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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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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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
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
