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A train X departs from station A at 11.00 am for station B, which is

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A train X departs from station A at 11.00 am for station B, which is 180 km away. Another train Y departs from station B at 11.00 am for station A. Train X travels at an average speed of 70 kms/hr and does not stop any where until it arrives at station B. Train Y travels at an average speed of 50 kms/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B en route to station A. Ignoring the lengths the train, what is the distance, to the nearest km, from station A to the point where the trains cross each other?

A. 112
B. 118
C. 120
D. 122
E. Cant be determined
[Reveal] Spoiler: OA

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A train X departs from station A at 11.00 am for station B, which is [#permalink]

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anupamdev wrote:
A train X departs from station A at 11.00 am for station B, which is 180 km away. Another train Y departs from station B at 11.00 am for station A. Train X travels at an average speed of 70 kms/hr and does not stop any where until it arrives at station B. Train Y travels at an average speed of 50 kms/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B en route to station A. Ignoring the lengths the train, what is the distance, to the nearest km, from station A to the point where the trains cross each other?

A. 112
B. 118
C. 120
D. 122
E. Cant be determined


Time taken by train Y to travel a distance of 60km\(= \frac{60}{50}*60=72\) minutes

Train Y was stationed at point C for 15 minutes, hence total time elapsed to reach point C \(= 72+15=87\) minutes.

During this period train X covered a distance of \(= 70*\frac{87}{60} = 101.5\) or approx \(102\)km

so distance left between these two train to cover \(= 180-102-60=18\)km

as the trains are moving in opposite direction, relative speed \(= 70+50=120\) km/h

time taken to cover 18km \(= \frac{18}{120}\)

so distance traveled by train X \(= 70*\frac{18}{120}=10.5\)

Hence train X's distance from A \(= 102+10.5=112.5 =112\) (as rounding was done earlier)

Option A

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A train X departs from station A at 11.00 am for station B, which is [#permalink]

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anupamdev wrote:
A train X departs from station A at 11.00 am for station B, which is 180 km away. Another train Y departs from station B at 11.00 am for station A. Train X travels at an average speed of 70 kms/hr and does not stop any where until it arrives at station B. Train Y travels at an average speed of 50 kms/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B en route to station A. Ignoring the lengths the train, what is the distance, to the nearest km, from station A to the point where the trains cross each other?

A. 112
B. 118
C. 120
D. 122
E. Cant be determined



hi..

Combined speed = 70+50 = 120..

so both travel for 60/50 hour before Y halts....
distance traveled in \(\frac{60}{50} = 120*\frac{60}{50} = 144\) kms
During 15 min halt of Y, X travels \(70*\frac{15}{60}=17.5\)

remaining distance to be covered = 180-144-17.5 = 18.5
this distance will be covered by both trains in \(\frac{18.5}{120}*60 = 9.25\) minutes
distance that will be traveled by Y in \(9.25 = 50*\frac{9.25}{60} = 7.7\)

so Distance traveled by Y from B = \(60+7.7= 67.7\)..
distance from A = \(180-67.7 = 112.3\)

Another Logical way..
It can be seen from the distance that X and Y meet after Y starts after the halt..
so Distance traveled in those 15 minutes is \(70*\frac{1}{4}=17.5\)
the remaining distance \(180-17.5=162.5\) in ratio 70:50..
so X travels \(162.5*\frac{7}{7+5} = 94.8\)

total distance = 94.8+17.5=112.3

A
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Re: A train X departs from station A at 11.00 am for station B, which is [#permalink]

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Quickest method: Assume X runs for 't' hrs. to meet Y. Then Y runs for (t-1/5)hrs. to meet X (since Y makes a 15 min. stopover). So, in order to arrive at the meeting point, distance covered by A is 70t kms and that covered by Y is 50(t-1/5). The sum of these two distances is the distance between Stations A & B which is 180kms. So, 180=70t+50(t-1/5). Thus, t=(77/48)hrs. The distance of the meeting point from Station A is the distance traveled by X which is 70(77/48)=112.3
Note: The fact that Station C is 60kms from B is irrelevant. In fact, C could have been anywhere along the route as long as it was less than (180-112.3)kms from St. B.

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Re: A train X departs from station A at 11.00 am for station B, which is [#permalink]

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New post 21 Oct 2017, 02:00
Sorry, I made a mistake in my post. Should have been Y runs for (t-1/4)hrs., not (t-1/5)hrs.

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Re: A train X departs from station A at 11.00 am for station B, which is [#permalink]

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New post 21 Oct 2017, 20:05
anupamdev wrote:
A train X departs from station A at 11.00 am for station B, which is 180 km away. Another train Y departs from station B at 11.00 am for station A. Train X travels at an average speed of 70 kms/hr and does not stop any where until it arrives at station B. Train Y travels at an average speed of 50 kms/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B en route to station A. Ignoring the lengths the train, what is the distance, to the nearest km, from station A to the point where the trains cross each other?

A. 112
B. 118
C. 120
D. 122
E. Cant be determined


Train X has a speed of 70 km/hr
Train Y has a speed of 50 Km/hr

Lets solve this problem step by step :
Step 1 : If there is no stoppage at C, time taken by both trains to cross each other = 180/(70+50) = 180/120 = 1.5 hrs
Step 1: Y reaches station C
Time taken by Y = 60/50 = 6/5 hrs. So, till this time trains will not cross each other as 6/5 < 1.5 hrs.
Distance traveled by Y = 60 Km
Distance traveled by X in this time = 70*6/5 = 84 Km
Total distance traveled by both trains = 60+84 = 114 km.

Step 2 : Y is at station C for 15 minutes
Distance traveled by x till this time = 70 (6/5+1/4) = 70 *1.45 = 101.5
Total distance traveled by both trains = 101.5 +60 = 161.5 . So, even after 15 minutes of stoppage of Y, the trains didn't cross each other.

Step 3: Y starts running again after 15 minutes stoppage at station C.
Now, total distance left to travel = 180 - 161.5 = 18.5 km
Time taken by both trains to cover this distance = 18.5/120
Distance traveled by train X in this 28.5 km = 70/120 * 18.5 = 10.791 km
Total distance traveled by train X = 101.5 +10.791 = 112.291 km

So, distance from station A to the point where trains cross each other = 112.291 km ~ 112 km

Answer A
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Re: A train X departs from station A at 11.00 am for station B, which is   [#permalink] 21 Oct 2017, 20:05
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