Can anyone help me solve this? Two taps A and B can fill a swimming pool in $3$ hours. If turned on alone, it takes tap A $5$ hours less than tap B to fill the same pool. How many hours does it take tap A to fill the pool?
\begin{align*}
3(A+B) & = x\\
\frac{x}{B} - \frac{x}{A} & = 5
\end{align*}
Can anyone show me some hints?
 A: First of all, if you have a tap that fills the pool in $t_1$ hours and another than fills the pool in $t_2$ hours, how much time will they take to fill the pool if they're turned on together? This is the heart of the problem, work that out as its own problem and then use what you find to solve the original problem. You should find that with both taps running, you'll take
$$t=\frac {t_1t_2} {t_1 + t_2}$$
hours to fill up the pool.
You're also falling into a common trap for beginners. The problem mentions several different quantities, so you instinctively write down variable names for all of them, and make the problem more complicated than it needs to be. This problem only requires one variable (and only one equation), not two and certainly not three. In the second sentence, they say, if $t_1$ is the time tap A takes and $t_2$ is the time tap B takes, that $t_1=t_2-5$. So you don't need both variables, any time you would normally write down $t_1$ just replace it with $t_2 - 5$.

To derive the relation $t=\frac {t_1t_2} {t_1 + t_2}$, let $V$ be the volume of the pool, and let $v_1$ and $v_2$ be the volume pumped out by each tap per hour. Then we have $t_1v_1=V$, $t_2v_2=V$. If you turn on both taps at once, you're getting $v_1+v_2$ water per hour, so
$$t=\frac V {v_1 + v_2}$$
Except that we don't want $t$ expressed using $v_1$, $v_2$ and $V$, we want to express it in terms of $t_1$ and $t_2$. Here you can use the relations $t_1v_1=V$, $t_2v_2=V$.
A: Ah, the inverse proportion problems: a pain, indeed. Denote by $\;a,b\;$ the times it takes taps $\;A,B\;$ resp. each to fill the pool alone .
Were given that
$$\begin{align}&3\left(\frac1a+\frac1b\right)=1\iff a+b=\frac{ab}3&\\{}\\&b=a+5\end{align}$$
Solve the above system ( the answers are ugly)
