D01-25

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
_________________

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!

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

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)
_________________

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

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!
_________________

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 \(4+3+2+5+1=15\).

Please give Kudos if it helps.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
_________________