alanforde800Maximus wrote:
Both runners and walkers are participating in a marathon on exactly the same course. The walkers start at 6:45 AM, and the runners begin at 8:00 AM. Bill is entered as a walker and is going to walk at a constant rate of 4 miles per hour. The fastest runner is going to be running at a constant rate of 9 miles per hour. How far will Bill have walked before he is passed by the fastest runner?
a) 4 miles
b) 5 miles
c) 8 miles
d) 9 miles
e) 10.5 miles
We can classify this problem as a “catch-up” rate problem, in which we use the formula:
distance of Bill = distance of fastest runner
We are given that Bill starts at 6:45 AM and walks at a constant rate of 4 mph and that the fastest runner starts at 8 AM and runs at a constant rate of 9 mph.
Since Bill started an hour and 15 minutes, or 5/4 hours, before the fastest runner, we can let the fastest runner’s time = t hours and Bill’s time = t + 5/4 hours. Since distance = rate x time, we can calculate each person’s distance in terms of t.
Bill’s distance = 4 x (t + 5/4) = 4t + 5
fastest runner’s distance = 9t
We can equate the two distances and determine t.
4t + 5 = 9t
5 = 5t
t = 1 hour
Now we can determine how far Bill will have walked by the time he is passed by the fastest runner.
Bill’s distance = 4(1) + 5 = 9 miles
Answer: D
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