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Math Help - Algebra

  1. #1
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    Smile Algebra

    A ship's hull was punctured and filled with water before being sealed off. A pump removed water at a steady rate of 5 gallons per minute, but after 75% of the water was removed, the pump slowed. If the water was removed at an average rate of 4 gallons per minute, at what rate was the remaining 25% of the water removed ?
    Thank you.
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  2. #2
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    Hello, sahip!

    An interesting problem . . .
    I had to baby-talk my way through it.


    A ship's hull was punctured and filled with water before being sealed off.
    A pump removed water at a steady rate of 5 gallons per minute,
    but after 75% of the water was removed, the pump slowed.
    If the water was removed at an average rate of 4 gallons per minute,
    at what rate was the remaining 25% of the water removed ?
    Let V = orginal amount of water (gallons).

    Let R = slower rate of pumping (gallons per minute).


    The first 75% of the water was removed at 5 gallons per minute.
    That is, \tfrac{3}{4}V gallons were removed at 5 gal/min.
    . . This took: . \frac{\frac{3}{4}V}{5} minutes.

    The last 25% of the water was removed at R gallons per minute.
    That is, \tfrac{1}{4}V gallons were removed at R gal/min.
    . . This took: . \frac{\frac{1}{4}V}{R} minutes.
    The total time to empty the water was: . T \:=\: \frac{\frac{3}{4}V}{5} + \frac{\frac{1}{4}V}{R} \:=\:\frac{(3R+5)V}{20R} minutes.


    Recap: They pumped V gallons of water in T minutes
    . . . . . and the average rate was 4 gallons per minute.


    There is our equation: . \frac{V}{\frac{V(3R+5)}{20R}} \:=\:4 \quad\Rightarrow\quad V \:=\:4\cdot\frac{V(3R+5)}{20R}\quad\Rightarrow\qua  d 20RV \:=\:4V(3r+5)

    Divide by 4V\!:\;\;5R \:=\:3R + 5 \quad\Rightarrow\quad 2R \:=\:5 \quad\Rightarrow\quad R \:=\:\frac{5}{2}


    The last 25% of the water was pumped out at 2\tfrac{1}{2} gallons per minute.

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