Problema Solution
a 9:00 on Saturday morning,two bicyclists heading in opposite direction pass each other on a bicycle path. the bicyclist heading north is riding 6 km/hour faster than the bicyclist heading south. at 10:15 they are 42.5 km apart. find the two bicyclists rates.
Answer provided by our tutors
let
x = the rate of the first bicyclist heading north
y = the rate of the second bicyclist heading south
d = 42.5 km the distance between them after period of time t
the period of time t = the hours between 9:00 and 10:15 that is 1 hour 15 min = 1 15/60 hours =1.25 hours
t = 1.25 hours
since the rate = distance / time => distance = speed * time
the distance of the first bicyclist + the distance of the second bicyclist = the total distance
t*x + t*y = d
1.25*x + 1.25*y = 42.5
the bicyclist heading north is riding 6 km/hour faster than the bicyclist heading south
x = y + 6
by solving the system of equations
x = y + 6
1.25*x + 1.25*y = 42.5
we find
x = 20 km/h
y = 14 km/h
The rates of the two bicyclist are are 20 km/h and 14 km/h.