# Question #a8a49

Apr 13, 2017

I normally use a "strange" approach in this cases...anyway I got 1 hour and 45 minutes.

#### Explanation:

Let us call the quantity of water in the pool $x$ and the rates (in liter per hour) of the two pumps ${r}_{1}$ and ${r}_{2}$ so we get:
${r}_{1} \cdot 6 = x$
${r}_{2} \cdot 10 = x$
meaning that the rate times the time will give us the amount of water pumped out.
Let us use the two pumps together; we get:
$\left({r}_{1} + {r}_{2}\right) t = x$
where $t$ will be the requred time.
From the two previos equations we get:
${r}_{1} = \frac{x}{6}$
${r}_{2} = \frac{x}{10}$
that substituted into: $\left({r}_{1} + {r}_{2}\right) t = x$ gives us:
$\left(\frac{x}{6} + \frac{x}{10}\right) t = x$
$x \left(\frac{1}{6} + \frac{1}{10}\right) t = x$
or:
$\left(\frac{1}{6} + \frac{1}{10}\right) t = 1$
$t = 3.75$ hours corresponding to 1 hour and 45 minutes.