# Keg and taps problem.

• Dec 26th 2012, 05:40 AM
hisajesh
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?
• Dec 26th 2012, 05:55 AM
skeeter
Re: Keg and taps problem.
Quote:

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

Quote:

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

Quote:

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

solve the system ...
• Dec 26th 2012, 06:49 AM
HallsofIvy
Re: Keg and taps problem.
Skeeter's point is that the rates add. Let $\displaystyle t_1$, $\displaystyle t_2$, and $\displaystyle t_3$ 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 $\displaystyle t_1+ t_2+ t_3$ "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": $\displaystyle t_1+ t_2+ t_3= 2/3$ 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: $\displaystyle t_2+ t_3= 1/2$ 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: $\displaystyle t_1+ t_3= \frac{13}{30}$ which is the same as his last equation.

Now, can you solve those three equations? You might find it simplest first to let $\displaystyle x= 1/t_1$, $\displaystyle y= 1/t_2$, and $\displaystyle z= 1/t_3$ so your equations are x+ y+ z= 2/3, y+ z= 1/2, and x+ z= 13/30.
• Dec 26th 2012, 09:51 AM
emakarov
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.
• Feb 14th 2013, 01:10 PM
hisajesh
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.