# SOLUTION: Can someone please help me with this "setting fares" word problem? a bus company has 3,000 passengers daily, paying a 25 cent fare. For each nickel increase in fare, the company

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 Click here to see ALL problems on Money Word Problems Question 211589This question is from textbook : Can someone please help me with this "setting fares" word problem? a bus company has 3,000 passengers daily, paying a 25 cent fare. For each nickel increase in fare, the company projects that it will lose 80 passengers. What fare increase will produce \$994 in daily revenue? Please show me all the steps and get back to me before Tues 10am. Thanks so much This question is from textbook Answer by ankor@dixie-net.com(15649)   (Show Source): You can put this solution on YOUR website!a bus company has 3,000 passengers daily, paying a 25 cent fare. For each nickel increase in fare, the company projects that it will lose 80 passengers. What fare increase will produce \$994 in daily revenue? : Use the given information to make two sets of coordinates Let x = no. of .05 increases above the .25 fare, and x = no. of 80 passenger decrease from 3000 let y = daily revenue : The equation: Revenue = fare * no. of passengers y = (.25 + .05x) * (3000 - 80x) FOIL 750 - 20x + 150x - 4x^2 = y : -4x^2 + 130x + 750 = y : "What fare increase will produce \$994 in daily revenue?+ y = 994, find x -4x^2 + 130x + 750 = 994 : -4x^2 + 130x + 750 - 994 = 0 : -4x^2 + 130x - 244 = 0 Simplify, divide by -2 2x^2 - 65x + 122 = 0 Factor (2x - 61) (x - 2) = 0 The reasonable solution x = 2 ea .05 increases (10 cents) in fare will produce \$994 revenue : : Check solution in the original equation y = (.25 + .05(2)) * (3000 - 80(2)) y = (.25 + .10) * (3000 - 160) y = .35 * 2840 y = \$994