Let \( d \) be the distance in miles that you drove to the beach. We can set up the problem using the formula for distance, which is given by
\[ \text{distance} = \text{rate} \times \text{time}. \]
The time it takes to travel a distance at a given speed is calculated by rearranging the formula:
\[ \text{time} = \frac{\text{distance}}{\text{rate}}. \]
- Calculate the time taken at 40 miles per hour:
The time taken to drive \( d \) miles at 40 miles per hour is:
\[ t_1 = \frac{d}{40}. \]
- Calculate the time taken at 30 miles per hour:
The time taken to drive the same distance \( d \) at 30 miles per hour is:
\[ t_2 = \frac{d}{30}. \]
- Set up the equation based on the time difference:
According to the problem, if you drove at 30 miles per hour, you would arrive 25 minutes later than if you drove at 40 miles per hour. We need to convert 25 minutes into hours:
\[ 25 \text{ minutes} = \frac{25}{60} \text{ hours} = \frac{5}{12} \text{ hours}. \]
Now we can express the relationship between \( t_1 \) and \( t_2 \):
\[ t_2 = t_1 + \frac{5}{12}. \]
Substituting the expressions for \( t_1 \) and \( t_2 \) into this equation gives us:
\[ \frac{d}{30} = \frac{d}{40} + \frac{5}{12}. \]
- Clear the fractions by multiplying through by a common denominator:
The least common multiple of 30, 40, and 12 is 120. We multiply the entire equation by 120 to eliminate the denominators:
\[ 120 \cdot \frac{d}{30} = 120 \cdot \frac{d}{40} + 120 \cdot \frac{5}{12}. \]
This simplifies to:
\[ 4d = 3d + 50. \]
- Solve the equation for \( d \):
Subtract \( 3d \) from both sides:
\[ 4d - 3d = 50, \] \[ d = 50. \]
- Conclusion:
The distance driven to the beach is
\[ \boxed{50} \text{ miles}. \]
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