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October 13th, 2014, 10:30 PM   #1
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How long would it take with the third pipe alone?

A water tank can be emptied by 3 pipes in 3 hours; one alone would take 6 hours, another alone 9 hours; how long would it take with the third pipe alone? How do I solve this?
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October 13th, 2014, 10:58 PM   #2
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$\displaystyle 3\left(\frac19+\frac16+\frac1x\right)=1 \Rightarrow\,x=18$
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October 13th, 2014, 11:25 PM   #3
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Why did you do this?
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October 13th, 2014, 11:41 PM   #4
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Let's call them pipe A, pipe B and pipe C. Pipe A drains the tank in 9 hours and pipe B drains
the tank in 6 hours. Pipe C drains the tank in x hours. This means that, in 1 hour, Pipe A
drains 1/9 of the tank and pipe B drains 1/6 of the tank. Pipe C drains 1/x of the tank. The
sum of these fractions is what fraction of the tank will be drained in 1 hour. Multiplying these
by 3 should give us 1 - the tank is empty after 3 hours.
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October 14th, 2014, 12:10 AM   #5
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Quote:
Originally Posted by greg1313 View Post
Let's call them pipe A, pipe B and pipe C. Pipe A drains the tank in 9 hours and pipe B drains
the tank in 6 hours. Pipe C drains the tank in x hours. This means that, in 1 hour, Pipe A
drains 1/9 of the tank and pipe B drains 1/6 of the tank. Pipe C drains 1/x of the tank.
Why did you say that:
Pipe A
drains 1/9 of the tank and pipe B drains 1/6 of the tank. Pipe C drains 1/x of the tank? And why must we find the amount drained in 1 hour?
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October 14th, 2014, 01:47 AM   #6
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There's nothing wrong with Greg's solution, but here's an equivalent solution without algebra just in case you don't like letters! The key is to think in terms of fractions.

Let's say the total amount of water in the tank when is full is 1 whole. The fraction of tank leaking out each hour is given by 1 divided by the time taken (in hours) to empty the tank. You could use minutes or seconds if you wanted to, but you'd just be making more work for yourself since everything is already given in hours. You should get the same answer either way.

If the tank of water empties in 3 hours by all 3 pipes, then the fraction emptied by all three pipes each hour = $\displaystyle \frac{1}{3}$
If the tank of water empties in 9 hours by pipe A, then the fraction emptied by pipe A each hour = $\displaystyle \frac{1}{9}$
If the tank of water empties in 6 hours by pipe B, then the fraction emptied by pipe B each hour = $\displaystyle \frac{1}{6}$

The total fraction emptied by all three pipes per hour = fraction emptied by A per hour + fraction emptied by B per hour + fraction emptied by C per hour

So...

fraction emptied by C per hour = total fraction emptied by all three pipes per hour - fraction emptied by A per hour - fraction emptied by B per hour

fraction emptied by C per hour = $\displaystyle \frac{1}{3} - \frac{1}{6} - \frac{1}{9}$
fraction emptied by C per hour = $\displaystyle \frac{6}{18} - \frac{3}{18} - \frac{2}{18}$
fraction emptied by C per hour = $\displaystyle \frac{6-3-2}{18}$
fraction emptied by C per hour = $\displaystyle \frac{1}{18}$

This means that the total time taken to empty the tank by pipe C alone would be $\displaystyle 1 \div \frac{1}{18} = 18$ hours.

Also, see my other post on the salary problem.

Last edited by Benit13; October 14th, 2014 at 02:42 AM.
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October 15th, 2014, 01:57 AM   #7
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What if the question is phrased this way?

A water tank can be emptied by 3 pipes in 5 hours; one alone would take 6 hours, another alone 9 hours; how long would it take with the third pipe alone?
Is it right to write
$\displaystyle 5\left(\frac19+\frac16+\frac1x\right)=1$

And if they are two tanks emptying at the same rate, it would be

$\displaystyle 5\left(\frac29+\frac26+\frac2x\right)=2$

Last edited by Chikis; October 15th, 2014 at 02:15 AM.
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October 15th, 2014, 04:44 AM   #8
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Quote:
Originally Posted by Chikis View Post
What if the question is phrased this way?

A water tank can be emptied by 3 pipes in 5 hours; one alone would take 6 hours, another alone 9 hours; how long would it take with the third pipe alone?
Is it right to write
$\displaystyle 5\left(\frac19+\frac16+\frac1x\right)=1$
Yes. Because the fraction removed by all three pipes per hour is now $\displaystyle \frac15$. Think of it as:

$\displaystyle \frac19+\frac16+\frac1x=\frac15$

However, in this particular case the tank can never take as long as 5 hours to empty because pipes A and B alone can empty the tank in less than 5 hours. i.e. x would have to be negative, which makes no physical sense.

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And if they are two tanks emptying at the same rate, it would be
$\displaystyle 5\left(\frac29+\frac26+\frac2x\right)=2$
No. If you double the volume of water to leak out, the tank is going to take twice as long to empty for the individual pipes also, so you have
$\displaystyle \frac{1}{18}+\frac{1}{12}+\frac1x=\frac{1}{y}$

where $\displaystyle y$ is whatever time it takes to empty both tanks with all three pipes.
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October 15th, 2014, 06:28 AM   #9
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Quote:
Originally Posted by Chikis View Post
What if the question is phrased this way?

A water tank can be emptied by 3 pipes in 5 hours; one alone would take 6 hours, another alone 9 hours; how long would it take with the third pipe alone?
Is it right to write
$\displaystyle 5\left(\frac19+\frac16+\frac1x\right)=1$
No, because 5/9 + 5/6 > 1 - the tank would be drained in less than 5 hours.
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October 15th, 2014, 09:43 AM   #10
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If we look at Op, am thinking that the pipes have different diameters. If they all have the same diameter. It would have been more easier to handle.
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