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Math Help - Keg and taps problem.

  1. #1
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    Keg and taps problem.

    There are 3 taps in a keg. If all three are open the keg is drained in one and a half-minute. If tap 1 is closed and the other two are opened the keg drains in 2 minutes. If tap 2 is closed and the others are open it drains in 2 4/13 (two and 4 over 13)minutes. How long will it take to drain the keg if tap 3 is closed and the other two are open?
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  2. #2
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    Re: Keg and taps problem.

    If all three are open the keg is drained in one and a half-minute.
    \left(\frac{1}{t_1} + \frac{1}{t_2} + \frac{1}{t_3}\right) \cdot \frac{3}{2} = 1


    If tap 1 is closed and the other two are opened the keg drains in 2 minutes.
    \left(\frac{1}{t_2} + \frac{1}{t_3}\right) \cdot 2 = 1


    If tap 2 is closed and the others are open it drains in 2 4/13 (two and 4 over 13)minutes.
    \left(\frac{1}{t_1} + \frac{1}{t_3}\right) \cdot \frac{30}{13} = 1


    solve the system ...
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  3. #3
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    Re: Keg and taps problem.

    Skeeter's point is that the rates add. Let t_1, t_2, and t_3 be the rates at which the three taps drain the keg (in units of "kegs per minute"). Then all three drain the keg at the rate of t_1+ t_2+ t_3 "keg per minute". Saying that they will drain the keg in one and a half minutes means that their rate is "1 keg per 3/2 minutes" or 1/(3/2)= 2/3 "keg per minute": t_1+ t_2+ t_3= 2/3 is the same as Skeeter's first equation.

    Saying that taps 2 and 3 will drain it is 2 minutes says that they drain it at the rate of "one keg per 2 minutes" or 1/(2)= 1/2 keg per minute: t_2+ t_3= 1/2 which is the same as his second equation.

    Finally, saying that taps 1 and 3 will drain it in "2 and 4/13 minutes"= 30/13 minutes means their rate is "one keg per 30/13 minutes" or 1/(30/13)= 13/20 keg per minute: t_1+ t_3= \frac{13}{30} which is the same as his last equation.

    Now, can you solve those three equations? You might find it simplest first to let x= 1/t_1, y= 1/t_2, and z= 1/t_3 so your equations are x+ y+ z= 2/3, y+ z= 1/2, and x+ z= 13/30.
    Last edited by HallsofIvy; December 26th 2012 at 06:52 AM.
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  4. #4
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    Re: Keg and taps problem.

    Note also that problems about draining reservoirs usually cannot, strictly speaking, be solved using simple arithmetic because the draining rate is not constant. See this post.
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  5. #5
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    Re: Keg and taps problem.

    @ Skeeter: Can you show solving the system of equation without using the method 1/t1=x , 1/t2=y., 1/t3=z.
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