crejoc wrote:
At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?
A. 6:00 p.m.
B. 6:30 p.m.
C. 8:00 p.m.
D. 8:30 p.m.
E. 10:00 p.m
We can classify this problem as a “catch-up” rate problem, for which we use the formula:
distance of Peter = distance of John
We are given that at 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph, and that at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph.
Since Peter started
4 hours before John, we can let Peter’s time = t + 4 hours, and John’s time = t.
Since distance = rate x time, we can calculate each person’s distance in terms of t.
Peter’s distance = 10(t + 4) = 10t + 40
John’s distance = 15t
We can equate the two distances and determine t.
10t + 40 = 15t
40 = 5t
t = 8 hours
Since John left at 2 p.m. and caught up to Peter 8 hours later, he caught up with Peter at 10 p.m.
Answer: E
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