Question

a tank is filled by three pipes which uniform flow the first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone the second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe find the time taken by the first pipe alone to fill the tank

Answer 1

Refer to the above attachment.

Answer 2

The time taken by the first pipe alone to fill the tank is 15 hours.

Solution:

Let us assume the first pipe takes x hours to fill the tank

So the second pipe will fill the tank in (x-5) hours

And the third pipe will fill the tank in (x-9) hours.

Now,  $\frac{1}{x}+\frac{1}{x-5}=\frac{1}{x+9}$

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

$\Rightarrow (2 x-5)(x-3)=x(x-5)$

$\Rightarrow x^{2}-18 x+45=0$

$\Rightarrow (x-15)(x-3) = 0$

x = 15 or 3

As x =3 is not possible so x =15

So, the first pipe will take 15 hours.