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Peter leaves Riverdale at 10:00 a.m. and starts pedaling his

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Hermione
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Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink] New post   07 Dec 2006, 09:07
Peter leaves Riverdale at 10:00 a.m. and starts pedaling his bicycle at 10 mph. If John starts on the same path at 2:00 p.m., when will he catch Peter if he pedals at 15 mph?

10t = 15(t – 4)
t=12 hours --> which is the correct answer

But why is that when i use the alternative scenario where

Peter = t+4 (instead of t)
John = t (instead of t-4)

we have the equation 10(t+4) = 15t

t = 8 hours --> which is wrong

shouldn't the two methods yield the same answer?



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Re: Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink] New post   07 Dec 2006, 09:15
:) It is yielding the additional time in the second equation. That means we need to add 8+4 = 12 for the second one.
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Re: Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink] New post   07 Dec 2006, 09:21
Ohhh yea.... :-D Thanks hsampath!
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Re: Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink] New post   07 Dec 2006, 11:43
The First solution gives the additional time required,where as the first gives the original time.
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Re: Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink] New post   07 Dec 2006, 16:11
He'll catch him at 10pm

s=dt ----> d=ts
The two will meet when the two distances equal eachother.

Since Peter set off 4 hours before then he has travelled 4*10 = 40 miles
so:

40 + 10t = 15t

t=8

8+2 = 10pm



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Re: Peter leaves Riverdale at 10:00 a.m. and starts pedaling his [#permalink]
07 Dec 2006, 16:11
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