Question

At 9:00 on Saturday morning, two bicyclists heading in opposite directions 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

A. Northbound bicyclist= 20 km/h; southbound bicyclist= 14 km/h
B. Northbound=20 km/h; southbound= 11km/h
C. Northbound= 23 km/h; southbound= 17km/h
D. Northbound= 20km/h; southbound= 13 km/h

Answer 1

(A) Northbound bicyclist=20 km/hr, southbound bicyclist=14 km/h.

Explanation:

The total time taken is= 1 hour 15 min=1.25 hours. Let the distance traveled by bicyclist heading in the south direction is represented by
`D_(S)` and the distance traveled by the bicyclist heading in north direction be
`D_(N)`, then according to question
`D_(N)`+
`D_(S)`=42.5km. (1)

Therefore, using speed=
`(Distance)/(time)`,

Distance traveled by bicyclist heading in the south direction is
`D_(S)` = speed × time= 1.25×S, where S is the speed of the bicyclist heading in south........(2)

Since, The bicyclist heading in north direction is 6km/hr faster than the bicyclist heading in south direction, therefore,

Speed of bicyclist heading in north direction will be, N=S+6...........(3)

Let the distance travelled by the bicyclist heading in north direction is represented as
`D_(N)`.

Therefore,
`D_(N)`= 1.25N

=1.25(S+6)

=1.25S+ 7.5 (4)

Now, subtracting (2) from (4), we have

`D_(N)`-
`D_(S)`=1.25S+7.5-1.25S

`D_(N)`-
`D_(S)`=7.5 (5)

Now, from equation (1), we have
`D_(N)`+
`D_(S)`=42.5, solving equation (1) and equation (5), we get
`D_(N)`=25km.

Now, substituting this value of
`D_(N)` in equation (1),

`D_(N)`+
`D_(S)`=42.5

`D_(S)`=42.5-25.0

`D_(S)`=17.5km

Now,speed of the bicyclist heading in north direction will be:
`(Distance)/(Time)`

=
`(25)/(1.25)`

=
`20km/hr`

Also, Speed of bicyclist heading in south direction=
`(17.5)/(1.25)`

=
`14km/hr`