Bunuel
A cruise ship traveled for 3 hours. In the first hour, the ship sailed at a speed of 25 Km/h, which was 25% faster than the speed in the third hour. In the middle hour the ship sailed at the average speed of the first and third hours. What was the total distance of the ship during the 3 hours of sailing?
A. 65.
B. 66.5.
C. 67.5.
D. 70.
E. 72.5
This question has a lot of words, but its premise is
very simple. The speed of each segment is the distance of each segment.
Each segment is 1 hour. Speeds are in km per HOUR. Hours cancel. Speed = distance.
\((r*t)=D\)
\((X\frac{km}{hour}*1 hr)=X\) km = \(D\)
Leg 1 speed: 25 kmh
Leg 3 speed. Leg 1 speed is 25% faster than Leg 3 speed. 25% faster =\(1.25=1\frac{1}{4}=\frac{5}{4}\)
Leg 3 speed: \((25=\frac{5}{4}L_3)\) => \((\frac{4}{5}*25=L_3)\)
Leg 3 speed: 20 kmh
Leg 2 speed: ave. of 25 and 20 = 22.5 kmh
Each distance equals each leg's speed.
Leg 1, distance: \(25\frac{km}{hour}*1hr=\) 25 km
Leg 2, distance: 22.5 km
Leg 3, distance: 20 km
Total distance: \((25+22.5+20)=67.5\) km
Answer C