Let the speed of the current be $x$ km/h.

• The speed of the motorboat in still water is 18 km/h.
• The distance to travel upstream and downstream is 24 km.

Upstream:

• The effective speed when going upstream (against the current) is $18 - x$ km/h.
• Time taken to travel upstream is given by: $t_{\text{upstream}} = \frac{24}{18 - x}$

Downstream:

• The effective speed when going downstream (with the current) is $18 + x$ km/h.
• Time taken to travel downstream is: $t_{\text{downstream}} = \frac{24}{18 + x}$

Given Condition:

It is given that it takes 1 hour more to go upstream than downstream. This translates to the equation: $t_{\text{upstream}} = t_{\text{downstream}} + 1$ Substituting the expressions for the times: $\frac{24}{18 - x} = \frac{24}{18 + x} + 1$ Now, let's solve this equation step by step.

$\frac{24}{18 - x} - \frac{24}{18 + x} = 1$ Taking the LCM of the left-hand side: $\frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1$ Simplifying the numerator: $\frac{24(18 + x - 18 + x)}{(18 - x)(18 + x)} = 1$ $\frac{24 \times 2x}{(18 - x)(18 + x)} = 1$ Simplify further: $\frac{48x}{18^2 - x^2} = 1$ $48x = 324 - x^2$ Rearranging the terms: $x^2 + 48x - 324 = 0$ Now, solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 48$, and $c = -324$.

$x = \frac{-48 \pm \sqrt{48^2 - 4(1)(-324)}}{2(1)}$ $x = \frac{-48 \pm \sqrt{2304 + 1296}}{2}$ $x = \frac{-48 \pm \sqrt{3600}}{2}$ $x = \frac{-48 \pm 60}{2}$ So, we have two possible solutions: $x = \frac{-48 + 60}{2} = 6 \quad \text{or} \quad x = \frac{-48 - 60}{2} = -54$ Since the speed of the current cannot be negative, the speed of the current is: $x = 6 \, \text{km/h}$

Conclusion:

The speed of the current is $6 \, \text{km/h}$.

Would you like further details or have any questions?

Here are 5 related questions to expand on this:

1. How would the problem change if the boat's speed in still water was different?
2. What if the distance traveled was 30 km instead of 24 km?
3. How would you solve the problem if the time difference was 2 hours instead of 1 hour?
4. What is the effect of increasing the current speed on the time taken for upstream travel?
5. How does the quadratic formula help in finding the speed of the current?

Tip: When working with speed and current problems, always set up the time equations carefully to reflect the upstream and downstream differences clearly.