A water tank has three taps attached, $A,B$ and $D$.
$A$ and $B$ fill the water tank completely in $\displaystyle\frac{25}{3}$ minutes and $\displaystyle\frac{25}{2}$ minutes, respectively.
$D$ drains water at the rate of $162$ $\text{L/min}$.
If the tank is initially full, and then all three taps are opened simultaneously, the tank is empty in $4$ minutes.
What is the capacity of water tank?
My try was the following:
Let capacity of tank be $x$.
Then Tap $A$ fills $\displaystyle\frac{12}{25}x$ litres in $4$ minutes and tap $B$ fills $\displaystyle\frac{8}{25}x$ litres in four minutes.
Thus total water filled in $4$ minutes is $\displaystyle\frac{4}{5}x$.
And tap $D$ will drain $162\times 4=648 $ litres in four minutes, thus $648=\displaystyle\frac{4}{5}x$, which implies $x=810$, but this is not the correct answer.
Where did I go wrong?
Take $x$ as the capacity, the rate $a$ for $A$ is $\frac{3}{25}x$, the rate $b$ for $B$ is $\frac{2}{25}x$. The outflow rate for $D$ is $162$. So the net outflow rate with all taps open is $162-a-b$. Hence $x=4(162-a-b)=648-4a-4b$.
Thus we have $x=648-\frac{12}{25}x-\frac{8}{25}x$, so $x=\frac{25}{45}648=360$.