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Two Pipes and a Drain


Date: 01/07/98 at 09:15:17
From: andrei
Subject: 2 pipes

A swimming pool has 2 inlet pipes. One fills the pool in 4 hours, the 
other in 6 hours. The outlet pipe empties the pool in 5 hours. 

Once the outlet pipe was left open when the pool was being filled. In 
how many hours was the pool full?


Date: 01/07/98 at 12:48:48
From: Doctor Anthony
Subject: Re: 2 pipes

In 1 hour the first inlet pipe fills 1/4 of the pool.

In 1 hour second inlet pipe fills 1/6 of the pool.

In 1 hour the drain empties 1/5 of the pool.

If all the inlets and the drain are open, then in 1 hour the pool 
fills by

        1/4 + 1/6 - 1/5  = 13/60  of its volume.

Time to fill the pool =  60/13 hours =  4 hours 36 minutes 55 secs.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 01/07/98 at 09:17:16
From: andrei
Subject: 2 pipes

A hot water faucet fills a tub in 30 minutes, and the cold water 
faucet in 20 minutes. The tub can be drained in 15 minutes. 

If both faucets are open while the drain is open, how soon will the 
tub be full?


Date: 01/09/98 at 09:25:19
From: Doctor Ezick
Subject: Re: 2 pipes

Good Question!

The trick to this question is to take advantage of the fact that the 
problem leaves the size of tub unspecified. The fact that the size of 
the tub is unspecified means one of two things.

1. It doesn't matter - the fill time is the same no matter how large  
   the tub is.

2. We can't answer the question absolutely - the best we can do is to 
   give an expression that depends upon the size of the tub.

For the moment take my word for it that we are dealing with case one.

Since the size of the tub does not make any difference, let's say that 
it's 60 gallons (60 is the least common multiple of 15, 20 and 30, 
which is going to make for some nice divisions in a second).

What does this mean?

If the hot water faucet fills the tub in 30 minutes it must be pumping 
in 60 gallons / 30 minutes = 2 gallons per minute.

Likewise, the cold water faucet is pumping in 60 / 20 = 3 gallons per 
minute.

This means that 2 + 3 = 5 gallons per minute of water are flowing into 
the tub.

Meanwhile, if the drain can drain a full tub in 15 minutes, that means 
that 60 / 15 = 4 gallons per minute are being drained from the tub.

The result: Every minute 5 gallons of water go in and 4 go out.  
This means that 1 gallon of water per minute is going into the tank.  
Since the tank holds 60 gallons, it takes 60 gallons / 1 gallon per 
minute = 60 minutes to fill the tank!

The question now is: How do we know that the size of the tank does not 
matter?  We could solve this problem the same way letting the size be 
unspecified by setting up an equation as follows:

                      Size
  ------------------------------------------------ = Time to Fill
  (Size/20 mins) + (Size/30 mins) - (Size/15 mins) 
    (Hot Water)     (Cold Water)       (Drain)

This is the same equation we just solved with 60 gallons in the place 
of Size.

Notice that we can factor out Size from the denominator and then 
divide it by the Size in the numerator to get 1.

  Size                       1
  ---- x ---------------------------------------  = Time to fill
  Size   (1/20 mins) + (1/30 mins) - (1/15 mins)

Using 60 as the common denominator.
  
                      1                            1
1x---------------------------------------  =  ------------- = 60 mins
  (3/60 mins) + (2/60 mins) - (4/60 mins)     (1/60 mins)

Notice how the value Size got factored out of the problem - this 
demonstrates that what I asked you to take my word for is in fact 
true: that size doesn't matter in this problem. Any value you choose 
to use for Size, from a thimble to the Grand Canyon, produces the same 
answer - Cool, huh?

Hope this helps,

-Doctor Ezick,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
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