fozzzy wrote:
Trains A and B left stations R and S simultaneously on two seperate parallel rail tracks that are 350 miles long. The trains passed each other at point X after travelling for a certain amount of time. How many miles of the rail tracks had train A travelled when the two trains passed each other?
1) up to point X, the average speed of train B was 25% less than the average speed of train A.
2) up to point X, the average speed of train B was 60 mph and it took two and a half hours for train B to arrive at point X.
First of all we have to make an inference here, that the 2 trains are running opposite to each other. Since they meet up at some point X. Otherwise they wouldn't have ever met if they ran in the same direction.
Statement 1:-Let Speeds be B and A.
Therefore \(B = \frac{3A}{4}\)
\(Speed = \frac{Distance}{Time}\)
\(A+B = \frac{350}{T}\) Lets take T as time taken by the trains to meet at point X. A+B is the effective speed of A and B.
Therefore\(A+\frac{3A}{4} = \frac{350}{T}\)
\(\frac{7A}{4} = \frac{350}{T}\)
Therefore \(A = \frac{200}{T}\)
Now A's distance will be \(Distance= Speed * Time\)
i.e. Distance covered by\(A = \frac{200}{T} * T\)
i.e. 200
hence Sufficient.
Statement 2 :-\(Speed = \frac{Distance}{Time}\)
\(A+B = \frac{350}{T}\)
if we know B and we know T then we can get value of A from above equation.
And then using \(Distance = Speed * Time\)(where time will again be two and half hours) we can calculate distance covered by A
Hence Sufficient.
Thus D is the answer.
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