|
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(22740)   (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
|
|
|
| |
