SOLUTION: A is 78 km distant from B. A cyclist leaves A in the direction of B. One hour later, another cyclist leaves B in the direction of A and cycles 4km/h faster than the first one. The

Question 1114183: A is 78 km distant from B. A cyclist leaves A in the direction of B. One hour later, another cyclist leaves B in the direction of A and cycles 4km/h faster than the first one. They meet 36 km from B. How long is each one enroute prior to the encounter and what are their speeds? (ans:14 km/h , 18 km/h ; 3 hours , 2 hours)
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.

Let  r  be the rate of the first cyclist, im km/h.

Then the rate of the second cyclist is (r+4) km/h.


36 km from B is the same as 78-36 = 42 km from A.


Time equation is 


 -  = 1.


42(r+4) - 36r = r*(r+4)

42r + 168 - 36r = r^2 + 4r


r^2 - 2r - 168 = 0

(r-14)*(r+12) = 0


We look for the positive root which is  r= 14.


Answer.  The 1-st cyclist speed was 14 km/h.   The 2-nd cyclist speed was  14+4 = 18 km/h.

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