Let the speed of the faster boat in still water be $x \, \text{km/h}$, and let the speed of the slower boat in still water be $y \, \text{km/h}$. The speed of the current is $v \, \text{km/h}$.

Downstream Scenario:

• When the boats go downstream (with the current), their effective speeds are:
  • Faster boat: $x + v \, \text{km/h}$
  • Slower boat: $y + v \, \text{km/h}$

We are given that the faster boat covers the distance from A to B $1.5$ times as fast as the slower boat, which means the time taken by the faster boat is $\frac{2}{3}$ of the time taken by the slower boat.

Let the distance from A to B be $D \, \text{km}$. Then, using the formula for time $t = \frac{\text{distance}}{\text{speed}}$, we have:

$\frac{D}{x + v} = \frac{2}{3} \times \frac{D}{y + v}$

Simplifying this equation:

$\frac{1}{x + v} = \frac{2}{3} \times \frac{1}{y + v}$

$3(y + v) = 2(x + v)$

$3y + 3v = 2x + 2v$

$3y + v = 2x \quad \text{(Equation 1)}$

Also, it is given that for every hour, the slower boat lags behind the faster boat by $8 \, \text{km}$. This means:

$(x + v) - (y + v) = 8$

$x - y = 8 \quad \text{(Equation 2)}$

Upstream Scenario:

• When the boats go upstream (against the current), their effective speeds are:
  • Faster boat: $x - v \, \text{km/h}$
  • Slower boat: $y - v \, \text{km/h}$

We are given that the faster boat covers the distance from B to A in half the time taken by the slower boat. So:

$\frac{D}{x - v} = \frac{1}{2} \times \frac{D}{y - v}$

Simplifying this:

$\frac{1}{x - v} = \frac{1}{2} \times \frac{1}{y - v}$

$2(y - v) = x - v$

$2y - 2v = x - v$

$x = 2y - v \quad \text{(Equation 3)}$

Solving the System of Equations:

We now have the system of three equations:

1. $3y + v = 2x$
2. $x - y = 8$
3. $x = 2y - v$

Substitute $x = 2y - v$ from Equation 3 into Equation 1:

$3y + v = 2(2y - v)$

$3y + v = 4y - 2v$

$3v = y$

So, $v = \frac{y}{3}$.

Now, substitute $v = \frac{y}{3}$ into Equation 3:

$x = 2y - \frac{y}{3}$

$x = \frac{6y}{3} - \frac{y}{3} = \frac{5y}{3}$

Finally, substitute $x = \frac{5y}{3}$ into Equation 2:

$\frac{5y}{3} - y = 8$

$\frac{5y - 3y}{3} = 8$

$\frac{2y}{3} = 8$

$y = 12 \, \text{km/h}$

Now, substitute $y = 12 \, \text{km/h}$ into $x = \frac{5y}{3}$:

$x = \frac{5 \times 12}{3} = 20 \, \text{km/h}$

Answer:

The speed of the faster boat in still water is $20 \, \text{km/h}$. Therefore, the correct option is:

$\boxed{20 \, \text{km/h}}$

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Would you like a detailed explanation of any part, or have further questions?

Here are 5 related questions you might find useful:

1. What is the formula for effective speed when moving with or against the current?
2. How do you set up equations for relative speeds of two moving objects?
3. How does the current's speed affect the overall travel time?
4. How can you solve linear equations involving two variables?
5. What are some other examples of word problems involving boats and currents?

Tip: When solving word problems, carefully translate the problem into mathematical equations before attempting to solve them!