Answer:
The smaller pipe can fill the tank in 6 hours alone.
Step-by-step explanation:
Let us assume that the smaller pipe can fill the tank alone in x hours and the larger pipe fills it in (x - 3) hours.
Therefore, the smaller pipe in 1 hour can fill [tex]\frac{1}{x}[/tex] part of the tank.
Again the larger pipe in 1 hour can fill [tex]\frac{1}{x - 3}[/tex] part of the tank.
So, if both the pipes are open then, in 1 hour they can fill ([tex]\frac{1}{x} + \frac{1}{x - 3} = \frac{2x - 3}{x(x - 3)}[/tex] part of the tank.
Therefore, they can fill the full tank in [tex]\frac{x(x - 3)}{2x - 3}[/tex] hours.
As per given condition, we can write
[tex]\frac{x(x - 3)}{2x - 3} = 2[/tex]
⇒ x² - 3x = 4x - 6
⇒ x² - 7x + 6 = 0
⇒ (x - 6)(x - 1) = 0
⇒x = 6 or x = 1
But x can not be less than 2 hours.
So, x = 6 hours.
Therefore, the smaller pipe can fill the tank in 6 hours alone. (Answer)
