MATH SOLVE

3 months ago

Q:
# 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.

Accepted Solution

A:

Answer: The rates of bicyclists of north and south are 20km/hr and 14km/hr respectively.Step-by-step explanation:Since we have given that Let the speed of bicyclist heading south be 'x'.Let the speed of bicyclist heading north be 'x+6'.Since they are moving in opposite directions,So, relative speed = x+x+6=2x+6Time taken = 10: 15 to 9:00 = 1 hour 15 minutes = [tex]1\dfrac{15}{60}=1\dfrac{1}{4}=\dfrac{5}{4}[/tex]Distance between them = 42.5 kmAccording to question, it becomes,[tex]\dfrac{Distance}{time}=Speed\\\\\dfrac{42.5}{\dfrac{5}{4}}=2x+6\\\\42.5=\dfrac{5}{4}(2x+6)\\\\4\times 42.5=5(2x+6)=10x+30\\\\170=10x+30\\\\170-30=10x\\\\140=10x\\\\x=\dfrac{140}{10}\\\\x=14\ km/hr[/tex]Speed of bicyclist heading north would be 14+6=20 km/hrSo, the rates of bicyclists of north and south are 20km/hr and 14km/hr respectively.