Official Solution:

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


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 at city B).

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 \(r*(t-3)=2r*(t-5)\), which gives \(t=7\).

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


Answer: B
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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)

My solution:
Let D be the distance and r be the rate:
As given: D/r + 3 (three one-hour stops) = D/2r + 5 ( ten thirty minutes stops = 5 hour)
=> D/r - D/2r = 2
=> D=4r => 4r/r + 3 = 7
=> Total time: 7+7+1=15
Is it right!

Hello Bunuel,

Could you please explain why we are not considering the total waiting time (during the trip) in total trip time?

Thanks

I think this is a high-quality question and I agree with explanation.

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.

If the above solution helped you. Hit Kudos like a Hero

Bunuel or other experts IanStewart EMPOWERgmatRichC VeritasKarishma chetan2u

I would like a reply to this as well... why are we subtracting the wait time when setting up the equation?

I get that rate of 0 * time = 0 distance covered, since the train is standing still ... but shouldn't this time be ADDED to the final amount, rather than subtracted? I feel like I am missing something obvious... but how is the time waiting not part of the trip? Is the problem supposed to imply that the "trip" is only the time spent moving?

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
Since both trips took same time, it means trip from A to B took 120 mins 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 120 mins so total travel time = 3*120 = 360 mins = 6 hrs

Total Time for Return Trip = Total travel time + Stoppage Time + Break Time at B = 6 hrs+ 8 hrs + 1 hr = 15 hrs

Answer (B)
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Time taken for return trip includes all stoppage time and all actual travel time. If time taken for return trip is 15 hrs, it means if the train starts from A at 12 noon, it reaches back to A at 3 am.
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Please let me know if this works, experts? Bunuel VeritasKarishma chetan2u


Time from A to B: T + 3
Where T= time taken from A to B

Time from B to A: X + 5
Where X= time taken from B to A

Since the speed from B to A is double the speed from A to B, the time from B to A must be half the time from A to B

Therefore, 2X= T

Total time is T + 3 + X + 6 ( 1 hr stop included)
= T + X + 9
= 2X + X + 9
= 3X + 9

We know that 2X = T
So, 2X + 3 = X + 5 ( given)
X=2

Therefore, total time = 3*2 + 9
= 15

I think this is a high-quality question and I agree with explanation.

From \(A\) to \(B\) at a rate of \(R\)

From \(B\) to \(A\) at a rate of \(2R\)

That also means that from \(A\) to \(B\) took twice as long as from \(B\) to \(A\) in raw (ignoring the stops for now) driving time

Including the stops, we can therefore formulate the following

\(\frac{t}{2}+5=3+t \implies t=4\)

Therefore, including the layover, the total time was

\( raw \ driving \ time + stops + layover =t+\frac{t}{2}+5+3+1=4+2+5+3+1\)
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Can we solve this using the below method but might be little time consuming

T1 + 5 +T2 +3 + 1
Since T1+5 = T2 +3, then T1 =2 and T2=4 makes both the time equal and has the option as 15

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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Bunuel, if the speed is two times in the return journey shouldn't time decrease?

Notice that we are directly told in the stem that both trips took the same amount of time.

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?

Yes, during the return trip, the speed of the train was twice that of during the first round but notice that during the return trip the train made 5 hour stop (Along this return trip the train makes ten thirty minutes stops) while during the first round it made only 3 hours stop (Along this trip the train makes three one-hour stops)
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Bunuel
Please let me know what I am missing here :
Let the time taken from A to B is t hours ( without delays)
Stoppages : 3 + 1 = 4 hours
so total time taken = (t + 4) hours

Return :
Return speed is twice , so time taken will be half = t/ 2
Stoppages : 10 * 0.5 hrs = 5 hrs
But we are told that both journey took equal time , so we can set up following equation : t+ 4 = t/2 + 5 , solving for t we get t= 2
Total time = 12 hrs , which is not there in choices. I am sure there is something I am missing here ?

Let the time taken from B to A be t hrs

Since the distance from A to B was covered in half the speed as B to A, therefore the time taken from A to B will be 2t hrs

Total time taken from B to A with stops = t + 0.5(10) = t + 5 hrs --- (1)

Total time taken from A to B with stops = t + 3 hrs --- (2)

Since given that both trips take same amount of total time, therefore (1) = (2)

2t + 3 = t + 5
gives t=2

Now, calculating total time by replacing value of t in eq (1) and (2)

i.e Total time = Time from A to B + 1 hour stop at B + Time from A to B
which gives, 4 + 3 + 1 + 2 + 5 = 15

Therefore, ans = 15

Time from A to B: T + 1*3 = T+3 hrs
Time from B to A: T/2 + 10*(1/2) = T/2 + 5. {Since the speed doubled in the return journey, the Time reduced by 1/2}

The two travel times were same

--> T + 3 = T/2 + 5 ---> T = 4
Total time: T + 3 + T/2 + 5 + 1 = 15hrs

(D) is the answer