how would you solve:

the current in the river flowed at 3 mph. The boat could travel 24 mph downstream in half the time it took to travel 12 mi upstream. what was the speed of the boat in still water?

please help.

The time required to go any distance upstream a distance L is L/(12 - V) and the time to go downstream is
L/(12 + V),
where V is the speed of the water in the river.

Since L/(12-V) = 2 L/(12+V),
then 12 + V = 2 (12 - V)

Solve for V. L cancels out.

To solve this problem, let's break it down step by step:

Step 1: Define the variables:
Let's define the speed of the boat in still water as "B" and the speed of the river current as "C".

Step 2: Set up the equation for the upstream journey:
When traveling upstream, the effective speed of the boat is the difference between its speed in still water and the speed of the current. So, the effective speed is "B - C". The distance traveled upstream is 12 miles. We can calculate the time it takes to travel upstream using the formula: time = distance/speed. Therefore, the time taken to travel upstream is 12/(B - C).

Step 3: Set up the equation for the downstream journey:
When traveling downstream, the effective speed of the boat is the sum of its speed in still water and the speed of the current. So, the effective speed is "B + C". The distance traveled downstream is also 12 miles (half the distance of the upstream journey). We can calculate the time it takes to travel downstream using the same formula: time = distance/speed. Therefore, the time taken to travel downstream is 12/(B + C).

Step 4: Write the equation and solve for B:
According to the problem, the time taken to travel downstream is half the time taken to travel upstream, so we can set up the equation: 12/(B + C) = 2*(12/(B - C)).

To simplify the equation, we can cross multiply and solve for B:

12 * (B - C) = 2 * 12 * (B + C)
12B - 12C = 24B + 24C
12B - 24B = 24C + 12C
-12B = 36C
B = -3C

Step 5: Find the speed of the boat in still water:
From the equation obtained in step 4, we know that B = -3C. This means the speed of the boat in still water, "B," is equal to negative three times the speed of the river current, "C."

So, the speed of the boat in still water is three times the speed of the river current, in the opposite direction.

In conclusion, the speed of the boat in still water is three times the speed of the river current.