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

So it took an hour an 15 minutes which means the distance traveled is their speed+1/4 of their speed, also any answers that have northbound going faster than 6km/h above south is not right so B and D seems to be out of the picture

Answer 2

Answer:

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

Step-by-step 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= $\frac{Distance}{time}$,

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: $\frac{Distance}{Time}$

=$\frac{25}{1.25}$

=$20km/hr$

Also, Speed of bicyclist heading in south direction=$\frac{17.5}{1.25}$

=$14km/hr$