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September 10th, 2018, 08:50 PM   #1
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How to solve this emptying pipe problem?

Pipe A and Pipe B can fill a tank in 12 mins and 6 mins respectively, whereas pipe C can empty it completely in 5 mins B is opened to fill an empty tank After some time, A and C are also opened, after which it takes 15 min to fill the tank. How long was B open alone?

"x" is the volume of the tank. Q denotes the flow rate of each respective pipe.

Qa = x/12
Qb = x/6
Qc = -x/5


After this step, how to proceed ?
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September 10th, 2018, 10:37 PM   #2
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Let $t$ be the amount of time pipe B was open by itself and say the tank can hold $x$ litres.

In the time pipe B was on alone, it fills the tank with $\frac{tx}{6}$ litres of water. When pipe A and pipe C are also turned on for 15 minutes, the pipes combined fill the tank with $15(\frac{x}{6} + \frac{x}{12} - \frac{x}{5})$ litres of water.

Since the tank is filled up in this time, we can form the equation
$$\frac{tx}{6} + 15(\frac{x}{6} + \frac{x}{12} - \frac{x}{5}) = x.$$

Now solve for $t$.
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September 10th, 2018, 11:00 PM   #3
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Quote:
Originally Posted by Azzajazz View Post
In the time pipe B was on alone, it fills the tank with $\frac{tx}{6}$ litres of water.
How you obtain this equation: $\frac{tx}{6}$?
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September 11th, 2018, 02:25 AM   #4
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As B can fill the tank in 6 mins, B provides x/6 in each minute, and therefore tx/(6 min) l in time t.
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September 11th, 2018, 10:20 AM   #5
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Look at it as the job being travelling distance 1:

A:@a.................1...................>12
B:@b.................1...................>6
C:@c.................1...................>5

B:@b....>t;(A,B,C)@a+b-c........>15

a=1/12, b=1/6, c=1/5

bt + 15(a + b - c) = 1
t/6 + 15(1/12 + 1/6 - 1/5) = 1

Solve for t : t = 2/3

Same as Azzajazz's downunder way !
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September 11th, 2018, 12:47 PM   #6
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However, t = 2/3 is incorrect.
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September 11th, 2018, 02:39 PM   #7
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Quote:
Originally Posted by skipjack View Post
However, t = 2/3 is incorrect.
BUT 3/2 is! 1st typo this year
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