Question 123826: A bus company has 4000 passengers daily, each paying a fare of $2. For each $.015 increase, the company estimates that it will lose 40 passengers per day. If the company needs to take in $10,450 per day to stay in business, what fare should be charged?
The answer supposedly should be $2.75 ???
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A bus company has 4000 passengers daily, each paying a fare of $2. For each $.015 increase, the company estimates that it will lose 40 passengers per day. If the company needs to take in $10,450 per day to stay in business, what fare should be charged?
:
Let x = no. of .15 fare increases to attain a revenue of $10,450
:
No. of passengers: (4000-40x)
:
Fare charged: (2+.15x)
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The equation:
No. of pass * fare charge = $10450
:
(4000 - 40x) * (2 + .15x) = 10450
FOIL
8000 + 600x - 80x - 6x^2 = 10450
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Arrange as a quadratic equation
-6x^2 + 520x + 8000 - 10450 = 0
:
-6x^2 + 520x - 2450 = 0
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Simplify, divide equation by -2
+3x^2 - 260x + 1225 = 0
Factor this to:
(3x - 245) (x - 5) = 0
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x = +5 is the only reasonable solution
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That means five .15 increases
:
5 * .15 = .75 increase above $2 = $2.75
:
:
We can check this: 5 * 40 = 200 decrease in passengers, from 4000
3800 * 2.75 = 10450
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