Pipes and Tanks Problem Overview

This problem involves understanding the work rates of pipes that fill a tank (inlet pipes) and pipes that empty a tank (outlet pipes). We are given the individual filling times for pipes X and Y, and information about the combined operation with an outlet pipe Z. Our goal is to determine how long pipe Z takes to empty a specific portion of the tank that was previously filled by pipes X and Y.

Individual Pipe Filling Efficiency

To begin, let's calculate how much of the tank each inlet pipe, X and Y, can fill in one hour. This is also known as their individual work rate.

• Pipe X fills the entire tank in 10 hours. Therefore, in 1 hour, Pipe X fills \(\frac{1}{10}\) of the tank.
• Pipe Y fills the entire tank in 15 hours. Therefore, in 1 hour, Pipe Y fills \(\frac{1}{15}\) of the tank.

Combined Filling Rate of Pipes X and Y

When pipes X and Y are opened together, their combined filling rate per hour is the sum of their individual rates. This combined rate tells us how much of the tank they can fill together in one hour.

Combined rate of X and Y (per hour) = (Rate of X) + (Rate of Y)

Combined rate of X and Y (per hour) = \(\frac{1}{10} + \frac{1}{15}\)

To add these fractions, we find the least common multiple (LCM) of 10 and 15, which is 30.

Combined rate of X and Y (per hour) = \(\frac{3}{30} + \frac{2}{30} = \frac{3+2}{30} = \frac{5}{30} = \frac{1}{6}\) of the tank.

This means that pipes X and Y together can fill \(\frac{1}{6}\) of the tank in 1 hour. Consequently, the total time taken by X and Y together to fill the entire tank is the reciprocal of their combined rate, which is 6 hours.

Combined Work Rate with Outlet Pipe Z

The problem states that when all three pipes (X, Y, and Z) are opened together, it takes \(2\frac{1}{3}\) hours more than the time taken by X and Y together to fill the tank. Since X and Y together take 6 hours, let's calculate the total time with Z.

• Time taken by X and Y together = 6 hours.
• Additional time due to pipe Z = \(2\frac{1}{3}\) hours = \(\frac{7}{3}\) hours.
• Total time taken by X, Y, and Z together = \(6 + \frac{7}{3} = \frac{18}{3} + \frac{7}{3} = \frac{25}{3}\) hours.

Now, we can find the combined work rate of all three pipes (X, Y, and Z) when operating together. This is the reciprocal of the total time they take to fill the tank.

Combined rate of X, Y, and Z (per hour) = \(\frac{1}{\frac{25}{3}} = \frac{3}{25}\) of the tank.

Calculating Outlet Pipe Z's Rate

Pipe Z is an outlet pipe, meaning it empties the tank. Therefore, its contribution to the tank's filling is negative. The combined rate of X, Y, and Z is the sum of the inlet rates minus the outlet rate.

Combined rate (X + Y + Z) = (Rate of X + Rate of Y) - Rate of Z

We know the combined rate of X and Y is \(\frac{1}{6}\) and the combined rate of X, Y, and Z is \(\frac{3}{25}\).

\(\frac{3}{25} = \frac{1}{6} - \text{Rate of Z}\)

Now, we can solve for the emptying rate of pipe Z:

Rate of Z = \(\frac{1}{6} - \frac{3}{25}\)

To subtract these fractions, we again find the LCM of 6 and 25, which is 150.

Rate of Z = \(\frac{25}{150} - \frac{18}{150} = \frac{25-18}{150} = \frac{7}{150}\) of the tank (emptying per hour).

This means that pipe Z can empty \(\frac{7}{150}\) of the tank in one hour. If Z were to empty the entire tank alone, it would take \(\frac{150}{7}\) hours.

Amount Filled by X and Y Before Z is Opened

The problem states that pipes X and Y are opened together for \(3\frac{1}{2}\) hours, and then they are closed. We need to calculate how much of the tank they filled during this period.

• Time X and Y worked = \(3\frac{1}{2}\) hours = \(\frac{7}{2}\) hours.
• Combined filling rate of X and Y = \(\frac{1}{6}\) of the tank per hour.
• Amount filled = (Combined rate of X and Y) \(\times\) (Time they worked)
• Amount filled = \(\frac{1}{6} \times \frac{7}{2} = \frac{7}{12}\) of the tank.

So, \(\frac{7}{12}\) of the tank is filled before pipes X and Y are closed and pipe Z is opened.

Time for Z to Empty the Filled Portion

Finally, we need to determine how long it takes pipe Z, working alone, to empty the \(\frac{7}{12}\) portion of the tank that was filled. We use the emptying rate of Z calculated earlier.

• Amount to be emptied = \(\frac{7}{12}\) of the tank.
• Rate of Z to empty = \(\frac{7}{150}\) of the tank per hour.
• Time to empty = \(\frac{\text{Amount to be emptied}}{\text{Rate of Z}}\)
• Time to empty = \(\frac{\frac{7}{12}}{\frac{7}{150}}\)
• Time to empty = \(\frac{7}{12} \times \frac{150}{7}\)
• Time to empty = \(\frac{150}{12}\) hours.

Simplifying the fraction \(\frac{150}{12}\):

\(\frac{150}{12} = \frac{75}{6} = \frac{25}{2}\) hours.

Converting this improper fraction to a mixed number:

\(\frac{25}{2} = 12\frac{1}{2}\) hours.

Final Answer

Pipe Z can empty the tank in \(12\frac{1}{2}\) hours after pipes X and Y have filled it for \(3\frac{1}{2}\) hours.