Question

Question #a8a49

Answer 1

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*6=x`
`r_2*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:
`(r_1+r_2)t=x`
where `t` will be the requred time.
From the two previos equations we get:
`r_1=x/6`
`r_2=x/10`
that substituted into: `(r_1+r_2)t=x` gives us:
`(x/6+x/10)t=x`
`x(1/6+1/10)t=x`
or:
`(1/6+1/10)t=1`
`t=3.75` hours corresponding to 1 hour and 45 minutes.