Date: Thu, 22 Apr 1999 10:23:01 -0600 (CST)
Who is asking: Student
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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