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  1. #1
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    Help (story problems)

    1. The length of a rectangular playing field is 5 ft less than twice its width. If the perimeter of the playing field is 230 ft, find the length and width of the field.





    2. A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi apart?
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  2. #2
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    Quote Originally Posted by Patience View Post
    1. The length of a rectangular playing field is 5 ft less than twice its width. If the perimeter of the playing field is 230 ft, find the length and width of the field....

    Hi,

    let l be the length of the rectangle and
    let w be the width of the rectangle. Then
    the perimeter of a rectangle is calculated by:

    p = 2l + 2w

    You know that l = 2w - 5 and that p = 230'
    Now substitute these term into the first equation:

    2(2w - 5) + 2w = 230. Expand the bracket

    6w - 10 = 230
    6w = 240

    w = 40 feet and l = 75 feet

    p = 2*40 + 2*75

    p = 80 + 150 = 230 feet. Seems to be OK

    EB

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  3. #3
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    Quote Originally Posted by Patience View Post
    ...
    2. A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi apart?


    Hi,

    at 2:00 pm the distance between the 2 buses is 44 mi. From now on the distance increases at (44 + 48) mi/h.

    44 + (44 + 48)*t = 274

    92t = 230

    t = 230/92 h

    After 230/92 h after 2:00 pm the buses have a distance of 274 mi

    That is at 4:23:28.7 pm

    EB
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