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

Question

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

TimeTraveller · Accepted Answer

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.

Lucy Phuong · Answer

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.

pushpitkc · Answer

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.

arvind910619 · Answer

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

GMATinsight · Answer

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!!!

effatara · Answer

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.

permination · Answer

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

--> this is where they meet! 830AM --> C is your correct answer

Manishcfc8 · Answer

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.

kunalcvrce · Answer

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

gracie · Answer

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
C

Fdambro294 · Answer

We’re given information in which both trains cover the same distance.

Given the same constant distance, the Ratio of time traveled is inversely Proportional to the Ratio of Speeds

Time for train 1 from P : Time for train 2 from Q = 5 hours : 3 hours

Speed for train 1 : Speed for Train 2 = 3 : 5

Let train 1 leaving from P have a Speed = 3 mph

Distance from P to Q = (3 mph) * (5 hours to reach destination) = 15 m

Train 1 gets a 1 hour headstart when it leaves at 6 AM. 3 miles of the Gap Distance is closed.

At 7 AM ——-> 12 miles is between the 2 trains

Since train 1 Speed = 3 mph

Train 2 speed = 5 mph

Time to meet = (Gap Distance) / (Speed of 1 + Speed of 2)

Time = 12 / (3 + 5) = 12/8 = 3/2 hours

The trains will meet 1.5 hours after 7 AM

Answer: 8:30 AM

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Kinshook · Answer

Given: 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.

Asked: At what time did the two trains pass one another?

Let the distance between P & Q be D km.

P-Q:- 
Time = 5 hours
Speed = D/5 kmh

Q-P:-
Time = 3 hours
Speed = D/3 kmh

From 6-7 am
First train travelled = D/5 km

Remaining Distance = D - D/5 = 4D/5 km
Relative speed = D/3 + D/5 = 8D/15
Time taken = (4D/5)/(8D/15) = 4*15/5*8 = 3/2 = 1.5 hours

Time when they meet = 7 + 1.5 = 8.5 hours = 8:30 am

IMO C

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

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