Posted by Anonymous on Monday, April 18, 2011 at 3:02pm.
Rate of 2nd cyclist = X mi/h.
Rate of 1st cyclist = (x+10) mi/h.
Eq1: d2 = Xt = 210mi.
X = 210 / t.
Eq2: d1 = (X+10)(t-2.4) = 210mi.
In Eq2, substitute 210/t for X:
(210/t+10)(t-2.4) = 210,
210-504/t+10t-24 = 210,
-504/t+10t-24 = 210-210,
-504/t+10t-24 = 0,
Multiply both sides by t:
10t^2-24t-504 = 0,
Divide both sides by 2:
5t^2-12t-252 = 0.
Solve using Quadratic Formula and get:
t = 8.4, and -6. Select positive t.
t = 8.4h.
X = 210/t = 210 / 8.4 = 25mi/h.
X+10 = 25+10 = 35mi/h = Rate of 1st cyclist.
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