Final answer:
Train A had traveled for a total of 390 minutes before being caught up by train B. Train A's 90-minute lead and the relative speed between train A and B, which is 15 mph, were used to calculate the time it took for train B to catch up to train A.
Option c is correct.
Step-by-step explanation:
To solve the problem of determining how many minutes train A had traveled before train B meets up with it, we can use the concept of relative motion. The given information allows us to understand that train A leaves 1 hour and 30 minutes before train B and that train A travels at 50 mph while train B travels at 65 mph.
First, we convert Train A's lead time from hours and minutes to just minutes: 1 hour and 30 minutes is equivalent to 90 minutes. In that 90 minutes, train A would have covered (50 miles/hour) × (1.5 hours) = 75 miles.
Next, when train B starts moving, it needs to cover the distance already traveled by train A plus keep up with train A's ongoing journey. So, the relative speed of train B with respect to train A is (65 mph - 50 mph) = 15 mph. The distance train B needs to cover to catch up with train A is 75 miles, which was covered by train A in the lead time.
Using the speed-distance-time relationship, time = distance / speed, the catch-up time can be calculated by dividing the 75 miles by the relative speed of 15 mph. This results in a catch-up time of 75 miles / 15 mph = 5 hours.
Converting this time back to minutes, we get 5 hours × 60 minutes/hour = 300 minutes. So, train B would take a total of 300 minutes to meet train A. Now we need to add the lead time of train A to this duration: 300 minutes + 90 minutes = 390 minutes.
Therefore, train A travels for a total of 390 minutes before train B meets up with it, which means the answer is c) 390 minutes.