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jet travels 4 hours at avg speed 250mph returning the same way it can go 200mph how many hrs is the trip back?

July 14th, 2008 · 3 Comments

basic algebra which normally needs a 'let statement' using the distance formula which is distance=rate times time.
help!

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    Tags: AVG


    3 responses so far ↓

    • 1 visram007 // Jul 14, 2008

      5 hours,becoz
      distance=4*250=1000
      on return journey,time=distance/speed
      time=1000/200=5hours

    • 2 jmwalle21 // Jul 14, 2008

      i agree with the first answer

    • 3 fabregas4 // Jul 14, 2008

      (250m/h*4h) / 200m/h=5h

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