SOLUTION: This is driving me crazy. I finally solved my problem, but I can’t figure out why my ‘tweaking’ made a difference. The problem reads: “A bus traveled at an average rate of 50 m

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: This is driving me crazy. I finally solved my problem, but I can’t figure out why my ‘tweaking’ made a difference. The problem reads: “A bus traveled at an average rate of 50 m      Log On

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Question 290439:
This is driving me crazy. I finally solved my problem, but I can’t figure out why my ‘tweaking’ made a difference. The problem reads:
“A bus traveled at an average rate of 50 mph and then reduced its average rate to 40 mph for the rest of the trip. If the 220-mile trip took 5 hours, determine how long the bus traveled at each rate.”
So I set up my little table: (apparently there are spacing issues and I can't lay the table out here.) In short I used 'x' for the 'time' on the 50 mph leg of the trip, and '5-x' for the 'time' on the 40 mph leg of the trip.

And then I did the math:
50x + 40(5-x) = 220
50x + 200 -40x = 220
10x = 20
X = 10
I knew that couldn’t be right as the answer made no sense, but I couldn’t (and still can’t) figure out what I did wrong. I finally switched things around in my table changing the time of the 40mph leg to 'x' and the 50mph leg to '5-x'. It would be easier if I could show you my table...hopefull you understand what I'm trying to say.
All I did was switch the factors around in the “time” row and when I did the math that way the answer came out correctly. Why did that make a difference? I would have thought it would have worked either way. Or maybe I made my mistake the first time elsewhere? (I sure hope I copied everything correctly). I have a similar problem (except it involves interest rates and bank accounts) where the same thing happened. I thought as long as I made one account "x" and the other "Blah-x" it didn't matter which I used for which. Maybe I've just made simple math errors when I did them the first time?
Thanks a million...you guys rock!

Found 2 solutions by kensson, MathTherapy:
Answer by kensson(21) About Me  (Show Source):
You can put this solution on YOUR website!
I think you're going to kick yourself.
10x=20
x = 2.

Answer by MathTherapy(10549) About Me  (Show Source):
You can put this solution on YOUR website!
This is driving me crazy. I finally solved my problem, but I can’t figure out why my ‘tweaking’ made a difference. The problem reads:
“A bus traveled at an average rate of 50 mph and then reduced its average rate to 40 mph for the rest of the trip. If the 220-mile trip took 5 hours, determine how long the bus traveled at each rate.”

Let the time at 50 mph be T

Then the time at 40 mph = 5 - T, since the trip took 5 hours

We're now going to set up an equation based on the 2 distances. In other words:

Distance on 1st leg of trip + Distance on 2nd leg of trip = 220 miles.

Distance on 1st leg of trip = 50T, and distance on 2nd leg = 40(5 - T)

We now have 50T + 40(5 - T) = 220

50T + 200 - 40T = 220

10T = 20

T+=+20%2F10+=+2

This means that time traveled at 50 mph = highlight_green%282%29 hours, and time traveled at 40 mph = 5 - T, or 5 - 2 = highlight_green%283%29 hours.

You did everything correctly except when you got to: 10x = 20. This should give you a value of 2 for x, NOT 10.