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Let's see this problem with a diagram:

Train A left 40 mins before Train B at a speed of 60 mph
So distance traveled by Train A in 40 mins= 40 miles
Let's say the total distance at the point of overtaking is x
So, the time taken for covering the remaining distance x-40 by Train A=x-40/60
Time taken for covering the distance x at 75 mph by Train B=x/75

Both Time should be equal which gives us,
x-40/60=x/75
x=160 mins=2 hr 40 mins
Time when train B overtake train A = 3:30+40 mins+ 2 hr 40 mins= 6:50

Correct Answer is D.
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Train A starts at 3:30 at a speed of 60mph
Train B starts at 4:10 at a speed of 75mph

When train B started train A would have already covered 40 miles, since A travels 60mph which means 1 mile per minute, therefore it will travel 40 miles in 40 minutes

Now lets assume both trains meet in T hours
Therefore in the next T hours
Train A would have travelled 60T
Train B would have travelled 75T+40(extra miles covered by Train A due to starting early)

Since Distance travelled by both trains in T hours will be the same
we can say
60T=75T+40 ---(1)
Solving equation (1) we get T=160 minutes which is 2 hours and 40 minutes
Adding this to 4:10 we get 6:50
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Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

4:10
5:40
6:10
6:50
7:30

t*75=(t+2/3)*60
t=8/3 hours=2 hours 40 minutes
4:10+2:40=6:50
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For the two trains to meet, they must travel the same distance [say 'd']. Let 't' be the time taken by B to cover this distance ['d' miles]. Since A has a head-start of 40 mins. [2/3 hrs], it takes A [t+2/3] hrs. to travel d miles.
Now, from the data we have on B, d=speed*time=75*t
Also, ....................................... A, d=..................=60*[t + 2/3]
So, 75*t=60*[t + 2/3]
or t=8/3 hrs=2 hrs 40 mins
Since B started at 4:10, it catches up with A at 6:50. Ans: D
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60T=(T-40)75
T=200 minutes
3:30 + 200 minutes= 6:50ANS
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Solution:
Using the relative speed concept, the problem can be solved in the following way:
Train A and Train B travelled in the same direction, their relative speed will be the difference between the speeds of the two trains.

Here,
Speed of Train A = 60 mhp
Speed of Train B = 75 mph
Their relative speed = (75- 60) mph = 15 mph = 15/60 mile per minute = 1/4
Distance between the two trains = 40 miles
Time taken by train B to overtake Train A = Distance between the two trains / Relative speed=(40 * 4/1) minutes = 160 minutes = 2 hour and 40 minutes
Answer: 6:50
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Train A left station T at 3:30 p.m. and traveled on straight tracks at a constant speed of 60 miles per hour. Train B left station T on adjacent straight tracks in the same direction that train A traveled on. Train B left station T 40 minutes after train A, and traveled at a constant speed of 75 miles per hour. Train B overtook train A on these straight tracks. At what time did train B overtake train A?

Distance Travelled by Train A in 40 minutes = (60/60)*40=40 miles
Relative speed of Train B w.r.t Train A = 75-60=15
Time taken by Train B to cross Train A = 40/15 hours = (40/15)*60=160 minutes

The Time Train B crosses Train A= 3:30pm + 40 minutes + 160 minutes = 6:50 pm

Hence D
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