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Math Word Problem

  1. Oct 3, 2006 #1
    Bickford traveled twice as fast as Shawn traveled. Thus, Bickford could travel the 320 miles to the reef in only 2 hours less than it took shawn to travel the 240-miles to Jane's house. Find the rates and times of both boys.

    Just having problems figuring out the formula and how to set it up.

    So Shawn is x
    Which would make Bickford 2x
  2. jcsd
  3. Oct 3, 2006 #2
    Use [tex] d = rt [/tex]. So for Bickford, [tex] 320 = 2v(t-2) [/tex] and for Shawn [tex] 240 = vt [/tex].
    Last edited: Oct 3, 2006
  4. Oct 3, 2006 #3
    Please explain how you set that up and the what the variables represent
  5. Oct 3, 2006 #4
    Distance is defined as speed [tex]\times [/tex] time . Let [tex] v [/tex] be the speed of Shawn. Then [tex] 2v [/tex] is the speed of Bickford. Also, let [tex] t [/tex] be the time it takes Shawn to travel to Jane's house. Using the equation [tex] d = vt [/tex] (where v is the speed) we can set up two equations with two unknowns.

    Bickford's speed is [tex] 2v [/tex] and his time is [tex] t-2 [/tex] (two hours less time than Shawn) and he travels 320 miles. So we have [tex] 320 = 2v(t-2) [/tex].

    Shawn's speed is [tex] v [/tex] and his time is [tex] t [/tex] and he travels 240 miles. So we have [tex] 240 = vt [/tex]. Can you solve for [tex] v [/tex] and [tex] t [/tex]?
  6. Oct 3, 2006 #5
    thanks for the help, I should be able to get it now
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