Question

two taps x and y can fill a tank in 9 hours and 6 hours respectively. if both of them are opened together, how long will they take to fill the tank completely.

Answer 1

Let the Tank to be Filled be : T

Given that the Tap 'x' fills the Tank in 9 Hours

⇒ Tap 'x' fills $\frac{T}{9}$ of the Tank in One Hour

Given that the Tap 'y' fills the tank in 6 Hours

⇒ Tap 'y' fills $\frac{T}{6}$ of the Tank in One Hour

If both of them are Opened Together :

In One Hour they Fill : $(\frac{T}{9} + \frac{T}{6})$ of the Tank

$\Longrightarrow \frac{T}{9} + \frac{T}{6} = \frac{2T + 3T}{18} = \frac{5T}{18}$

⇒ In One Hour Both Taps fill $\frac{5T}{18}$ of the Tank

⇒ In $\frac{18}{5}$ Hours Both Taps fill $(\frac{5T}{18}\times\frac{18}{5})$ of the Tank

⇒ In $\frac{18}{5}$ Hours Both Taps fill the Tank Completely

$\Longrightarrow \frac{18}{5} = 3.6\;Hour$

So, Both Tanks Fill the Tank in 3.6 Hours

Answer 2

Answer:

Answer: Both the tap on opening together can completely fill the tank in 2 hours.

Step-by-step explanation:

Tap A fills the tank in 3 hours

Thus it can fill 1/3 of the tank in 1 hour,

Tap B fills the tank in 6 hours.

This it can fill 1/6 of the tank in 1 hour

In how much time will the tank be filled if both the taps are opened together?

If both the tap opened together they can fill the tank

= 1/3 + 1/6

= (2 + 1)/6 in one hour

= 3/6 or 1/2 of the tank in one hour

Thus, in 2 hours both the tap can completely fill the tank.