Question

Two pipes are connected to the same tank. When working together, they can fill in 2hrs. The larger pipe, working alone, can fill the tank in 3 hrs less time than the smaller one. How long would the smaller one take, working alone, to fill the tank?

Answer 1

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 $\frac{1}{x}$ part of the tank.

Again the larger pipe in 1 hour can fill $\frac{1}{x - 3}$ part of the tank.

So, if both the pipes are open then, in 1 hour they can fill ($\frac{1}{x} + \frac{1}{x - 3} = \frac{2x - 3}{x(x - 3)}$ part of the tank.

Therefore, they can fill the full tank in $\frac{x(x - 3)}{2x - 3}$ hours.

As per given condition, we can write

$\frac{x(x - 3)}{2x - 3} = 2$

⇒ 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)