Fred arrives at the subway station every day at a random time and takes the first train that arrives. If he takes the A train, which arrives each hour on the hour, he ends up at the fitness center. If he takes the B train, which arrives every hour 15 minutes after the A train, he ends up at the park. If Fred plays this game over a long period of time, about what percent of the time can he expect to end up at the park?
15%
75%
50%
25%
25
To determine the percent of the time Fred can expect to end up at the park, we need to compare the frequencies of the A train and the B train.
The A train arrives each hour on the hour, so it arrives 24 times in a 24-hour day. This means that out of every 24 train arrivals, the A train arrives 24/24 = 1 time.
The B train arrives every hour 15 minutes after the A train. Since there are 60 minutes in an hour, the B train arrives 60/15 = 4 times in the time it takes for the A train to arrive once.
Therefore, out of every 24 train arrivals, the B train arrives 4 times.
To calculate the percentage, we add up the total number of times the B train arrives in a day (4), divide it by the total number of train arrivals in a day (24), and then multiply by 100 to get the percentage.
(4 / 24) * 100 = 16.67%
Therefore, Fred can expect to end up at the park approximately 16.67% of the time.