Question 429237: A doctor drove 200 miles to attend a conference. Bad weather on the return trip caused her average speed to be 10 mph less than it was going to the convention. If the return trip took 1 hour longer, how fast did she drive in both directions?
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! Let X represent the doctor's original speed. Then his time taken to attend the conference is:200/x. Likewise, his return trip took 200/x-10 hours. So, our equation looks like:
200/X + 1=200/X-10
Multiplying through by (X)(X-10), we get:
200X-2000+X^2-10X=200X
X^2-10X-2000=0
(X-50)(X+40)=0
Solving for X, and tossing out the negative result (-40), we get his speed going there to be 50 mph. His speed coming back was 50-10 mph, or 40 mph.
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