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March 24, 2017

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An employee drove to work on Monday at 45 mi/h and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mi/h and arriving one minute late.

a. How far does the employee live from work?

b. At what speed should the employee drive to arrive five minutes early?

  • Algebra 2 - ,

    Using V = D/T

    60D/45 = T - 1
    60D/40 = T + 1
    Solving for T and equating yields
    60(40D - 45D)/1800 = 2 or 5D = 60 making D = 12 miles

    In order to arrive in 17 minutes,
    60(12)/V = 17 yielding = 42.35mph.

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