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!


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
{{{D/9 = D/12 + 75}}}


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


4D – 3D = 2,700


D, or distance traveled by Mark to catch up to Chris = {{{highlight_green(2700)}}} feet


You can do the check!!


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