Two cyclist leave towns 76 mi apart at the same time and travel towards each One cyclist travels 10mi/h slower than the other. If the meet in 2 hours, what is the rate of each cyclist?

Thank you for the opportunity to help you with your question!

Let the speed of faster one be x mph and the speed of the other slower cyclist be y mph.

We have been given that the speed of one cyclist is 10 mph slower than the other.

Then the equation is ,x=y+10 ........eqn(1)

Also the distance traveled by cyclist of speed x mph =Speed*time ,d1 =2x,as the time taken is 2 hour.

Distance traveled by the cyclist of speed y mph ,d2=2y

As the two cyclist meet at some point after 2 hour therefore the sum of the distance must be equal to 76 mile

2x+2y+76

x+y=38 ...............eqn(2)

Solving both equation using substitution.Substitute x=y+10 in eqn(2),we get

y+10+y=38

2y=28

y=14 mph=>Speed of the slower cyclist(answer)

Speed of the faster cyclist=10+14=24 mph (answer)

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