The excursion boat on the river takes 2½ hours to make the trip to a point 12 miles upstream and to return. If the rate at which the boat travels in still water is 5 times the rate of the river current, what is the rate of the current?

Answer 1

Let's denote the rate of the boat in still water as B and the rate of the river current as R.

When the boat is traveling upstream, the effective speed is reduced by the rate of the current, so the boat's speed is B - R. Similarly, when the boat is traveling downstream, the effective speed is increased by the rate of the current, so the boat's speed is B + R.

We are given that it takes 2½ hours to make the round trip, which consists of the trip upstream and the trip downstream. Since the distance traveled upstream and downstream is the same, we can set up the following equation:

(12 miles) / (B - R) + (12 miles) / (B + R) = 2½ hours.

Now we need to represent the relationship between the boat's speed in still water and the rate of the current. The problem tells us that the boat's speed in still water is 5 times the rate of the river current, i.e., B = 5R.

Substituting this expression for B into the earlier equation:

(12 miles) / (5R - R) + (12 miles) / (5R + R) = 2½ hours.

Simplifying the denominators:

(12 miles) / (4R) + (12 miles) / (6R) = 2½ hours.

Combining the fractions:

(36R + 24R) / (24R) = 2½ hours.

Simplifying the numerator:

60R / (24R) = 2½ hours.

Simplifying the fraction:

60R / 24R = 2½ hours.

Cancelling out the common factor of 12:

5 / 2 = 2½ hours.

Now we can solve for the unknown, which is the rate of the river current, R:

5 / 2 = 5/2 hours.

Dividing both sides by 2:

5 / 2 = 2½ hours.

Simplifying the fraction:

5 = 5 hours.

Therefore, the rate of the river current is 5 miles per hour.