SOLUTION: can you please help me solve this problem? and thank you. A number of persons agreed to raise $1200, but 5 of them failed to pay their share. As a result, each of the other had to

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: can you please help me solve this problem? and thank you. A number of persons agreed to raise $1200, but 5 of them failed to pay their share. As a result, each of the other had to       Log On


   

Question 356730: can you please help me solve this problem? and thank you. A number of persons agreed to raise $1200, but 5 of them failed to pay their share. As a result, each of the other had to pay $1 more. How many were in the party?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A number of persons agreed to raise $1200, but 5 of them failed to pay their share.
As a result, each of the other had to pay $1 more.
How many were in the party?
:
Let x = number originally in the party
then
(x-5) = actual number that paid
:
1200%2Fx = original amt each should pay
And
1200%2F%28%28x-5%29%29 = actual amt paid by each
:
Original amt for each = actual amt for each - $1
1200%2Fx = 1200%2F%28%28x-5%29%29 - 1
Multiply by x(x-5), results
1200(x-5) = 1200x - x(x-5)
:
1200x - 6000 = 1200x - x^2 + 5x
Arrange as a quadratic on the left
x^2 - 5x + 1200x - 1200x - 6000 = 0
:
x^2 - 5x - 6000 = 0
factor to
(x-80)(x+75) = 0
x = 80 people originally in the group
:
:
See if that's true, (75 people actually paid)
1200/75 = $16
1200/80 = $15