Here's an algebraic approach:

We know the good 'ole formula d = rt (distance equals rate times time). This is a good time to use that one.

Here are our variables:

d = distance the cyclists have traveled at the point cyclist 2 passes cyclist 1
r1 = rate of cyclist 1
r2 = rate of cyclist 2
t1 = time that cyclist 1 rides
t2 = time that cyclist 2 rides

Here are the formulas we can infer from the problem statement:
d = r1*t1
d = r2*t2
t1 = t2 + 3 hrs

Let's set the first two equations equal. This gives us:

r1*t1 = r2*t2

Plugging in the numbers from the problem statement as well as the third formula, I get:

6 mi/hr (t2 + 3 hr) = 10 mi/hr * t2

Multiplying I get:

(6 mi/hr * t2) + 18 mi = 10 mi/hr * t2
18 mi = 4 mi/hr * t2
18 mi/(4 mi/hr) = t2
4.5 hr = t2

Hope that helped,

Tatiana

As I was doing this problem, I usually am confused whenever to add the 3 or subtract it from time. For instance, wouldn't 6t = 10(t-3) be mathematically equivalent b/c the second cyclist travels 3 hours after the first cyclist?

but the answer obviously yields a different result. Is the rule of thumb to add the time instead of subtracting it?

Sorry if this is a dumb question.

Hi islands,

Not a dumb question at all!! It doesn't matter whether you add 3 hours to one side or subtract 3 from the other. If you do the work properly, the answer will be the same. The important thing is that the object that traveled longer have the longer time. So we could have t for the shorter time and t+3 for the longer time, or we could have t-3 for the shorter time and t for the longer time.

Just know what you're solving for. If you want the shorter time, you would be solving for t in the first case, but you would be solving for t-3 in the 2nd case.

If you have trouble with advanced Rate questions, use the Drill Generator to create and take timed drills, and set topics='Rates & Work' and difficulty='600-700 & 700+'

Hope that helps,
-Patrick