# Two water taps together can fill a tank in

Question:

Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution:

Let the first water tape takes $x$ hours to fill the tank. Then the second water tape will takes $=(x+10)$ hours to fill the tank.

Since, the faster water tape takes $x$ hours to fill the tank.

Therefore, portion of the tank filled by the faster water tape in one hour $=\frac{1}{x}$

So, portion of the tank filled by the faster water tape in $9 \frac{3}{8}$ hours $=\frac{75}{8 x}$

Similarly,

Portion of the tank filled by the slower water tape in $9 \frac{3}{8}$ hours $=\frac{75}{8(x+10)}$

It is given that the tank is filled in $9 \frac{3}{8}$ hours.

So,

$\frac{75}{8 x}+\frac{75}{8(x+10)}=1$

$\frac{75(x+10)+75 x}{8 x(x+10)}=1$

$75 x+750+75 x=8 x^{2}+80 x$

$8 x^{2}+80 x-150 x-750=0$

$8 x^{2}-70 x-750=0$

$4 x^{2}-35 x-375=0$

$4 x^{2}-60 x+25 x-375=0$

$4 x(x-15)+25(x-15)=0$

$(x-15)(4 x+25)=0$

But, $x$ cannot be negative.

Therefore, when $x=15$ then

$(x+10)=15+10$

$=25$

Hence, the first water tape will takes 15 hours to fill the tank, and the second water tape will takes 25 hours to fill the tank.