dourtuntellorvl

2022-06-24

One tap fills a pool. The other one empties it. It's a word problem.
In a pool there are two taps, one for filling and one for emptying. Once, when the pool was empty they opened the filling tap for 4 hours. Afterwards, they opened by mistake the emptying tap and after 2 hours the pool was filled 80% from its volume. When the pool was completely filled it turned out that the filling tap filled like water amount of $1\frac{1}{2}$ pools and the emptying tap emptied water amount of $\frac{1}{2}$ pool.
So what I did is
$\overline{)\begin{array}{cccc}& \text{power / rate}& \text{time (hours)}& \text{Total}\\ \text{Filling tap until 80%}& \frac{1}{x}& 6& \frac{1}{x}×6=\frac{6}{x}\\ \text{Emptying tap until 80%}& \frac{1}{y}& 2& \frac{1}{y}×2=\frac{2}{y}\\ \text{Filling tap until the end}& \frac{1}{x}& \frac{1.5}{x}& 1.5\\ \text{Emptying tap until the end}& \frac{1}{y}& \frac{0.5}{y}& 0.5\end{array}}$
I can write that:
$\frac{6}{x}-\frac{2}{y}=0.8$
And here I'm stuck. I don't know how to use the information of the last two rows of my chart. If I would know the time it took to fill 100% the pool then I could sum up the two last rows from the middle column. Or the time it took to fill up the remaining 20%.
Sorry if my translation of the question wasn't so good.

Rebekah Zimmerman

Expert

Let's say the time between when the pool was 80% filled and when the pool was completely filled is $z$ hours. The filling tap had been running for $6+z$ hours and the emptying tap had been running for $2+z$ hours. Thus, we have the two following equations:
$\frac{6+z}{x}=1.5\to z=1.5x-6$
$\frac{2+z}{y}=0.5\to z=0.5y-2$
Set both equations equal to each other using the Transitive Property of Equality:
$1.5x-6=0.5y-2\to 1.5x=0.5y+4\to 3x=y+8\to x=\frac{y+8}{3}$
Substitute this into your first equation:
$\frac{6}{\frac{y+8}{3}}-\frac{2}{y}=0.8\to \frac{18}{y+8}-\frac{2}{y}=0.8$
Multiply everything by $5y\left(y+8\right)$
$\frac{90y\left(y+8\right)}{y+8}-\frac{10y\left(y+8\right)}{y}=4y\left(y+8\right)\to 90y-10\left(y+8\right)=4{y}^{2}+32y\to 80y-80=4{y}^{2}+32y$
Put everything on one side of the equation and then divide by $4$:
$4{y}^{2}-48y+80=0\to {y}^{2}-12y+20=0$
This quadratic can be factored into $\left(y-10\right)\left(y-2\right)$

Thus, $y=10$ or $y=2$. However, if $y=2$, then, when we plug back into our $z=0.5y-2$ formula, we get $z=0$, which doesn't make any sense since $z$ is the amount of time between the pool being 80% filled and the pool being completely filled. Therefore, $y=10$
Finally, substitute this into $x=\frac{y+8}{3}$
$x=\frac{10+8}{3}=\frac{18}{3}=6$
Therefore, the filling tap takes 6 hours to fill a pool while the emptying tap takes 10 hours to fill a pool.
(Also, if you're curious, you can plug $y=10$ back into the $z=0.5y-2$ formula or plug $x=6$ back into the $x=1.5x-6$ formula to find that there were 3 hours between the pool being 80% filled and the pool being completely filled, meaning this whole process took $4+2+3=9$ hours overall.)

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