Filling pool with 3 hoses puzzle!

Question

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 :)

Answer 1

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?

Answer 2

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.

Answer 3

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 ...

Answer 4

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.