Solving the Tank Filling Pipe Problem

This problem involves understanding the concept of work rate. When pipes fill a tank, their rate is the fraction of the tank they can fill in a unit of time (usually an hour). The faster the pipe, the higher its rate.

Understanding Pipe Rates and Relationships

Let's denote the rate at which each pipe fills the tank:

• Rate of pipe A = \(R_A\) (tank per hour)
• Rate of pipe B = \(R_B\) (tank per hour)
• Rate of pipe C = \(R_C\) (tank per hour)

The problem gives us the relationships between their rates:

• Pipe B is 3 times as fast as A: \(R_B = 3 \times R_A\)
• Pipe C is \(1\frac{1}{2}\) times as fast as B: \(R_C = 1.5 \times R_B\)

Let's express all rates in terms of \(R_A\):

• We already have \(R_B = 3 R_A\).
• Substitute \(R_B\) into the relationship for \(R_C\):
  \(R_C = 1.5 \times (3 R_A) = 4.5 R_A\)

So, we have the rates as:

• \(R_A\)
• \(R_B = 3 R_A\)
• \(R_C = 4.5 R_A\)

Calculating the Combined Rate

When pipes A, B, and C work together, their rates add up. The combined rate (\(R_{ABC}\)) is:

\(R_{ABC} = R_A + R_B + R_C\)

Substitute the expressions in terms of \(R_A\):

\(R_{ABC} = R_A + 3 R_A + 4.5 R_A\)

\(R_{ABC} = (1 + 3 + 4.5) R_A\)

\(R_{ABC} = 8.5 R_A\)

Using the Total Time to Find the Combined Rate

We are given that the three pipes together fill the tank in 4 hours. The total work (filling one tank) is done in 4 hours. The formula relating work, rate, and time is:

Work = Rate × Time

In this case, Work = 1 tank, Time = 4 hours, and Rate = \(R_{ABC}\).

So, \(1 = R_{ABC} \times 4\)

From this, we can find the numerical value of the combined rate:

\(R_{ABC} = \frac{1}{4}\) tank per hour

Finding the Rate of Pipe A

Now we have two expressions for the combined rate \(R_{ABC}\):

• \(R_{ABC} = 8.5 R_A\) (from the relationship between rates)
• \(R_{ABC} = \frac{1}{4}\) (from the total time taken)

Equating these two, we get:

\(8.5 R_A = \frac{1}{4}\)

To find \(R_A\), we can write 8.5 as a fraction: \(8.5 = \frac{85}{10} = \frac{17}{2}\).

So, \(\frac{17}{2} R_A = \frac{1}{4}\)

Multiply both sides by \(\frac{2}{17}\) to isolate \(R_A\):

\(R_A = \frac{1}{4} \times \frac{2}{17}\)

\(R_A = \frac{2}{68}\)

\(R_A = \frac{1}{34}\) tank per hour

Calculating Time Taken by Pipe A Alone

Pipe A alone fills the tank at a rate of \(R_A = \frac{1}{34}\) tank per hour.

To find the time taken by pipe A alone to fill 1 tank, we use the formula: Time = Work / Rate.

Time for A alone = \(\frac{1 \text{ tank}}{R_A}\)

Time for A alone = \(\frac{1}{1/34}\)

Time for A alone = \(1 \times 34\)

Time for A alone = 34 hours

Summary of Rates and Times

Based on \(R_A = \frac{1}{34}\), we can find the rates and times for B and C as well:

• \(R_B = 3 R_A = 3 \times \frac{1}{34} = \frac{3}{34}\) tank/hour. Time for B alone = \(\frac{1}{3/34} = \frac{34}{3}\) hours.
• \(R_C = 4.5 R_A = 4.5 \times \frac{1}{34} = \frac{9}{2} \times \frac{1}{34} = \frac{9}{68}\) tank/hour. Time for C alone = \(\frac{1}{9/68} = \frac{68}{9}\) hours.

Let's verify the combined rate using these values:

\(R_A + R_B + R_C = \frac{1}{34} + \frac{3}{34} + \frac{9}{68} = \frac{2}{68} + \frac{6}{68} + \frac{9}{68} = \frac{2+6+9}{68} = \frac{17}{68} = \frac{1}{4}\) tank/hour.

The combined rate is indeed \(\frac{1}{4}\), which means they fill the tank in \(1 / (1/4) = 4\) hours, matching the information given in the question.

Therefore, pipe A alone will take 34 hours to fill the tank.

Revision Table: Key Concepts

Additional Information on Tank Filling Problems

Tank filling and pipe problems are common in aptitude tests. They often involve rates of filling (inlet pipes) and rates of emptying (outlet pipes).

• Inlet Pipes: Have positive work rates (add water to the tank).
• Outlet Pipes: Have negative work rates (remove water from the tank).
• Net Rate: If both inlet and outlet pipes are open, the net rate is the sum of inlet rates minus the sum of outlet rates. If the net rate is positive, the tank fills; if negative, it empties.
• Efficiency and Rate: The term "fast" or "efficient" directly relates to the rate. If something is twice as fast, its rate is twice as high. Time taken is inversely proportional to the rate (or efficiency). If A is twice as fast as B, A takes half the time B takes to do the same amount of work.

These problems often require expressing rates in terms of a common variable, as we did with \(R_A\) in this problem, to solve for unknown times or rates.