Q:

USE DISTANECE= RATE * TIME 1- Anjali traveled to teh recycling plant and back. The trip there took 5.8 hrs and the trip back took 5.1 hrs. She averaged 7 mph faster on the return trip than the outbound trip. What was Anjali's average speed on the outbound trip?2- A fishing boat left Hawaii traveling west 0.5 hrs before a cruise ship. the cruise ship traveled in the opposite direction going 12.5km/h faster than the fishing boat for 11 hrs after which the time the ships were 322 km apart. What was te fishing boat's speed.

Accepted Solution

A:
a]
D=R×t
d=5.8r
d2=5.1(r+7)
0.7r=35.7
r=51 mi/hr
thus the average speed on the outbound trip would be:
51+7=58 mi/hr

b]
let speed of fishing boat=x km/h
speed of cruise ship=(x+12.5) km/h
distance traveled by fishing boat=11.5×x=11.5x km
distance traveled by cruise ship=(x+12.5)×11=(11x+137.5) km
Total distance covered by the ships:
11.5x+11x+137.5=322
22.5x=322-137.5
22.5x=184.5
thus
x=184.5/22.5
x=8.2 km/h
The speed of fishing boat is 8.2 km/hr