Question
How do you solve word problems dealing with algebra equations? What is the procedure to solve these can of problems? Where do you start? ex. A bus company in a small town has an average of riders of 800 per day. The bus company charges $2.25 for a ride. They conducted a survey of their customers and found that they will lose approximately 40 customers per day for each $.25 increase in fare.
Where will I start to even figure out the problem?
Answer
<P>Hmmm...I am surprised that you did not learn this sort of thing in middle school, or at least high school, let alone college. Your professor / teaching assistant is no help? Exactly where are you attending college? Let the rest of the world know so they can avoid the place.<BR><BR>800 riders per day * (2.25 dollars per rider) = 1800 dollars<BR><BR>The fare collected per rider = 2.25 + 0.25x (Each 25 cent increase in fare -- can take on values of 2.25, <BR>2.50, 2.75, etc.)<BR>The riders per day = 800 - 40x (For each increment the passengers decrease by 40) <BR><BR>total fare = (800 - 40x)*(2.25 + 0.25x) = 1800 + 200x - 90x - 10x^2 <BR> = 1800 + 110x - 10x^2<BR><BR>Joe McCollum<BR>FIA<BR>Knoxville, TN<BR><BR><BR><BR><BR></P>
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