Question

Three taps of different diameters can fill a tank in 40 minutes when opened
simultaneously. Tap 1 and tap 2 can fill the same tank in 120 minutes and 150
minutes, respectively, when opened alone. How much time will tap 3 take to fill
the same tank if it is opened alone?​

Answer 1

only plus all numbers

and answer is answer

Answer 2

Given,

The time required to fill a tank by three taps = 40 minutes

The time required to fill the tank by Tap 1 alone = 120 minutes

The time required to fill the tank by  Tap 2 alone = 150 minutes

To Find,

The time required to fill the tank by  Tap 3 alone.

Solution,

The filling capacity and time required to fill are inversely proportional to each other.

Let the time required to fill the tank when only Tap 3 is open be $x$.

∴ $\frac{1}{40} = \frac{1}{120} +\frac{1}{150} + \frac{1}{x}$

⇒$\frac{1}{40} = \frac{5+4}{600} + \frac{1}{x}$

⇒$\frac{1}{40} = \frac{3}{200} + \frac{1}{x}$

⇒$\frac{1}{x} =\frac{1}{40} -\frac{3}{200} = \frac{5-3}{200}$

⇒$\frac{1}{x} = \frac{1}{100}$

∴ $x = 100$

Henceforth, the time required to fill the tank when only Tap 3 is open is 100 minutes.