Let the speed of the fish be \( F \) and the speed of the current be \( C \).
When swimming upstream, the effective speed of the fish is \( F - C \), and when swimming downstream, the effective speed is \( F + C \).
We are given that the fish swims 12 miles upstream in 4 hours, so we have the equation:
\[ 4(F - C) = 12 \]
And we are also given that the return trip (downstream) takes only 3 hours, so:
\[ 3(F + C) = 12 \]
Solving these two equations simultaneously:
From the first equation:
\[ F - C = 3 \]
From the second equation:
\[ F + C = 4 \]
Adding the two equations together:
\[ 2F = 7 \]
Therefore, \( F = \frac{7}{2} = 3.5 \) mph
Substitute back to find \( C \):
\[ 3.5 - C = 3 \]
\[ C = 0.5 \]
So, the rate of the current is 0.5 mph.
A fish swims 12 miles upstream in 4 hours. The return trip takes only 3 hours. Find the rate of the current.
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