Posted by Heidi on Tuesday, November 2, 2010 at 4:08pm.
A cyclist and a jogger start from a town at the same time and head for a destination 11 mi away. The rate of the cyclist is twice the rate of the jogger. The cyclist arrives 1.1 h before the jogger. Find the rate of the cyclist.

Math  Henry, Thursday, November 4, 2010 at 11:16pm
Jogger's Rate = r mi/h.
Cyclist Rate = 2r mi/h.
Cyclist time = t hrs.
Jogger's time = (t + 1.1) hrs.
Eq1: 2r*t = 11 mi.
Eq2: r(t + 1.1) = 11 mi.
Solve Eq1 for r:
2r*t = 11,
r = 11 / 2t.
In Eq2, substitute 11/2t for r:
11/2t(t + 1.1) = 11,
(5.5/t)(t + 1.1) = 11,
5.5 + 6.05 / t = 11.
Common denominator = t.
(5.5t + 6.05) / t = 11,
Cross multiply:
5.5t + 6.05 = 11t,
11t  5.5t = 6.05,
5.5t = 6.05,
t = 6.05 / 5.5 = 1.1 h.
Solve Eq1 for 2r;
2r * 1.1 = 11,
2r = 11 / 1.1 = 10 mi / h = cyclist rate.
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