Date: Thu, 22 Apr 1999 10:23:01 -0600 (CST)
Sender: bill

Who is asking: Student
Level: Middle

Bill was in a bike race and his friend kyle calculated that if he went 15mph that he would cross the noontime checkpoint one hour early but if he rode 10mph he would arrive one hour late.
How far away is the checkpoint?

Hi Bill

In many of these rate problems the only "formula" you have is that distance = time x rate. In this problem you are given the rates and you want to find the distance. For time what you see is "one hour early" and "one hour late". Let D be the distance and t be the normal time it takes. Then when the rate is 15mph the time is one hour less than t, ie. D = (t-1) x 15. Also when the rate is 10mph the time is one hour more than t, ie. D = (t+1) x 10. Since D is the same for both trips, 15 (t-1) = 10 (t+1). Now solve for t and then use this value of t to find D.


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