Let the speed of the current be \( c \) miles per hour. The speed of the motorboat relative to the bank (ground speed) will then be:
- Upstream: \( 14 - c \) miles per hour
- Downstream: \( 14 + c \) miles per hour
We know that:
- The time taken to go upstream is 25 minutes, which is \( \frac{25}{60} \) hours or \( \frac{5}{12} \) hours.
- The time taken to return downstream is 10 minutes, which is \( \frac{10}{60} \) hours or \( \frac{1}{6} \) hours.
Let \( d \) be the distance to the point upstream. We can set up equations based on the formula \( \text{distance} = \text{speed} \times \text{time} \).
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For the upstream trip: \[ d = (14 - c) \cdot \frac{5}{12} \]
-
For the downstream trip: \[ d = (14 + c) \cdot \frac{1}{6} \]
Since both expressions represent the same distance \( d \), we can set them equal to each other:
\[ (14 - c) \cdot \frac{5}{12} = (14 + c) \cdot \frac{1}{6} \]
Now, we will eliminate \( d \) and solve for \( c \).
First, clear the fractions by multiplying both sides by 12:
\[ 5(14 - c) = 2(14 + c) \]
Expanding both sides gives:
\[ 70 - 5c = 28 + 2c \]
Now, combine like terms:
\[ 70 - 28 = 5c + 2c \]
This simplifies to:
\[ 42 = 7c \]
Dividing both sides by 7:
\[ c = 6 \]
Thus, the speed of the current is \( \boxed{6} \) miles per hour.