Two water taps together can fill a tank in

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.