David W. answered 06/07/16
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This is a D-I-R-T (Distance-Is-Rate-times-Time) problem. In such problems one variable is often the same and another is related somehow so we can solve the problem. In this problem, the distance is equal and one time is 1 hour less than the other.
Let T = time in hours since noon
Sam’s distance: D = 55T
Will’s distance: D = 75(T-1)
Same distance: 55T = 75(T-1)
55T = 75T – 75 [distribute]
-20T = -75 [subtract 75T from both sides]
T = 3.75 [divide both sides by (-20)]
Will catches up to Sam at 3.75 hours after noon (note: that’s 3 hours 45 minutes, or 3:45 PM).
Note that we first had to calculate the time. Now, we can find the distance using either formula:
D=55(3.75)
D= 206.25 miles