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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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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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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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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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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
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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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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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D - 7pm

Both meet at 500km
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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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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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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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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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