To solve this problem, we can use the concept of relative speed.
Let's assume the speed of the boat in still water is "b" units (e.g., kilometers per hour), and the speed of the current is "c" units.
When the boat is traveling upstream (against the current), the effective speed of the boat is reduced by the speed of the current. Therefore, the speed of the boat relative to the ground is (b - c) units.
When the boat is traveling downstream (with the current), the effective speed of the boat is increased by the speed of the current. Therefore, the speed of the boat relative to the ground is (b + c) units.
According to the problem, the boat takes 2 hours to travel upstream (Town A) and 1 hour and 20 minutes (1.33 hours) to travel downstream (back to the starting point).
Let's set up the equation for the upstream trip:
Distance = Speed × Time
Since the speed of the boat relative to the ground is (b - c) units and the time taken is 2 hours, we have:
Distance upstream = (b - c) × 2
Now, let's set up the equation for the downstream trip:
Distance = Speed × Time
Since the speed of the boat relative to the ground is (b + c) units and the time taken is 1.33 hours, we have:
Distance downstream = (b + c) × 1.33
Since the distance traveled upstream is the same as the distance traveled downstream (the boat returns to the starting point), we can equate the two distances:
(b - c) × 2 = (b + c) × 1.33
Simplifying the equation:
2b - 2c = 1.33b + 1.33c
0.67b = 3.33c
b = (3.33c) / 0.67
b = 4.97c
Therefore, the speed of the boat in still water is approximately 4.97 times the speed of the current.