Reply To: Puzzles
To find the distance of Ms. Warne’s trip in nautical miles, we can use the information provided about the speeds and the time difference between the two legs of the journey.
First, let’s denote:
( D ) as the distance of one leg of the journey in nautical miles.
( V_1 = 12.5 ) knots as the speed from west to east.
( V_2 = 10 ) knots as the speed from east to west.
The time taken for the west to east journey is ( \frac{D}{V_1} = \frac{D}{12.5} ) hours.
The time taken for the east to west journey is ( \frac{D}{V_2} = \frac{D}{10} ) hours.
We know the east to west journey takes two and a half days longer than the west to east journey. In hours, this is ( 2.5 \times 24 = 60 ) hours.
So, we have the equation:
\frac{D}{10} = \frac{D}{12.5} + 60
To solve for ( D ), first eliminate the fractions by multiplying the entire equation by 125:
12.5D = 10D + 750
Simplifying gives:
2.5D = 750
Solving for ( D ):
D = \frac{750}{2.5} = 300
Thus, the distance of Ms. Warne’s trip in nautical miles is 300 nautical miles for one leg of the journey, making the round trip 600 nautical miles.
WHERE DID I GO WRONG @robb?