Posted by **Amanda** on Monday, February 16, 2009 at 5:02pm.

a bus company has 4000 passengers daily, each paying a fare of $2. for each $0.15 increase, the company estimates that it will lose 40 passengers per day. if the company needs to take in 10 450$ per day to stay in business, what fare should be charged?

- Math -
**Damon**, Monday, February 16, 2009 at 5:26pm
increase fare by n 15 cent units

new fare = (2.00 + .15 n)

new passengers = (4000 - 40 n)

so

10450 = (2.00+.15n)(4000-40n)

10450 = 8000 + 520 n - 6 n^2

so

6 n^2 - 520 n + 2450 = 0

3 n^2 - 260 n + 1225 = 0

n = [260 +/- sqrt(67600-14700)]/6

n = [260 +/- sqrt(52900)]/6

n = [260 +/- 230]/6

n = [30]/6 or [490]/6

n = 5 or 81 2/3

well

2.00 + .15(5) = $2.75

which is a reasonable fare

however

2.00 +.15(81 2/3) = 14.25

and 4000 - 40 (81 2/3) = 733 passengers

Well, I guess not many people would ride for $14.25 per day so we better stick with the 75 cent fare increase

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