Finding Pipe Rate Ratio

Let the capacity of the tank be $V$ units.

Let the rate at which pipe A fills the tank be $R_A$ units per hour.

Let the rate at which pipe B fills the tank be $R_B$ units per hour.

Scenario 1: A Filling, B Filling

When both pipes A and B are used for filling, the combined filling rate is $(R_A + R_B)$.

The time taken to fill the tank is given as $t$ hours.

Therefore, the tank capacity $V$ can be expressed as:

$V = (R_A + R_B) \times t$

From this, we can write the time taken as:

$t = \frac{V}{R_A + R_B}$

Scenario 2: A Filling, B Emptying

When pipe A is filling and pipe B is emptying, the net filling rate is $(R_A - R_B)$. Note that this rate must be positive for the tank to fill.

The time taken to fill the tank in this case is given as $5t$ hours.

Therefore, the tank capacity $V$ can be expressed as:

$V = (R_A - R_B) \times 5t$

Calculating the Rate Ratio

We have two expressions for $V$ and $t$. Let's equate them:

Substitute the expression for $t$ from Scenario 1 into the equation from Scenario 2:

$V = (R_A - R_B) \times 5 \times \left( \frac{V}{R_A + R_B} \right)$

Assuming the tank capacity $V$ is not zero, we can cancel $V$ from both sides:

$1 = (R_A - R_B) \times \frac{5}{R_A + R_B}$

Rearrange the equation:

$\frac{R_A + R_B}{R_A - R_B} = 5$

Now, solve for the ratio $\frac{R_A}{R_B}$:

$R_A + R_B = 5(R_A - R_B)$

$R_A + R_B = 5R_A - 5R_B$

Group terms with $R_A$ and $R_B$:

$R_B + 5R_B = 5R_A - R_A$

$6R_B = 4R_A$

To find the ratio of the rates of A and B ($R_A : R_B$):

$\frac{R_A}{R_B} = \frac{6}{4}$

Simplify the fraction:

$\frac{R_A}{R_B} = \frac{3}{2}$

Thus, the ratio of the rates of A and B is $3 : 2$.