How do I solve this rate problem?

The trick to this problem is realizing that you really have TWO simultaneous rate / time problems.

What we know is that Alex drove 220 miles at one rate and 80 miles at another rate (15 mph slower) for two different lengths of time that together added up to 6 hours.

We can express this with our two rates R and r and times T and t, where R and T represent the 220 mile leg, and r and t represent the 80 mile leg:

R * T + r * t = 300
R * T = 220
r * t = 80

Also:
Since T + t = 6, then t = 6 – T
and
R = r + 15

Substituting for t:
r * (6 – T) = 80
6r – Tr = 80

Substituting for r:
6 * (R – 15) – T * (R – 15) = 80

And completing the operations:
6R – 80 – TR – 15T = 80

Then substituting for R (since R = 220 / T):
6 * (220 / T) – 80 – T * (220 / T) – 15T = 80

And completing the operations:
1320 / T – 80 – 220 – 15T = 80
1320 / T – 380 – 15T = 0

Then multiplying both sides by T:
1320 – 380T – 15T^2 = 0

Multiplying both sides by -1 and rearranging terms:
15T^2 + 380T – 1320 = 0

And simplifying:
3T^2 + 76T – 264 = 0
T^2 + 26T – 88 = 0

Solving for the roots:
T + 22 = 0 (irrational: no one could drive for -22 hours)
or
T – 4 = 0

The time at the higher speed was 4 hours. The rest is easy to solve.

220 miles in 4 hours = 55 mph
80 miles in 2 hours = 40 mph

Great answer. I knew it had to be a problem with two variables and couldn’t come up with the correct equation. I ended up with the wrong quadratic equation twice, and even graphed it to look for the zeroes, but got nonsense answers. Thanks for solving it.

Thanks, @cockswain, but I made two silly mistakes in the calculations. (The answer is still correct, but my mistakes appear to have cancelled each other to enable that.)

When I attempted to complete the operations as shown:
Substituting for r:
6 * (R – 15) – T * (R – 15) = 80

And completing the operations:
6R – 80 – TR – 15T = 80

I should have had:
6R – 90 – TR – 15T = 80

So that later I’d have had:
1320 / T – 390 – 15T = 0

And still later:
3T^2 + 78T – 264 = 0
T^2 + 26T – 88 = 0 (I finally got it right again, because I made another mistake above.)

This was where I made the cancelling error, allowing me to inexplicably arrive at the right answer… from the wrong calculations. I must lead a charmed life.

All I can say in my defense is that I was tired, half drunk, up way past my bedtime and impatient as hell… and frustrated that it took me as long to resolve as it did.

Yeah, I always get disappointed in myself when I can’t solve these sorts of problems.

@CyanoticWasp Nice one.

@chelle21689 A few general thoughts that might be useful (if they aren’t, feel free to not use them).
If your rate problem involves two different “things” happening, start out writing them seperately.
Start out using as many variables as you need. Once you write out the basic equations, try to connect the variables. For example, in this problem, @CyanoticWasp started with 4 variables (R, r, T, t) but was able to connect R to r and T to t. This meant there was only 2 variables instead of 4.
If you have 2 variables, you need two equations. If you can’t find 2 equations, that means either you can connect your two variables or you haven’t found all the equations you need.

Possibly Confusing Side Note: I’ve written this as if “connecting variables” and “finding equations” are different. I think its helpful to think of it like that at first, but in reality they are the same thing. If you relate R to r as “r = R – 15” that is really an equation…

Let t be the time required to drive the first 220 miles. The the original speed is 220/t, so the speed during the rain is (220 / t) – 15 = (220 – 15t) / t. But the speed during the rain is also 80 / (6 – t), so (220 – 15t) / t = 80 /(6 – t) and, multiplying by t(6 – t), (220 – 15t)(6 – t) = 80t or 1320 – 220t – 90t + 15t^2 = 80t. Thus 1320 – 390t + 15t^2 = 0. Divide by 5 and rearrange: 3t^2 – 78t + 264 = 0 and (3t – 12)(t – 22) = 0. Since 22 > 6, 3t – 12 = 0 and t = 4.