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At 9:00 a.m. train T left the train station and two hours later train

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At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 26 Aug 2019, 23:19
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At 9:00 a.m. train T left the train station and two hours later train S left the same station on a parallel track. If train T averaged 60 kilometers per hour and train S averaged 75 kilometers per hour until S passed T, at what time did S pass T?

(A) 2:00 p.m.

(B) 5:00 p.m.

(C) 6:00 p.m.

(D) 7:00 p.m.

(E) 9:00 p.m.

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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 26 Aug 2019, 23:41
2
train T speed = 60 km/hr
train s speed = 75 km/hr

train T left the station at 9:00 pm and 2 hours later train S left the same station

distance covered by train T in 2 hours = 60*2 = 120 km

distance between train T and train S = 120

total time taken by train S to overtake train T (from 11:00 am) = \(\frac{120}{75-60}\) = 8 hours

train S will overtake T at 7:00 pm

D is the answer
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 26 Aug 2019, 23:48
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At the time S passes T, they both would have covered the same distance, d.
We know Ss=75km/h and St=60km/hr
We also know that T departs two hours earlier (i.e. 9am) than S (i.e. 11am). Let the time taken by T to cover distance d be t, then S would have covered distance d within t-2

So equating the distances covered by both trains, we get 60t=75(t-2)
150=15t
t=10 hrs

10 hours after 9am is 7pm. Answer is therefore D imo.

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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 00:03
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From given data,

60(x) = 75(x-2) , x is total time taken
60x = 75x - 150
15x = 150
x = 10

so total time taken is 10 hrs i.e at 7pm
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 01:56
Train T has already traveled 60*2=120 km before train S started.

Their relate speed = 75-60=15 kmh

Time required for train S to cross train T= 120/15 = 8 hrs

Time T left the station = 9 AM

Time S crosses T = 9 AM + 2 + 8 hrs = 7 PM

IMO D

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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 03:05
When train s passes the train T, they have travelled the same distance, so:
60t=75(t-2)
15t=150
t=10
at 7 pm they meet each gether.
Option D

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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 03:15
Train T and Train S travel in the same direction, their speeds need to be subtracted: 75 - 60 = 15km/h

Train S leaves 2h after train T. At 11AM, Train T already travelled 60km/h x 2h = 120km.
How much time will it take to S to travel 120km with a speed that is 15km/h faster that T?
120 / 15 = 8h from the departure time of train S so, 11h + 8h = 19:00 or 7PM.

Answer D
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 03:27
At 9:00 a.m. train T left the train station and two hours later train S left the same station on a parallel track. If train T averaged 60 kilometers per hour and train S averaged 75 kilometers per hour until S passed T, at what time did S pass T?

(A) 2:00 p.m.

(B) 5:00 p.m.

(C) 6:00 p.m.

(D) 7:00 p.m.

(E) 9:00 p.m.

Speed of Train T = 60 kmph
Speed of Train S = 75 kmph

Train T starts at 9:00am and Train S starts after 2 hours i.e. 11:00am by which time Train T travels 60x2 = 120 km.
Thus to travel 120 km train S takes \(\frac{120}{(75 - 60)} = 8\) hours. Hence adding 8 hours to 11:00 am gives 7:00 pm.

Alternatively:
Distance traveled by both trains as follows:

Time Train T Train S
9am 0 -
10am 60 -
11am 120 0
12pm 180 75
1pm 240 150
2pm 300 225
3pm 360 300
4pm 420 375
5pm 480 450
6pm 540 525
7pm 600 600
8pm..

At 7pm they meet after which train S overtakes train T.

Answer (D).
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post Updated on: 28 Aug 2019, 00:24
At 9:00 a.m. train T left the train station and two hours later train S left the same station on a parallel track. If train T averaged 60 kilometers per hour and train S averaged 75 kilometers per hour until S passed T, at what time did S pass T?

(A) 2:00 p.m.

(B) 5:00 p.m.

(C) 6:00 p.m.

(D) 7:00 p.m.

(E) 9:00 p.m.

Relative speed = 75-60 ; 15kmph
and distance covered by train T in 2 hrs ; 2*60 ; 120 km
so relative time for train S ; 120/15 ; 8 hrs
total time S passes T ; 9am + 8 hrs+ 2hrs ; 7 PM
IMO D

Originally posted by Archit3110 on 27 Aug 2019, 03:30.
Last edited by Archit3110 on 28 Aug 2019, 00:24, edited 1 time in total.
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 04:22
Trains will be parallel when each of the train traveled same distance.

Train T departs at 9 am:
At 2 pm = 5 hours *60 = 180 km (distance covered)
At 5 pm = 8 hours *60 = 480 km
At 6 pm = 9 hours *60 = 540 km
At 7 pm = 10 hours *60 = 600 km
At 9 pm = 12 hours *60 = 720 km

Now train S which departs at 11 am (2 hours later):
At 2 pm = 3 hours *75 = 225 km (distance covered)
At 5 pm = 6 hours *75 = 450 km
At 6 pm = 7 hours *75 = 525 km
At 7 pm = 8 hours *75 = 600 km
At 9 pm = 10 hours *75 = 750 km

At 6 pm both trains will be parallel
Answer: D
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 06:29
D is the right answer. 7 PM
Let the distance traveled is in km.

Then according to given data in the question stem,
At 11 am, T and S would have traveled 120 and 0km
At 1 pm, 240 and 150km
At 3 pm, 360 and 300km
At 5 pm, 480 and 450km
At 7 pm, 600 and 600km.

But I have one doubt. If the question does not mention that they are traveling at constant speeds then can we apply this logic and get to any answer?
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 10:53
D - 7pm

Both meet at 500km
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 27 Aug 2019, 12:52
Let it takes time t for train T to travel certain distance after which Train S overtakes it.
So time taken by train S : t-2 : as it travels 2 hours after it

60*t =75* (t-2)
, Solving for t= 10 : 9 a.m + 10 hours= 7 p.m
IMO: D
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 28 Aug 2019, 00:04
Suppose S took h hours to pass T
Distance traveled by T = Distance traveled by S in h hours

Since T started 2 hours earlier, total distance traveled by T in h hours
=60*2 + 60*h (distance= speed*time)

Distance traveled by S in h hours
= 75*h

=> 120+ 60h = 75h

=> 120= 15h

or h = 8 hours

Since S started at 11 a:m
time at which S will cross T is

11 a:m + 8 hours
i.e. 7 p:m

correct answer choice is (D)
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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 28 Aug 2019, 00:40
The 1st train would have travelled 2*60 =120 km when 2nd train strted.
Since the difference of speed is 75-15 km/h
The 2nd train will cross 1st in 120/15= 8 hours.
So time will be 11+8= 7 pm.
D imo is correct choice.

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Re: At 9:00 a.m. train T left the train station and two hours later train  [#permalink]

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New post 28 Aug 2019, 02:12
let T travelled 120 km in 2 hrs .
time needed for S to travel 120 km with (75-60)km/hr = 120/15 = 8
train S will pass train T at 11:00 + 8 hrs = 07:00 pm
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Re: At 9:00 a.m. train T left the train station and two hours later train   [#permalink] 28 Aug 2019, 02:12
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