Awater tank can be filled in 4 hours by an inlet pipe and enptied in 6 hours by outlet pipe how long will it take to fill the tank if both pipes are left open?

Nice Question. Thanks for posting.

Figure out what will happen in 1 hour.  

The input pipe will fill 1/4 of a tank if left to it's own devices. You can just reason this. It he fills 1 tank in 4 hours it stands to reason that 1 tank / 4 hours will be what is filled up in 1 hour.

The problem is, there is a pipe that takes the water out. It takes 6 hours to empty the tank.

So in one hour, the tank will empty out 1 tank / 6 hours or 1/6 of a tank.

The difference is how much the tank will actually get filled in 1 hour.

1/4 - 1/6 = 3/12 - 2/12 = 1/12. So every hour, 1/12 of the tank will get filled. Now you can set up a proportion.

1 hour : (1/12) tank = x: 1 tank Set up the proportion.

1/(1/12) = x / 1 Cross multiply

1*1 = (1/12) * x Combine the left

1 = (1/12)x Multiply both sides by 12

12 * 1 = 1 * x

x = 12 hours.

You can do this question with decimals. You have to take it out 4 places to get close to 12.

Net = In - out  

Net = 1/4 - 1/6

Net = 0.25 - 0.16667

Net = 0.083333 This is how much you will get in during an hour.

1 hour / 0.08333 tank = x hours / 1 tank Cross multiply

1 tank hour = 0.083333 tanks x divide by 0.083333

1/0.083333 = x

This is good enough to give 12 = x

12 hours to fill the tank under the given conditions.

x = 3 ± √(5+y)