let's say the pipes are "a", to drain it in 6hours, "b" to fill it up by itself in 4hours and "c" to fill it up in 5 hours.
so..."a" can drain it in 6hours, that means, in 1 hour, "a" has only done 1/6 of the job, because in 6 hours, it would have done 6/6 or 1 whole, the whole job, the cistern fully filled up, but, in 1 hour, it has only done 1/6 of that.
"b" in 1 hour, has only done 1/4 of the whole job then.
"c" in 1 hour has only done 1/5 of the whole job then.
now... let's say it took "x" hours, if we have "a", "b" and "c" running, a draining away whilst b and c filling it up.
since it takes "x" hours to fill the cistern, after 1 hour, all three have only done 1/x of the whole job.
well, let's add their rates, to see what we get then.
[tex]\bf \begin{array}{llllllll}
\cfrac{1}{4}&+&\cfrac{1}{5}&-&\cfrac{1}{6}&=&\cfrac{1}{x}\\\\
\uparrow &&\uparrow &&\uparrow &&\uparrow \\
b&&c&&a&&x
\end{array}\\\\
-------------------------------\\\\
\textit{let's add the left-hand-side, using an LCD of 60}
\\\\\\
\cfrac{1}{4}+\cfrac{1}{5}-\cfrac{1}{6}=\cfrac{1}{x}\implies \cfrac{15+12-10}{60}=\cfrac{1}{x}\impliedby \textit{now cross-multiplying}
\\\\\\
x(15+12-10)=60(1)\implies 15x+12x-10x=60[/tex]
how long does it take to fill the cistern? well, just solve for "x".
