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Train A leaves New York for Boston at 3 PM and travels at [#permalink]

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15 Sep 2006, 13:51

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Train A leaves New York for Boston at 3 PM and travels at the constant speed of 100 mph. An hour later, it passes Train B, which is making the trip from Boston to New York at a constant speed. If Train B left Boston at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in New York?

(1) Train B arrived in New York before Train A arrived in Boston.

(2) The distance between New York and Boston is greater than 140 miles.

I think D is the answer.
This is the way I went about it and I am not sure if I am right (I am an absolute novice here and this is my first post ever!)
Let the distance between NY and Boston be y miles
Let the speed of train B be x miles/hr.
We know speed of A is 100 miles/hr.

Since combined travel time of both trains to travel to their destination is 2 hours,
y/x + y/100 =2

1 hour after A has left NY(i.e 4 pm), both the trains meet each other.
In 1 hour, distance covered by A is 100 miles.
At 4pm, Train B has travelled 10 minutes only(since it starts from Boston at 3.50 pm), so distance covered by B is x/6 miles.

When both the trains meet, together they have covered the distance between NY and Boston.

Hence, 100+ x/6 = y

Substituting Y in the prev equation,
x^2- 500x+60000=0
(x-300)(x-200)=0

x=300 for which y=150
OR
x=200 for which y=133.33

But since Statement 2 says distance y should be greater than 140 miles, statement 2 alone answers this query at this stage.

If we calculate further,
If y=150 miles, then train B will take 30 mins to reach NY which calculates to 4.20 pm and A will take 1.5 hours which calculates to 4.30 pm.

If y=133.33 miles, then train B will take apprx 40 mins to reach NY which calculates to 4.30 pm and A will take approx 1 hour 20 mins, which makes it 4.20 pm.

Statement 1 says train B arrived in NY before train A arrived in Boston and this is true for y=150 miles.

Thus each statement alone is sufficient to answer this query.