Question

Two water taps together can fill a tank in 9 hours 36 minutes. The tap of larger diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Answer 1

Let, tap A be of layer diameter.
If tap A takes x hrs to fill the tank,
$\Rightarrow$ Time taken by smaller tank to fill the tank =(x+8) hrs.
$\therefore$ Rate at which tap A fills the tank=$\left(\frac{1}{x}\right)$
Rate at which tap B fills the tank=$\left(\frac{1}{x+8}\right)$
two tap together, time taken = 9 hrs 36 min =$9+\frac{36}{60}=9+\frac{3}{5}=\frac{48}{5}hr$
$\therefore$ work done per hr $=\frac{1}{\left(\frac{48}{5}\right)}=\frac{5}{48}$
$\therefore (Rate{)}_A+(Rate{)}_B$ =total rate
$\Rightarrow \frac{1}{x}+\frac{1}{x+8}=\frac{5}{48}$
$\Rightarrow \frac{2x+8}{x(x+8)}=\frac{5}{48}$
$\Rightarrow 48(2x+8)=5(x^2+8x)$
$\Rightarrow 96x+384=5x^2+40x$
$\Rightarrow 5x^2-56x-384=0$
$\therefore x=\frac{5b\pm \sqrt{5b^2+\mathrm{4.5.384}}}{2.5}$
$=\frac{56\pm 104}{10}$
$=\frac{56+104}{10}$ or $\frac{56-104}{10}$
$\therefore x=16$ or $-$ (negative of possible)
$\therefore tapAcanfillthetankin16hrs$ and
tap B can fill In (16+8)=24 hrs .