Let's assume the speed of the slower cyclist is x mi/hr.
Since the faster cyclist is cycling twice as fast, their speed would be 2x mi/hr.
In 1 hour, the slower cyclist would have traveled x mi, and the faster cyclist would have traveled 2x mi.
Since they start 36 mi apart and meet after 1 hour, the distance traveled by the slower cyclist and the faster cyclist should add up to 36 mi.
So, x + 2x = 36
3x = 36
x = 12
Therefore, the slower cyclist is traveling at 12 mi/hr and the faster cyclist is traveling at 24 mi/hr.
Two cyclists, 36 mi apart, start riding towards each other at the same time. One cycles twice as fast as the other. If they meet 1 hr later, at what average speed is each cyclist traveling?
slower cyclist
mi/hr
faster cyclist
mi/hr
1 answer