Re: Keg and taps problem.

Quote:

If all three are open the keg is drained in one and a half-minute.

Quote:

If tap 1 is closed and the other two are opened the keg drains in 2 minutes.

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If tap 2 is closed and the others are open it drains in 2 4/13 (two and 4 over 13)minutes.

solve the system ...

Re: Keg and taps problem.

Skeeter's point is that the rates add. Let , , and 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 "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": 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: 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: which is the same as his last equation.

Now, can you solve those three equations? You might find it simplest first to let , , and so your equations are x+ y+ z= 2/3, y+ z= 1/2, and x+ z= 13/30.

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.

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.