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A train left a station P at 6 am and reached another station Q at 11

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A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Jul 2017, 03:32
1
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A
B
C
D
E

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A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Jul 2017, 04:32
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Let's call the train leaving P station is T1 and that leaving Q station is T2

Choose a number to be distance between P and Q, let's say 15 (miles). s1=s2=15
It took the train T1 5 hours (6am-11am) to travel: t1=5
It took the train T2 3 hours (7am-10am) to travel: t2=3

Formular: s = t x v => v = s/t

Speed of T1: v1= s1/t1 = 15/5 = 3 (miles/hour)
Speed of T2: v2= s2/t2 = 15/3 = 5 (miles/hour)

At 7am, train T1 already traveled: 1 x 3 = 3 (miles). At that time, the distance between 2 trains is: s= 15-3 = 12 (miles)

Because 2 trains take opposite directions: s = t(v1+v2)
Therefore, the number of hours it would take for both trains to pass one another is: t = s/(v1+v2) = 12/(3+5) = 1.5 (hours)
1.5 hours since 7am => 8:30am

Option (C) is correct.
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A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Jul 2017, 04:35
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Let the distance between the stations be x

First train(A) left station P- 6 AM and reached station Q - 11 AM, Time taken :5 hours.
Second train(B) left the station Q - 7 AM and reached station P - 10 AM, Time taken : 3 hours

Also since one of the trains left an hour later, the other train would have
traveled for an hour, covering \(\frac{x}{5}\) of the distance(Since Time: 5 hours)

For the remaining \(\frac{4x}{5}\) of the distance, the trains, together travel a distance of \(\frac{x}{3} + \frac{x}{5}(\frac{8x}{15})\) in an hour
Time taken for the trains to meet is \((\frac{4x}{5}/\frac{8x}{15}) = \frac{4*15}{5*8}= \frac{3}{2} hours\)

The trains meet at 8:30 am(Option C), one and an half hours after train B leaves the station Q.
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Jul 2017, 05:33
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Imo C
Let the distance be D
Speed of the first train be X
Speed of the second train be Y
Time taken by first train =5 hours
Time taken by second train=3 hours
X =D/5 and Y=D/3
first train travel for 1 hour from 6 to 7 am so distance covered is D-X*1=D-D/5=4*D/5
Now using concept of relative velocity we have
Time to meet =Distance traveled/Relative velocity
Relative velocity =D/5+D/3 as the trains are travelling in opposite direction
So we have (4*D/5)/(D/5+D/3) =(4/5)/(8/15)=3/2=1.5 hours
so the trains will meet at 8.30 am
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Jul 2017, 06:47
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Let the distance between the two stations be 150 kms.

Time taken by Train A = 5 hrs. Speed of Train A = 30 km/h.
Time taken by Train B = 3 hrs. Speed of Train B = 50 km/h.

Since they are travelling in opposite direction, their relative speed = 30+50 = 80. They must have passed one another at 120/80 = 1.5 hrs past 7 AM = 8:30 AM. Ans - C.
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Aug 2017, 08:15
Madhavi1990 wrote:
TimeTraveller wrote:
Let the distance between the two stations be 150 kms.

Time taken by Train A = 5 hrs. Speed of Train A = 30 km/h.
Time taken by Train B = 3 hrs. Speed of Train B = 50 km/h.

Since they are travelling in opposite direction, their relative speed = 30+50 = 80. They must have passed one another at 120/80 = 1.5 hrs past 7 AM = 8:30 AM. Ans - C.



What is 120 here?


As per the explanation here the first train starts at 6 AM i.e. 1 hour before the other train started journey. so the first train has already travelled 30 km till 7 AM wgen second train starts moving

hence, at 7 AM the distance left between two trains = 150-30 = 120
Relative speed at 7 AM = 30+50 = 80 km/h

Time = 120/80 = 1.5 hours


Hope this helps!!! :)
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 15 Nov 2017, 01:37
Let the distance between P and Q be 'd' miles and the time taken by T1 (which left P at 6:am) to travel to the point it meets T2 be 't' hrs.
T/4, distances covered by T1 and T2 until they meet is t(d/5)* miles and (t-1)**(d/3)* respectively. Obviously, the sum of these two distances is 'd'. Thus, t(d/5)+(t-1)(d/3)=d. So, t=2.5 hrs. Thus, T1 meets T2 2.5 hrs after it departs St.P, i.e at 8:30 am.

*Times taken by T1 and T2 to cover 'd' miles is 5 hrs and 3 hrs so their speeds are d/5 mph and d/3 mph respectively.
**T2 starts 1 hr after T1 so it travels 1 hr less.
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 15 Nov 2017, 04:47
Let's call the trains A and B. Note that both trains travel the same distance. (We'll get to that)

Train A's total time = 5 hours
Train B's total time = 3 hours

Let's choose an easy LCM between the two for distance. Let distance = 15 miles

Thus:
Train A's speed = 15/5 = 3 mph
Train B's speed = 15/3 = 5 mph

Now we know their speed! The question now asks when they meet up. Let's make a quick timetable for the time vs. distance travelled at each 1 hour interval: (We want their distances to match up!)

@ 6AM: Train A --> 0 miles || Train B --> 0 miles
@ 7AM: Train A --> 3 miles || Train B --> 0 miles
@ 8AM: Train A --> 6 miles || Train B --> 5 miles
@ 9AM: Train A --> 9 miles || Train B --> 10 miles

Oh! So we know they cross somewhere between 8AM and 9AM. Let's choose something in between:

@830AM: Train A --> 7.5 miles || Train B --> 7.5miles [notice it's halfway between 8AM and 9AM bc of constant speed]

--> this is where they meet! 830AM --> C is your correct answer
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 15 Nov 2017, 05:14
C

The ratio of speeds is 3:5 as the ratio of the time taken is 5:3.
Let's choose a distance that's divisively by both 3 and 5, say 300.

Speeds of the train are now 60 and 100.
Train at P had covered 60kms when The train at Q started.
After another hour, train P is at 120 km mark, while the other is at 200 mark. After another hour, at 180 Nd 100, so already crossed.
So, after 30 mins, they both at 150 mark.
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Re: A train left a station P at 6 am and reached another station Q at 11 a  [#permalink]

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New post 30 Mar 2018, 01:10
P(5hrs) R Q(3hrs)
|---------|---------------|
R - meeting point
p-q distance=D
v(p)=D/5,v(q)=D/3
Since P already started and traveled 1 hr when Q started
P traveled D/5 in one hour ..
Distance between them is now D-D/5=4D/5
New distance
P(5hrs) R Q(3hrs)
|----x-----|------(4d/5)-x---------|
They will reach R at same time.

so t1=t2,d1/v1=d2/v2
x/(D/5)=(4D/5-x)/(D/3)
x=3D/10
t=D/v=(3D/10)/(D/5)=1.5(1 hour 30 mins)
Total time =1+1.3=2.30
Time=6+2.30=8.30
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A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 30 Mar 2018, 10:48
Bunuel wrote:
A train left a station P at 6 am and reached another station Q at 11 am. Another train left station Q at 7 am and reached P at 10 am. At what time did the two trains pass one another?

(A) 7:50 am
(B) 8:13 am
(C) 8:30 am
(D) 8:42 am
(E) 9:03 am


at 8:00 am, train 1 has covered 2/5 of the distance
while train 2 has covered 1/3 of the distance
thus, 4/15 of the distance remains
4/15 distance/(1/5+1/3) combined rate=1/2 hour
8:00 am+1/2 hour=8:30 am
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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New post 01 Apr 2018, 09:25
I think the question should say that the trains are traveling at constant speeds.
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Re: A train left a station P at 6 am and reached another station Q at 11  [#permalink]

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Re: A train left a station P at 6 am and reached another station Q at 11   [#permalink] 20 Jun 2019, 20:27
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