the same amount of time, how many hours was the roundtrip?

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

Stoppage Time from A to B = 3*60 mins = 180 mins = 3 hrs
Stoppage Time from B to A = 10*30 mins = 300 mins = 5 hrs

Stoppage time from A to B is 2 hrs less than stoppage time from B to A. But both trips took the exact same total time.
This means that the actual travel time from A to B was 2 hrs extra.

Think about it this way:
Two people X and Y must take the same time to travel the same distance. Speed of X is less than the speed of Y. What will X do to cover the distance in same time as Y? X will take fewer breaks. X will continue moving for longer to make up for this low speed. So if X needed 4 hrs to actually cover the distance and Y needed only 2 hrs, X will take only a 3 hrs break compared with Y who plans to take a 5 hour break in between. So the shorter break that X takes is to compensate for the longer time he will need to travel. Since he will need 2 hrs extra to travel, he will take a break which is shorter by 2 hrs.

In the question we are given that A to B break time is 2 hrs less. It means that A to B travel time was 2 hrs extra.

Since speeds in the two trips are in the ratio 1:2, travel time taken would be in the ratio 2:1. The difference of 1 is actually 2 hrs so travel time from A to B is 2*2 = 4 hrs and travel time from B to A is 1*2 = 2 hrs.

Total Time for Return Trip = Total travel time + Stoppage Time + Break Time at B = (4 + 2) hrs + (3 + 5) hrs + 1 hr = 15 hrs

Answer (B)
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I think this is a high-quality question and I agree with explanation.
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Let’s assume two variables for the unknown quantities viz distance between A and B and the speed of the train.
Let distance between A and B = d miles.
Let speed of train = d miles per hour.

Then, total time taken by train for onward journey = \(\frac{d }{ s}\) + 3, taking into account the three one hour stops made on the way.

Total time taken by train for return journey = \(\frac{d }{ 2s}\) + 5 because the train travelled at twice the speed and also took 300 minutes i.e. 5 hours of stops.

Since both trips took the same amount of time,\(\frac{ d}{s} + 3 = \frac{d}{2s} + 5\). Solving the equation, we get d = 4s.

Therefore, time taken for onward journey = \(\frac{4s}{s}\) + 3 = 7 hours and,
Time taken for return journey = \(\frac{4s }{ 2s}\) + 5 = 7 hours.
The train also stopped for 1 hour at B.

Therefore, total time taken for the round trip = 7 + 7 + 1 = 15 hours.
The correct answer option is B.

Hope that helps!
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If this this helps:-

Total time for completion of trip = \(t1+3+t2+5+1\) = \(t1+t2+9\) [t1-time travelled with speed r from A to B, t2-time travelled with speed 2r from B to A, +3 stoppage time from A to B, +5 is stoppage time from B to A , and +1 is extra stoppage at station B]. -----(3)
We know distance from A to B is same as B to A so We get \(r*t1=2r*t2\)=\(t1=2*t2\). ------- (1)
Also given total time from A to B is same as time from B to A i.e \(t1+3=t2+5\) ------ (2).

From (1) and (2) , We get \(t1=4,t2=2\).
On putting t1 and t2 in (3) , We get total time as \(4+3+2+5+1=15\).
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I used an alternate solution.

The statement states that the ratio of the speed for onward journey to the speed for return journey = 1:2
Hence time for the same ratio = 2:1
Further, lets assume the time the train is moving(without stoppages) for onward journey = t. Therefore the time the train is moving for return journey = t/2.

Therefore, t+3=t/2+5. t=4.
Total time = t+3+5+t/2+1(stop at B) = 4+9+2 = 15.
Option B.
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My Approach

total time taken = (time from A to B) + (1hr waiting at B) + (time from B to A)

= (t1+3) + (1) + (t2+5)

( t1, t2 are time taken from A and B respectively without counting for stops );


It is given that

Time from A to B =Time from B to A

Therefore

t1 + 3 = t2+5 ;

t1=2*t2 since speed from B to A is double of that from A to B

2*t2 + 3 = t2 +5 ==> t2=2 hrs;

solving the equation t1 = 4 hrs; t2= 2 hrs

total time taken = (time from A to B) + (1hr waiting at B) + (time from B to A)

= (t1+3) + (1) + (t2+5)

= (4+3) + (1) + (2+5)

total time taken = 15 hrs
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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Given:
  • From A to B (forward trip): Let speed = r, distance = d
    • Stops: 3 one-hour stops during the forward trip
  • At B: 1-hour stop (this is between trips, not part of travel time)
  • From B to A: speed = 2r, distance = d
    • Stops: 10 thirty-minute stops = 5 hours during the return trip
  • Total forward time = total return time

To find: Total time of round trip

Solution:
Let’s break down travel time only (not including the 1-hour break at B, since it’s not part of either leg):

Forward trip:
  • Travel time = \(\frac{d}{r}\)
  • +3 hours of stopping during the trip
  • Total forward time =\(\frac{ d}{r }\) + 3

Return trip:
  • Travel time = \(\frac{d}{2r}\)
  • +5 hours of stopping during the trip
  • Total return time = \(\frac{d}{2r}\) + 5

Set both times equal:
  • \(\frac{d}{r}\) + 3 = \(\frac{d}{2r}\) + 5
    • \(\frac{d}{r}\) - \(\frac{d}{2r}\) = 2
    • \(\frac{d}{2r}\) = 2
    • \(\frac{d}{r}\) = 4 ---- This is the forward time without breaks.
  • Now plug back into total time:
    • Forward trip: 4 + 3 = 7
    • Return trip: 2 + 5 = 7
    • Add the 1-hour break at city B

Total round trip = 7 + 1 + 7 = 15 hours

Correct Answer: B

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my doubt is that in distance = speed * time formula : we consider the total trip time so here it should have been t+3 and t+5 respectively for to & fro journey, then why t-3 and t-5 instaed ?
TIA,
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
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anushree01
my doubt is that in distance = speed * time formula : we consider the total trip time so here it should have been t+3 and t+5 respectively for to & fro journey, then why t-3 and t-5 instaed ?
TIA,
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

The key is t represents the total duration of the trip, including stops.

So if the total trip took t hours, and there were 3 hours of stops, then the time the train was actually moving was t - 3 hours.

Same for the return trip: t total time, 5 hours of stops, so t - 5 hours of actual motion.

We subtract the stop time to get the time spent moving at constant speed.
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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?
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