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Question:

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?

A
2:21

Explaination

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^{1}/_{5}

26t = ^{26}/_{5}

t = ^{26}/_{5} ÷ ^{26}/_{1}

t = ^{26}/_{5} × ^{1}/_{26}

t = ^{1}/_{5}^{1}/_{5} of an hour = 12 minutes. The second bike left at 2:09, so both bikes will meet at 2:21.