Tuesday

September 23, 2014

September 23, 2014

Posted by **Heidi** on Tuesday, November 2, 2010 at 4:08pm.

- Math -
**Henry**, Thursday, November 4, 2010 at 11:16pmJogger'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.

**Answer this Question**

**Related Questions**

Math - A cyclist and a jogger start from a town at the same time and head for a ...

math - A cyclist and a jogger start from a town at the same time and headed for ...

maths - a cyclist and jogger are 20 miles apart. the cyclist rides at 17 mph and...

Physics - A cyclist starts from rest and coasts down a 6.5∘{\rm ^\circ} ...

Math word problem - Two cyclists start biking from a trail's start 3 hours apart...

bob help - how do u come up w/the answer? Two cyclists start biking from a trail...

algebra - Two cyclists start biking from a trail's start 3 hours apart. The ...

Math - A cyclist traveled at a rate of 32 mph to visit a nearby town. The ...

maths - Two cyclists start biking from a trail's start 3 hours apart. The second...

algebra - A cyclist traveled at a rate of 32 mph to visit a nearby town. The ...