Haji rows his canoe 10 mph in still water. He noticed one day that he could row 4 miles upstream in the same amount of time he could row 6 miles downstream. What was the approximate speed of the current that day?

Answer 1

Let \( c \) be the speed of the current in mph. When rowing upstream, the effective speed is \( 10 - c \) mph, and when rowing downstream, the effective speed is \( 10 + c \) mph.

Haji rows 4 miles upstream and 6 miles downstream. We can set up the time equations for both trips.

The time taken to row upstream is given by:

\[ \text{Time upstream} = \frac{4}{10 - c} \]

The time taken to row downstream is given by:

\[ \text{Time downstream} = \frac{6}{10 + c} \]

According to the problem, the time taken to row upstream is equal to the time taken to row downstream:

\[ \frac{4}{10 - c} = \frac{6}{10 + c} \]

To solve for \( c \), we can cross-multiply:

\[ 4(10 + c) = 6(10 - c) \]

Expanding both sides:

\[ 40 + 4c = 60 - 6c \]

Now, collect all terms involving \( c \) on one side:

\[ 4c + 6c = 60 - 40 \]

This simplifies to:

\[ 10c = 20 \]

Now, divide both sides by 10:

\[ c = 2 \]

Thus, the speed of the current that day is \( \boxed{2} \) mph.