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Thread: Advanced Algebra

  1. #1
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    Smile Advanced Algebra

    Can you help with this ? Thanks.

    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 ?
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  2. #2
    ux0
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    2.5 per minute, try to work out the details, if you cant figure it out post again
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  3. #3
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    Quote Originally Posted by sahip View Post
    Can you help with this ? Thanks.

    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 ?
    Suppose the water was removed at 5 gallons per minute for $\displaystyle T_1$ minutes. Then $\displaystyle 5T_1= .75V$ where V is the volume of the hull. Suppose further that the pump ran for $\displaystyle T_2$ minutes at r gallons per minute. Then $\displaystyle rT_2= .25V$.

    Divding the first equation by the second, $\displaystyle \frac{5T_1}{rT_2}= \frac {.75}{.25}= 3$ so $\displaystyle T_2= \frac{5}{3r}T_2$.

    Then entire volume, V, was emptied in time $\displaystyle T_1+ T_2= T_1+ \frac{5}{3r}T_1= \frac{5+ 3r}{3r}T_2$ so the rate was $\displaystyle \frac{V}{\frac{5+3r}{3r}T_1}= = \frac{3r}{5+ 3r}\frac{V}{T_1}= 4$.

    But $\displaystyle 5T_1= .75V$ so $\displaystyle \frac{V}{T_1}= \frac{5}{.75}= \frac{20}{3}$.

    That is, $\displaystyle \frac{20}{3}\frac{3r}{5+3r}= \frac{20r}{5+ 3r}= 4$.

    Solve that for r.
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