Question 304959: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
please give a complete step by step answer to this question - Thank You
I last took algebra in college back in the early 90s, so I know how to tackle the questions but I don't remember the correct way to solve many of the algebra questions I am facing. I am reviewing for the TEAS nursing aptitude entrance exam so, it very important that I review and re-master my weak subject areas like higher-math and Algebra specifically.
Thanks for you help in advance and I look forward to hearing back from you soon.
Matt
mcpcot@hotmail.com
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 miles per hour and starts 3 hours after the first cyclist who is traveling at 6 miles per hour. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
The first cyclist rides at 6 mph.
So in 3 hours he rides 18 miles
Let the distance where they meet be x+18
By the time second cyclist travels x+18 miles, first cyclist rides x miles
Time taken by fist cyclist = distance / rate = x/6 hours
time taken by second cyclist = x+18 / 10
The times are same in both the case
x/6 = x+18/10
6(x+18)=10x
6x+108= 10x
4x=108
x= 27 miles
Distance / rate = time
x+18 /10 is the time taken by second cyclist to cross the first cyclist
27+18 /10 = 45/10 = 4.5 hours
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