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.

Answer 1

rate of bicyclist heading north=20 km/hr

Answer 2

Answer:

the speed of the bicyclist heading south 14

the speed of the bicyclist heading north  20

Step-by-step explanation:

Let x is the speed of the bicyclist heading south

Let x+6 is the speed of the bicyclist heading north

Because, they are heading in opposite directions, so the relative speed is:

x + x + 6 = 2x+6

• Start time: 9:00 until 10:15 is 1 hour and 15 mins = 1$\frac{15}{60}$= $\frac{5}{4}$
• they are 42.5 km apart, that is the distance between them

As we know:

$\dfrac{Distance}{time}=Speed$

<=> $\dfrac{42.5}{\dfrac{5}{4}}=2x+6$

<=> 170 = 10x + 30

<=> x = 14 km/hour

=> Speed of bicyclist heading north would be 14+6=20 km/hr

Hope it will find you well.