precalculus question

Question

Two cyclists, 96 miles apart, start riding toward each other at the same time. One cycles 3 times as fast as the other. If they meet 4 hours later, what is the speed (in mi/h) of the faster cyclist?a. Write an equation using the information as it is given above that can be solved to answer this problem. Use the variable  to represent the speed of the slower cyclist.b. What is the speed of the faster cyclist?

Answer 1

r = rate of slower cyclist

3r = rate of faster cyclist

distance covered by slower cyclist = 4r (rate*time)

distance covered by faster cyclist = 12r

total distance = 96

4r + 12r = 96

r = 6

3r = 18 (faster cyclist rate is 18 mph)

Answer 2

Given:

Distance between two cyclists = 96 miles

r = rate of slower cyclist

3r = rate of faster cyclist

Cyclists are 96 miles apart and travel towards each other.

4 hrs = time both cyclists travel before meeting

Find: Speed of faster cyclist.

SOLUTION:

Use: Distance = (Rate)*(time)

D = (r)t + (3r)t

96 = (r)4 + (3r)4

96 = 4(r + 3r)

96 = 4(4r)

96 = 16r

6 = r

Faster cyclist speed:

3r = 3(6) = 18 miles/hr.