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I'm stumped on a math puzzle and I can't find an answer to it anywhere! A man is filling a pool from 3 hoses. Hose A could fill it in 2 hours, hose B could fill it in 3 hours and hose C can fill it in 6 hours. However, there is a blockage in hose A, so the guy starts by using hoses B and C. When the blockage in hose A has been cleared, hoses B and C are turned off and hose A starts being used. How long does the pool take to fill? Any help would be strongly appreciated :)

Karla
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  • How did you try to set up the problem? Use the information given in the problem to determine how much of the pool each hose can fill in one hour. These are the amounts that add up when more than one hose is use. – hardmath Oct 05 '18 at 15:30
  • Since the time that hose A is blocked is not specified, and math books don't tend to have unsolvable problems, that is a hint that the time is irrelevant. What remains is to show that that is true. What fraction of the pool does each hose fill in one hour? – user317176 Oct 05 '18 at 15:31
  • Does the fact that there is a blockage change the problem though? Since the person is starting with B and C and then switching to A? – Karla Oct 05 '18 at 15:49

4 Answers4

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Hose $A$ can fill half a pool per hour.
Hose $B$ can fill one third of a pool per hour.
Hose $C$ can fill one sixth of a pool per hour.

So how many pools per hour can hoses $B$ and $C$ fill, working together?

TonyK
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  • This makes so much sense when you put it like that...I can't thank everyone here enough! But the blockage in hose A is still bothering me, does it make a difference? – Karla Oct 05 '18 at 15:57
  • @Karla: the answer to my question is the reason it doesn't make a difference. – TonyK Oct 05 '18 at 16:59
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If the total volume of the pool is $x$, we can denote the rates as:

$r_A = \frac{x}{2}, r_B = \frac{x}{3}, r_C = \frac{x}{6}.$

From here you can see that:

$r_{B+C} = r_B + r_C = \frac{x}{3} + \frac{x}{6} = \frac{3x}{6} = \frac{x}{2} = r_A.$

So you can see that the two pipes fill up at the same rate as pipe $A$, your answer is simply 2 hours.

MRobinson
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  • You're amazing.... I never thought it could be that simple! I'm still stumped about the blockage though. Since B and C together can fill the pool in the same amount that A alone would, does it matter that there was a blockage? Would that change the 2 hours answer? – Karla Oct 05 '18 at 15:54
  • Nope, that's what the question wants you to realise – MRobinson Oct 05 '18 at 15:57
  • Awesome, I understand now. Thank you so much :) – Karla Oct 05 '18 at 15:58
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But the blockage in hose A is still bothering me, does it make a difference?

Finetuning MRobinson's solution.

Let the pool can fit $x$ units of water.

Let the rates of hoses be: $r_A=\frac{x}{2}; r_B=\frac x3; r_C=\frac x6$ per hour.

Assume the two hoses $B$ and $C$ worked $t_1$ hours and then only $A$ worked for $t_2$ hours. Then: $$\left(\frac x3+\frac x6\right)t_1+\frac x2\cdot t_2=x \Rightarrow \frac12(t_1+t_2)=1 \Rightarrow t_1+t_2=2.$$ Interpretation: Regardless of $t_1$ and $t_2$ hours, the total time is $2$ hours. For example, $B$ and $C$ could have worked for $0.5$ hours, then $A$ must have worked for $1.5$ hours, totalling $2$ hours.

farruhota
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Let's use meta-logic for this problem:

Assuming there is an answer to this problem at all, it must be true that it doesn't make a difference as to whether the guy uses both hoses B and C, or just hose A, for if there was a difference, then given that we are not told how long the blockage lasted, the problem would not be solvable.

Therefore, by meta-logic: we might as well assume that the guy was using just hose A, meaning it will take him 2 hours.

Of course, a true solution to the problem will have to include a verification that indeed hoses B and C together will indeed fill up the pool just as fast as hose A alone ...

Bram28
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