Questions: A rowing team rowed 50 miles while going with the current in the same amount of time as it took to row 25 miles going against the current. The rate of the current was 2 miles per hour. Find the rate of the rowing team in still water. Rate of the team in still water: miles/hour

A rowing team rowed 50 miles while going with the current in the same amount of time as it took to row 25 miles going against the current. The rate of the current was 2 miles per hour. Find the rate of the rowing team in still water.

Rate of the team in still water: miles/hour

Transcript text: A rowing team rowed 50 miles while going with the current in the same amount of time as it took to row 25 miles going against the current. The rate of the current was 2 miles per hour. Find the rate of the rowing team in still water. Rate of the team in still water: $\square$ miles/hour

Solution

Solution Steps

Step 1: Set Up the Equation

Let $ x $ be the rate of the rowing team in still water (in miles per hour). The rate with the current is $ x + 2 $ and the rate against the current is $ x - 2 $. The time taken to row 50 miles with the current is equal to the time taken to row 25 miles against the current. Therefore, we can set up the equation:

\[ \frac{50}{x + 2} = \frac{25}{x - 2} \]

Step 2: Solve the Equation

Cross-multiplying gives us:

\[ 50(x - 2) = 25(x + 2) \]

Expanding both sides:

\[ 50x - 100 = 25x + 50 \]

Rearranging the equation:

\[ 50x - 25x = 50 + 100 \]

This simplifies to:

\[ 25x = 150 \]

Dividing both sides by 25:

\[ x = 6 \]