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Given above is the schematic map of the metro lines in a city with rectangles denoting terminal stations (e.g. A), diamonds denoting junction stations (e.g. R) and small filled-up circles denoting other stations. Each train runs either in east-west or north-south direction, but not both. All trains stop for 2 minutes at each of the junction stations on the way and for 1 minute at each of the other stations. It takes 2 minutes to reach the next station for trains going in east-west direction and 3 minutes to reach the next station for trains going in northsouth direction. From each terminal station, the first train starts at 6 am; the last trains leave the terminal stations at midnight. Otherwise, during the service hours, there are metro service every 15 minutes in the north-south lines and every 10 minutes in the east-west lines. A train must rest for at least 15 minutes after completing a trip at the terminal station, before it can undertake the next trip in the reverse direction. (All questions are related to this metro service only. Assume that if someone reaches a station exactly at the time a train is supposed to leave, (s)he can catch that train.)
If Priya is ready to board a train at 10:25 am from station T, then when is the earliest that she can reach station S?
Priya can reach S from T via R or V.
Case 1:- T-V-S
In the east-west direction, the first train from P arrives at T at time = 6 am + $$\left(4\times2\right)+\left(3\times1\right)=\ 11$$ minutes = 6:11 am
Since T is a junction so this train will halt for 2 minutes at T and leave at 6:13.
Priya boards a east-west train then Priya will board a train for V from T at 10:33 am.
There are 9 stations between T and V
Travelling time between T and V = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes
Therefore, Priya will reach V latest by 10:33 am + 29 minutes = 11:02 am
In the north-south direction, the first train from D arrives at V at time = 6 am + $$\left(3\times3\right)+\left(2\times1\right)=\ 11$$ minutes = 6:11 am
Since V is a junction so this train will halt for 2 minutes at V and leave at 6:13.
Since every 15 minutes, a train starts from D in the north-south direction, so the latest by which Priya will be able to board such a train from V is at 11:13 am.
There are 3 stations between V and S
Travelling time between R and S =$$\left(4\times3\right)+\left(\left(3\times1\right)\right)=\ 15$$ minutes
Time by which she reaches S = 11:13 +15 minutes = 11:28 am
Case 2:- T-R-S
In the north-south direction, the first train from B arrives at T at time = 6 am + $$\left(3\times3\right)+\left(2\times1\right)=\ 11$$ minutes = 6:11 am
Since T is a junction so this train will halt for 2 minutes at T and leave at 6:13.
Since every 15 minutes a train starts from P in the east-west direction so the latest by which Priya will be able to board such a train is at 10:28 am.
Now since she will be able to board a north-south train earlier than the east-west train so Priya will board a train for R from T at 10:28 am.
There are 3 stations between T and R
Travelling time between T and R = $$\left(4\times3\right)+\left(\left(3\times1\right)\right)=\ 15$$ minutes
Therefore, Priya will reach R latest by 10:43 am
In the east-west direction, the first train from M arrives at R at time = 6 am + $$\left(4\times2\right)+\left(3\times1\right)=\ 11$$ minutes = 6:11 am
Since R is a junction so this train will halt for 2 minutes at R and leave at 6:13.
Since every 10 minutes, a train starts from M in the east-west direction, so the latest by which Priya will be able to board such a train is at 10:43 am.
There are 9 stations between R and S
Travelling time between R and S = $$\left(10\times2\right)+\left(9\times1\right)=29\ $$ minutes
Time by which she reaches S = 10:43 +29 minutes = 11:12 am
We are getting shorter time in case 2. So 11:12 am is the answer
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