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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