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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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!

hi guys i have a doubt,

question says
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.

now it means 3 hours+1 hours at station B?

now i was able to find running time as 7 hours

but than 7+5(stoppage time for B---------->A)+3(stoppage time for A-------->B)+1 hour for station B
that gives us 16 hours

Let's say that the trip on the way there took time T. Since distance is constant and we increase the speed two times, then we know that the time is cut in half. so T is the time to get there and .5T is the time to get back without considering any of the stops. Since we know that roundtrip is the same, then T + 3 = .5T + 5.

Solve for T = 4. so 4+3+ .5(5) + 5 + 1 (the extra hour one stayed).

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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