raleigh wrote:This is a problem where it helps to plug in a value. Since the first part of the trip is x percent of the distance, choosing the distance of the trip to be 100 miles means that the distance traveled at the first rate will be x.
So let D = 100 miles.
Then she traveled x miles at 50 miles per hour. D = RT gives us that the time of this portion of the trip is x/50 hours.
The second part of the trip was 100-x miles at 60 miles per hour. D = RT gives us that this portion of the trip took (100-x)/60 hours.
Average rate is distance/time. The distance is 100 miles, and the time is x/50 + (100-x)/60.
Let's work out the denominator first. x/50 + (100-x)/60 = [6x + 5(100-x)]/300 = (500-x)/300. you missed one zero
So the average rate is 100/[(500-x)/300] = 100*300/(500-x) = 30000/(500-x)
So the numerator of the average rate is 30,000. The trick is to plug in the appropriate distance. We choose 100 because they mention x percent of the total distance or (x/100)*distance.
Thanks raleigh, but I think you made some mistakes in the process, do it again.
Distance= 100
Time 1 = x/60 this is the first difference btw me and your approach.
Time 2= 100-x / 50
Ave = total distance / total time
Ave = 100/ x/60 + 100-x/50 = 100/ 50x + 6000 - 60x / 3000 this is the second difference.
Ave = 100/ 6000 - 10x / 3000 = 100 * 3000 / 6000 - 10x
= 300,000/ 6000 - 10x ###
so my answer will be 300,000 which is not correct .. why ?