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Math Help - distance rate time problem

  1. #1
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    distance rate time problem

    I can't figure out
    plane travels 680 mi one way with the wind at 300 mph plus tailwind
    plane traves 520 mi back against the wind at 300 mph less headwind
    the time to go and return are equal

    I keep getting 300 as wind but that can't be

    Thank you
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  2. #2
    Super Member bigwave's Avatar
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    since Distance = Rate x Time

    Quote Originally Posted by esther View Post
    I can't figure out
    plane travels 680 mi one way with the wind at 300 mph plus tailwind
    plane traves 520 mi back against the wind at 300 mph less headwind
    the time to go and return are equal

    I keep getting 300 as wind but that can't be

    Thank you
    since time is = then we have
    let w = wind

    since Distance = Rate x Time

    \frac{D_1}{R_1}= \frac{D_2}{R_2}


     <br />
\frac{680m}{300mph - w} = \frac{520m}{300mph + w}<br />

    680(300+w) = 520(300 - w)

    204000 + 680w = 156000 -520w

    48000 = -1200w

    |- 40| = w = 40mph
    Last edited by bigwave; December 11th 2009 at 09:47 AM. Reason: latex
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  3. #3
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    Hello, esther!

    A plane travels 680 mi one way with the wind at 300 mph plus tailwind.
    The plane travels 520 mi back against the wind at 300 mph less headwind.
    The time to go and return are equal
    Find the rate of the wind.

    Let x = rate of the wind.

    With the wind, the plane flies 680 miles at 300+x mph.
    . . This takes: . \frac{680}{300+x} hours.

    Against the wind, the plane flies 520 miles at 300-x mph.
    . . This takes: . \frac{520}{300-x} hours.


    Since the times are equal, we have: . \frac{680}{300+x} \;=\;\frac {520}{300-x}

    Divide by 40: . \frac{17}{300+x} \:=\:\frac{13}{300-x}

    Cross-multiply: . 5100 - 17x \:=\:3900 + 13x \quad\Rightarrow\quad 30x \:=\:1200\quad\Rightarrow\quad x \:=\:40


    The rate of the wind is 40 mph.

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