SOLUTION: I need help with this problem: Chris bikes a steady speed of 9 feet/second. Mark bikes a steady speed of 12 feet/second. If Chris gets a 75 second head start, how many feet will Ma

Question 691299: I need help with this problem: Chris bikes a steady speed of 9 feet/second. Mark bikes a steady speed of 12 feet/second. If Chris gets a 75 second head start, how many feet will Mark have to bike until he catches up to Chris? Type your answer as a decimal rounded to the nearest hundredth. --- I know how to start it with the equation chris: 9(timechris)= distance and mark: 12 (time mark)= d and then I get lost. Not sure if someone can help me or not, but I sure hope so!Thanks in advance!
Answer by MathTherapy(10553) (Show Source): You can put this solution on YOUR website!

I need help with this problem: Chris bikes a steady speed of 9 feet/second. Mark bikes a steady speed of 12 feet/second. If Chris gets a 75 second head start, how many feet will Mark have to bike until he catches up to Chris? Type your answer as a decimal rounded to the nearest hundredth. --- I know how to start it with the equation chris: 9(timechris)= distance and mark: 12 (time mark)= d and then I get lost. Not sure if someone can help me or not, but I sure hope so!Thanks in advance!

Let the distance that Mark has to travel to catch up to Chris be D
As Chris� speed is slower than Mark�s, then Chris will take 75 seconds more than Mark to get to the �catch up� point
Chris� time taken to travel to �catch-up� point equals Mark�s time to travel to the �catch-up� point, plus 75 seconds, OR

4D = 3D + 2,700 ----- Multiplying by LCD, 36

4D � 3D = 2,700

D, or distance traveled by Mark to catch up to Chris = feet

You can do the check!!

Send comments and �thank-yous� to �D� at MathMadEzy@aol.com
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