D01-25

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.

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

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

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.