Two bicyclists head toward each other from the opposite ends of Main Street, which is six miles long. The first biker started at 2:05 going 12 mph. The second biker began peddling 4 minutes later at a rate of 14 mph. What time will they meet?
Detailed Explanation
The first bike got a 4/5 mile head start (12 × 4/60).
Therefore, by the time the second bike leaves, there are 51/5 miles between them (6 – 4/5).
Their combined rate of travel is 12 + 14 = 26 mph.
Let t = the number of hours the second bike travels.
26t = \(5\frac{{1} }{{5}}\)
26t = \(\frac{{26} }{{5}}\)
t = \(\frac{{26} }{{5}} ÷ \frac{{26} }{{1}}\)
t = \(\frac{{26} }{{5}} × \frac{{1} }{{26}}\)
t = \(\frac{{1} }{{5}}\)
\(\frac{{1} }{{5}}\) of an hour = 12 minutes. The second bike left at 2:09, so both bikes will meet at 2:21.
Therefore, by the time the second bike leaves, there are 51/5 miles between them (6 – 4/5).
Their combined rate of travel is 12 + 14 = 26 mph.
Let t = the number of hours the second bike travels.
26t = \(5\frac{{1} }{{5}}\)
26t = \(\frac{{26} }{{5}}\)
t = \(\frac{{26} }{{5}} ÷ \frac{{26} }{{1}}\)
t = \(\frac{{26} }{{5}} × \frac{{1} }{{26}}\)
t = \(\frac{{1} }{{5}}\)
\(\frac{{1} }{{5}}\) of an hour = 12 minutes. The second bike left at 2:09, so both bikes will meet at 2:21.
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