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Hi, got stumped on this one. Any help would be really appreciated it.

2 Trains, X&Y, start simultaneously on opposite ends of a 100-mile route and travled toward each other on parallel tracks. Train X, traveling at a constant rate, completed the 100 mile trip in 5 hours. Train Y, also traveling at a constant rate completed the 100 mile trip in 3 hours. How many miles had train X traveled when it met Train Y?

Rate x Time = Distance
Train X 5 x t = d
Train Y 3 x t = 100-d

The only variable they have in common is that both trains have been travelling for the same amount of time...thus, set each equation equal to each other in terms of d.

so Train X = t = d/5
and Train Y = t = (100-d)/3

Set each equation equal to each other:
d/5 = (100-d)/3
Cross multiply and you get:
3d = 500 - 5d

Solve for d and you get 62.5. Thus the answer is 62.5

Clearly the speeds ratio of A and B is 3:5 So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,

Cicerone Where do you get 3/8 from? I like your ratio method of doing this.

Hey look at the bold part
Since the distance is shared in the raio of 3:5
A should cover 3/8 and in the same time B will cover 5/8 .............
_________________

Clearly the speeds ratio of A and B is 3:5 So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,

Yes, it works in this example, but it is better to have a clear concept of solving such problems. This ratio example won't work for other "rate" problems...

Clearly the speeds ratio of A and B is 3:5 So distance covered by them when they meet each other also will be in the ration 3:5

So distance covered by A is 3/8 x 100 = 37.5

Hence A.

Keep it simple....... Get the concepts of ratios.............Save time in the exam

Regards,

Yes, it works in this example, but it is better to have a clear concept of solving such problems. This ratio example won't work for other "rate" problems...

Hey SimaQ, please do not say this..............
Any question form Time and Distance out of Speed , Distance and Time if one of them is constant, i bet i will do it using ratios only..............
Why don't u try to test me........
_________________

cicerone's method is neater but this is how i did it. trick to remember with this type of question where 2 things travel towards each other is that together they travel the total distance, whether its lifts, trains, cars, water going down a pipe, frogs etc

so

where t = time

100/5 t + 100/3 t = 100
300/15 t + 500/15 t = 100
t = 1500/800 = 15/8