D01-25

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)

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

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)

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.

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

I like the solution - it’s helpful.

I like the solution - it’s helpful.

I like the solution - it’s helpful.

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

Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years

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,

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.

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?