A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe .

15 hrs

The correct answer is option (1) : 15 hrs

Let the tine required by 1st pipe to fill the tank alone be 't' hrs

∴ by second pipe $=(t-5)hrs$

and the time required by 3rd pipe $= (t-5-4)=(t-9)hrs$

According to given,

$\frac{1}{t}+\frac{1}{t-5}=\frac{1}{t-9}$

$\frac{t-5+t}{t(t-5)}=\frac{1}{t-9}$

$(2t-5)(t-9)=t^2-5t$

$t^2-18t+45=0$

$t^2-15t-3t+45=0$

$t(t-15)-3(t-15)=0$

$(t-15)(t-3)=0$

$t=3, 15$ But $t > 5$

$t = 15$

Hence, the time required by 1st pipe to fill the tank = 15 hrs