A train X starts from Meerut at 4 p.m. and reaches Ghaziabad at 5 p.m.

Great solutions above. Let us look at a solution without equations too.
X takes 1 hr to cover the distance that Y covers in 1.5 hrs. That is a ratio of 2:3.
So the ratio of their speeds is 3:2 since the distance they cover is the same.
Now imagine the situation when they meet. They start at 4 and meet after some time. In this time, they have covered distance in the ratio 3:2 since their speed is in the ratio 3:2.


Think of X now. It has covered 3/5th of the distance that it is supposed to cover to G. It needs to cover another 2/5th. So out of its 1 hr journey, it has covered 3/5th ie. 36 mins journey (because 3/5 * 60 min = 36 min)
So they met at 4:36.
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let d be the distance that both x and y travel. then X's speed=d and Y's speed=d/1.5
relative speed=d+d/1.5 = 5d/3
time when they will meet = T.Distance/ relative speed
d/5d/3 = 3/5 hours
3/5*60=36 mins hence A

We know that S = D/t , also D is fixed
Therefore, for train running from meerut to gazi… Sa = D / ta
but ta = 1 hr , therefore Sa = D
Now, for train running from Gazi… to meerut … Sb = D / tb
But tb = 1 ½ hr = 3/2 hr Therefore, Sb = 2D/3
However, D = Sa therefore Sb = 2Sa/3

Now, to calculate the time at which the trains will meet we need to calculate relative speed
Since, the trains are moving towards each other ,
Relative speed = Sa + Sb = Sa + 2Sa/3 = 5Sa /3
Distance will be “D” because it is fixed
And therefore Time at which the trains will meet i.e t = D / (5Sa /3)
but D = Sa therefore t = Sa / (5Sa /3) = 3 / 5 Hr = 3/5 * 60 = 36min

Thus, trains will meet at 4:36

Sol:

Let the distance between two points be "D miles"

Speed of X = D/1= D miles/h
Speed of Y = D/1.5= 2D/3 miles/h

Suppose X meets Y after completing "x miles" traveling for \(t_x\) hours and y traveled \(t_y\) hours

\(t_x=t_y\)

\(time=\frac{Distance}{Speed}\)

\(\frac{x}{D}=\frac{3(D-x)}{2D}\)

\(2x=3(D-x)\)

\(5x=3D\)

\(x=\frac{3}{5}D\)

Now,
x travels D miles in 1 hour
To travel (3/5)D miles, it would need 3/5 hours = 36minutes.

Thus; the trains meet at 4:36PM.

Going to explain how I calculated it.

The first train travels the entire distance in 1 hour. The other train travels only 2/3 of the distance in one hour. When the trains meet they will together have covered the entire distance, from each others ends. The speeds of the trains are 1 times the distance per hours and 2/3 of the distance per hour. After traveling at their respective speeds for time X both trains will together have covered the entire distance.

Therefore:

1X + (2/3)X = 1.

(5/3)X = 1

X = 1/(5/3)

X = 3/5

Both trains will meet after 3/5 hours, as they will have covered the entire distance together at this point in time. 1/5 of an hour is 12 minutes. 12 x 3 is 36. The trains will meet at 4:36.

For Train X =>

M (4 PM) ---------------------- G (5 PM) = 1hr

For Train Y =>

G (4 PM) ---------------------- M (5.30 PM) = 1.5 hrs

Now if Y is travelling at a speed of say 100 kms/hour.
100 km/hr means he can cover 100 kms in an hour. So in his travel time of 1.5 hrs he will cover 150 kms, his complete trip's distance.
Now X will also travel the same distance of 150 kms.
Now to cover 150 kms in 1 hour he will have to travel at a speed of 150 kms/hr.

So from assuming speed Y as 100 kms/hr, we have the following details -
Speed of Train Y = 100 kms / hr
Speed of Train X = 150 kms / hr

Now to find their meeting point, we have total speed = (100 + 150) kms/ hr [As trains are moving towards each other]
and we have total distance of 150 km covered by X and Y when they meet*.
*[As X will come from one direction and Y from other, and when they meet they would be covering the whole track in totality]

Hence,
\( Time = \frac{Distance }{ Speed} = \frac{150}{250}\)
Time = \(\frac{3}{5}\) hours = \(\frac{3}{5} * 60\) = 36 minutes

Hence answer is they will meet 36 mintues after their same starting time. Hence 4:36 pm
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we can solve it also using RTD chart. In this case we need to assign distance - let's say 4.5 miles (since it can be divisible by 1.5 and 1 hours). Then the speed of 1st train is 4,5 mph, the speed of second - 3 mph. Since the trains go to each other, then the relative speed will be 7.5 mph. Taking into account same distance 4.5 miles and new speed 7.5 mph we can find time - 0.6 hours which is 36 min.

TOTAL DISTANCE = D
T=D/S1 + S2

S1=D/1
S2=D/1.5

putting in values of S1and S2 in T

T=D/ (D/1 + D/1.5)

=3/5
=3/5*60 min = 36 min

This is a converging rate problem, so the two respective distances each train travels equals the whole distance (distance of x + distance of y = total distance at convergence point.).

First we need to identify the information already given to us, which is the rates of each train.

We can lay it out in a simple rate chart (pardon all of the dots, I couldn't figure out how to embed a table).

Train..........Rate..................Time..........Distance
X.................\(\frac{D}{1 hr.}\).........................t................Dt
Y.................\(\frac{ D}{3/2 hr.} = \frac{2D}{3 hr.}\).......t..............\(\frac{2Dt}{3}\)

Then we know adding the two respective distances equals total distance "D". So we can set up the formula Distance of X + Distance of Y = D.
\(Dt + \frac{2Dt}{3} = D\)
\(\frac{5Dt}{3} = D\)
\(t = \frac{3}{5} hr.\)

We can see the solutions ask for what time the trains converge at, so we need to convert the time into minutes.

\(\frac{3}{5} hr. = 36 minutes\)

Therefore the time the trains left, 4 PM, + 36 minutes is the time of convergence, aka (A) 4:36 PM.

Hello VeritasKarishma
The explanation looks great as always! But I have couple of doubts.
You mentioned - "So the ratio of their speeds is 3:2 since the distance they cover is the same."
so does that mean ratio of time can only be inverted to get ratio of speed when distance covered is same?

Also, you mentioned- "they have covered distance in the ratio 3:2 since their speed is in the ratio 3:2."
So does it always have to be that ratio of speed is same as ratio of distance covered? or Can it be that even ratio of time can tell us the ratio in which the distance covered is divided? If yes, then how are they related (i.e is ratio of time and ratio of distance same or inverse of each other) ?

SDW2 - Request you to check out both the links given. They explain exactly this - the relation between T, S and D and why this relation holds. They explain how ratio of one is related to another and under what constraints. You will find the posts useful.
Let me know if you still have questions after going through the posts.
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X ----------- (d/1)
y------------(d/1.5)
This is a catchup problem.

{(d/1)+(d/1.5)}t= d

t=3/5
(3/5)*60 = 36
therefore 36 mins
Answer is 4:36(Since both started at same time)

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