SOLUTION: The members of a flying club plan to share
equally the cost of a $200,000 airplane. The members
want to find five more people to join the club so that the
cost per person will d
Algebra.Com
Question 268715: The members of a flying club plan to share
equally the cost of a $200,000 airplane. The members
want to find five more people to join the club so that the
cost per person will decrease by $2000. How many members
are currently in the club?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The members of a flying club plan to share equally the cost of a $200,000 airplane.
The members want to find five more people to join the club so that the cost per person will decrease by $2000.
How many members are currently in the club?
:
Let x = no. of members currently
:
Let's do this in $1000's to avoid writing all these 0's; Plane cost 200, members want to save 2
:
Present cost per member:
Cost with 5 more members:
:
Present cost - cost with 5 more = 2
- = 2
Multiply equation by x(x+5), results
:
200(x+5) - 200x = 2x(x+5)
:
200x + 1000 - 200x = 2x^2 + 10x
:
1000 = 2x^2 + 10x
;
A quadratic equation
2x^2 + 10x - 1000 = 0
;
Simplify, divide by 2
x^2 + 5x - 500
;
Factor
(x+25)(x-20) = 0
:
Positive solution:
x = 20 members now
;
:
Check this by finding the actual cost per member
200000/20 = $10000
With 5 more
200000/25 = $8000, 2 thousand less
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