Question

There are two taps opening into a tank. If both are opened, the tank would be full in 18 minutes. The time taken for it to fill with only the large tap open in 15 minutes less than the time to fill with only the small tap open. What is the time taken to fill the tank only with the small tap open?

Answer 1

Let the smaller tap fill the tank alone in x minutes.
Then, the larger tap will fill the tank in (x - 15) minutes.
In 1 minute, the smaller tap can fill $\frac{1}{x}$ part of the tank and
The larger tap can fill $\frac{1}{x-15}$ part of the tank.
The tank is filled up in 18 minutes if both the taps are open.
Thus, in one minute, the taps fill $\frac{1}{18}$ part of the tank.
$\frac{1}{x}+\frac{1}{x-15}=\frac{1}{18}$
$\Rightarrow \frac{x-15+x}{x(x-15)}=\frac{1}{18}$
$\Rightarrow \frac{2x-15}{x^2-15x}=\frac{1}{18}$
$\Rightarrow 36x-270=x^2-15x$
$\Rightarrow x^2-51x+270=0$
$\Rightarrow x^2-45x-6x+270=0$
$\Rightarrow x(x-45)-6(x-45)=0$
$\Rightarrow (x-45)(x-6)=0$
$\Rightarrow x=45$ or $x=6$
Now, x = 6 is inadmissible as x - 15 = 6 - 15 = - 9 is not possible
Thus, the time taken by small tap alone to fill the tank is 45 minutes.