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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 $$\frac{{4} }{{5}}$$ mile head start (12 × $$\frac{{4} }{{60}}$$).
Therefore, by the time the second bike leaves, there are $$\frac{{51} }{{5}}$$ miles between them (6 – $$\frac{{4} }{{5}}$$).
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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