# 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 $1\frac12$ pools and the emptying tap emptied water amount of $\frac12$ pool.

So what I did is

$$\begin{array}{c|c|c|} & \text{power / rate} & \text{time (hours)} &\text{Total} \\ \hline \text{Filling tap until 80%} & \frac1x & 6 & \frac1x \times 6 = \frac6x\\ \hline \text{Emptying tap until 80%} & \frac1y & 2 & \frac1y \times 2 = \frac2y \\ \hline \text{Filling tap until the end} & \frac1x & \frac{1.5}{x} & 1.5\\ \hline \text{Emptying tap until the end} & \frac1y & \frac{0.5}{y} & 0.5 \\ \hline \end{array}$$

I can write that:

$$\frac6x - \frac2y = 0.8$$

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.

## 1 Answer

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:

$$\frac{6+z}{x}=1.5 \rightarrow z=1.5x-6$$ $$\frac{2+z}{y}=0.5 \rightarrow z=0.5y-2$$

Set both equations equal to each other using the Transitive Property of Equality:

$$1.5x-6=0.5y-2 \rightarrow 1.5x=0.5y+4 \rightarrow 3x=y+8 \rightarrow x=\frac{y+8}{3}$$

Substitute this into your first equation:

$$\frac{6}{\frac{y+8}{3}}-\frac{2}{y}=0.8 \rightarrow \frac{18}{y+8}-\frac{2}{y}=0.8$$

Multiply everything by $5y(y+8)$:

$$\frac{90y(y+8)}{y+8}-\frac{10y(y+8)}{y}=4y(y+8) \rightarrow 90y-10(y+8)=4y^2+32y \rightarrow 80y-80=4y^2+32y$$

Put everything on one side of the equation and then divide by $4$:

$$4y^2-48y+80=0 \rightarrow y^2-12y+20=0$$

This quadratic can be factored into $(y-10)(y-2)$:

$$(y-10)(y-2)=0 \rightarrow y-10=0 \ \text{OR} \ y-2=0 \rightarrow y=10 \ \text{OR} \ y=2$$

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.)

• What is Transitive Property of Equality? – Pichi Wuana Mar 26 '16 at 17:32
• We have $z=1.5x-6$ and $z=0.5y-2$. Therefore, since both $1.5x-6$ and $0.5y-2$ are equal to $z$, they are equal to each other, so we get $1.5x-6=0.5y-2$. That's how the Transitive Property of Equality works. – Noble Mushtak Mar 26 '16 at 17:33
• Oh! I didn't know it has a name! – Pichi Wuana Mar 26 '16 at 17:34