Kaplan-train departure (tough)

singhsa wrote:A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

OA -25 mins

Is the ime taken from Station X to Station Y, and Station Y to Station Z : "A hour later"

singhsa wrote:A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

OA -25 mins

The travel time between stations is not very clear from the problem so I will assume it 1-hour since it says "an hour later".

Train from X starting 9 a.m. -> reaching Y at 10 a.m.
Train from Y starting 10:30 a.m. -> reaching Z at 11:30 a.m. (At Y, waiting time 30 mins.)
Train from Z starting at 11:45 a.m. (At Z, waiting time 15 mins.)
Total waiting time 30+15 = 45 mins.

I can't figure out how the waiting time can be anything other than a multiple of 15 mins.
What is the source of this question?

euro wrote:

singhsa wrote:A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

OA -25 mins

The travel time between stations is not very clear from the problem so I will assume it 1-hour since it says "an hour later".

Train from X starting 9 a.m. -> reaching Y at 10 a.m.
Train from Y starting 10:30 a.m. -> reaching Z at 11:30 a.m. (At Y, waiting time 30 mins.)
Train from Z starting at 11:45 a.m. (At Z, waiting time 15 mins.)
Total waiting time 30+15 = 45 mins.

I can't figure out how the waiting time can be anything other than a multiple of 15 mins.
What is the source of this question?

Kaplan test

singhsa wrote:A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

OA -25 mins

can you recheck the PS

diebeatsthegmat wrote:

singhsa wrote:A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

OA -25 mins

can you recheck the PS

I did....OA is 25 mins only.

Maybe experts might be able to help here.

Departure Times (DT) & Time to Destination (TD)
DT: Station X: 6am, 7am, 8am, 9am...
TD: 1 hour and 45 min

DT: Station Y: 9am, 9:30am, 10:00am, 10:30am...
TD: after 2 hours and 20 min

DT:: 8am, 8:45am, 9:30am, 10:15am, 11:00am, 11:45am, 12:30nn, 1:15pm, 2:00pm, 2:45pm, 3:30

Option 1: take 7am departure from station x

Departure form X: 7am
Arrival at Y: 8:45am
Wait time: 15min.

Departure from Y: 9:00am
Arrival at Z: 11:20am
Wait time: 25min.

Departure from Z: 11:45am.

Total Wait Time = 15 +25 = 40

Option 2: take 8am departure from station x

Departure form X: 8am
Arrival at Y: 9:45am
Wait time: 15min.

Departure from Y: 10:00am
Arrival at Z: 12:20am
Wait time: 10min.

Departure from Z: 12:30am.

Total Wait Time = 15 +10 = 25

Answer: 25

The problem should read as follows:

A woman is planning a trip that involves 3 connecting trains that depart from Stations X, Y, and Z, respectively. The first train leaves Station X every hour, beginning at 6 a.m., and arrives at Station Y 1(3/4) hours later. The second train leaves Station Y every half hour, beginning at 9 a.m., and arrives at Station Z 2(1/3) hours later. The third train leaves Station Z every 45 minutes, beginning at 8 a.m. What is the least total amount of time the woman must spend waiting between trains if all trains depart and arrive on schedule, and if she arrives at Station Z no later than 3:30 p.m.?

15 minutes
25 minutes
1 hour 15 minutes
1 hour 40 minutes
4 hours 30 minutes

The woman must arrive at Station Z by 3:30pm.
Let DT = departure time and AT = arrival time.
Write out the schedule and WORK BACKWARDS from DT = 3:30pm for the third train.

Train 1:
AT: 06:45, 07:45, 08:45, 09:45, 10:45, 11:45, 12:45, 01:45, 02:45...

Train 2:
DT: 09:00, 09:30, 10:00, 10:30, 11:00, 11:30, 12:00, 12:30, 01:00.
AT: 11:20, 11:50, 12:20, 12:50, 01:20, 01:50, 02:20, 02:50, 03:20.

Train 3:
DT: 08:00, 08:45, 09:30, 10:15, 11:00, 11:45, 12:30, 01:15, 02:00, 02:45, 03:30.

The times in red yield a wait-time of 15 minutes between trains 1 and 2 (from 12:45 to 01:00) and a wait-time of 10 minutes between trains 2 and 3 (from 03:20 to 03:30), for a total wait-time of 25 minutes.

The correct answer is B.

This question has a strong east coast (or European) bias, as few people in the rest of car-crazy America are used to dealing with transit tables!

That said, here's a pretty quick and easy way of approaching the problem.

The first train ALWAYS arrives at 45 minutes past the hour (7:45, 8:45, etc).

The second train ONLY leaves at 30 past or on the hour (9:30, 10:00, 10:30, etc.)

So there is a MANDATORY wait of AT LEAST 15 minutes between the first and second trains. (If we arrive at :45 and leave at :00.)

Now we'll move to the third train. The second train only arrives at 20 or 50 minutes past the hour (11:50, 12:20, 12:50, etc.)

Since the third train CANNOT leave at 20 or 50 minutes past the hour (it leaves at 8, 8:45, 9:30, 10:15, 11, etc., so either :00, :45, :30, :15), we have AT LEAST a 10 minute wait. (Either :20 to :30, or :50 to :00.)

That gives us a minimum of 15 minutes + 10 minutes, or 25 minutes.

Matt@VeritasPrep wrote:This question has a strong east coast (or European) bias, as few people in the rest of car-crazy America are used to dealing with transit tables!

That said, here's a pretty quick and easy way of approaching the problem.

The first train ALWAYS arrives at 45 minutes past the hour (7:45, 8:45, etc).

The second train ONLY leaves at 30 past or on the hour (9:30, 10:00, 10:30, etc.)

So there is a MANDATORY wait of AT LEAST 15 minutes between the first and second trains. (If we arrive at :45 and leave at :00.)

Now we'll move to the third train. The second train only arrives at 20 or 50 minutes past the hour (11:50, 12:20, 12:50, etc.)

Since the third train CANNOT leave at 20 or 50 minutes past the hour (it leaves at 8, 8:45, 9:30, 10:15, 11, etc., so either :00, :45, :30, :15), we have AT LEAST a 10 minute wait. (Either :20 to :30, or :50 to :00.)

That gives us a minimum of 15 minutes + 10 minutes, or 25 minutes.

Finally a quick way to do it. Can all transit be problems be solved this way though?

francoimps wrote:Finally a quick way to do it. Can all transit be problems be solved this way though?

This is a rather unusual problem, so I wouldn't lose too much sleep over it! But yeah, I'd look for a clever solution on any such problem: there has to be a better way than compiling the whole timetable by hand.