Keg and taps problem.

**hisajesh** · December 26th 2012, 05:40 AM

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?

**skeeter** · December 26th 2012, 05:55 AM

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 ...

**HallsofIvy** · December 26th 2012, 06:49 AM

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.

**emakarov** · December 26th 2012, 09:51 AM

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.

**hisajesh** · February 14th 2013, 01:10 PM

@ 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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