SOLUTION: The distance between two points And B is 100km Two cyclists start simultaneously from A to B.The speed of first cyclist is 10km/hr greater than that of the other. The first cyclist

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Question 1064928: The distance between two points And B is 100km Two cyclists start simultaneously from A to B.The speed of first cyclist is 10km/hr greater than that of the other. The first cyclist stops on his way to B and rests for50 minutes but still he is the first to arrive at B. The speed of the first cyclist is
Answer by KMST(5328) About Me  (Show Source):
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At a speed of x kph (kilometers per hour),
the second cyclist cover 100km in
100%2Fx hours.
The second cyclist, at x%2B10 kph,
would cover the same distance in
100%2F%28x%2B10%29 hours,
but takes a break lasting 50%2F6=5%2F6 hours,
so his/her trip takes 100%2F%28x%2B10%29%2B5%2F6 hours.
The problem says the first cyclist trip took less time,
so 100%2F%28x%2B10%29%2B5%2F6%3C100%2Fx is our inequality.
Multiplying both sides of the inequality times
the positive number 6x%28x%2B10%29 , we get
600x%2B5x%28x%2B10%29%3C600%28x%2B10%29
600x%2B5x%5E2%2B50x%3C600x%2B6000%7D%7D%0D%0A%7B%7B%7B5x%5E2%2B50x%3C6000
5x%5E2%2B50x-6000%3C0
Dividing by 5 both sides of the inequality, we get
x%5E2%2B10x-1200%3C0
You know that the quadratic polynomial
x%5E2%2B10x-1200 is negative only between its zeros.
The solutions to
x%5E2%2B10x-1200=0 <---> %28x-30%29%28x%2B40%29=0
are x=30, and x=-40 ,
and since speeds should not be negative,
0%3C=x%3C30 ---> 10%3C=x%2B10%3C40 is the solution.
The second cyclist may be moving at 0 kph (sitting still),
and then the first one would be going at 10kph,
or they both may be moving,
but the first cyclist is moving at
10 kph or more, but less than 40 kph.