Questions: Pipe A can fill a tank in 2 hours less time it takes for Pipe B to fill the same tank. Pipe A starts filling a tank on its own for 5.5 hours, at which point Pipe B joins Pipe A in filling the tank. Together, the two pipes finish filling the tank in 5.5 additional hour(s). How long would it take each pipe to fill the tank on its own? Round your answers to three significant figures.

Solution Steps

To solve this problem, we need to set up equations based on the rates at which the pipes fill the tank. Let's denote the time it takes for Pipe B to fill the tank as \( t \) hours. Therefore, Pipe A takes \( t - 2 \) hours to fill the tank. We can use the information given to set up an equation and solve for \( t \).

1. Calculate the fraction of the tank filled by Pipe A in 5.5 hours.
2. Calculate the combined rate of both pipes working together.
3. Use the combined rate to determine how much of the tank is filled in the additional 5.5 hours.
4. Set up an equation to solve for \( t \).

Let \( t \) be the time it takes for Pipe B to fill the tank. Therefore, Pipe A takes \( t - 2 \) hours to fill the tank.

The fraction of the tank filled by Pipe A in 5.5 hours is: \[ \frac{5.5}{t - 2} \]

The combined rate of both pipes working together is: \[ \frac{1}{t - 2} + \frac{1}{t} \]

The fraction of the tank filled by both pipes in the additional 5.5 hours is: \[ 5.5 \left( \frac{1}{t - 2} + \frac{1}{t} \right) \]

The total fraction of the tank filled should be 1 (the whole tank): \[ \frac{5.5}{t - 2} + 5.5 \left( \frac{1}{t - 2} + \frac{1}{t} \right) = 1 \] Simplifying, we get: \[ \frac{11.0}{t - 2} + \frac{5.5}{t} = 1 \]

Solving the equation: \[ \frac{11.0}{t - 2} + \frac{5.5}{t} = 1 \] We get two solutions: \[ t \approx 0.6150 \quad \text{and} \quad t \approx 17.885 \]

Since \( t \) must be greater than 2 (because Pipe A takes \( t - 2 \) hours), we discard the first solution.

Final Answer