SOLUTION: I am trying to find the formula to use in order to solve this word problem.
The members of a flying club plan to share
equally the cost of a $200,000 airplane. The members
wan
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Question 286423: I am trying to find the formula to use in order to solve this word problem.
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?
Found 4 solutions by stanbon, richwmiller, josmiceli, unlockmath:
Answer by stanbon(75887) (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 the # of members be x.
Their cost per person is 200,000/x
------------------------------------------
Increase # of members by 5 so you have x+5 members.
Then cost per person is 200,000/(x+5)
-----------------------------------------
Equation:
original cost per person - new cost per person = 2000
(200,000/x) - (200,000/(x+5)) = 2000
-----
Solve for x to find the original number of members.
==========================================================
Cheers,
Stan H.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
200000/x=y
200000/(x+5)=y-2000
x=20 members now
y=10000
25 members would have a cost of 8000 each
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You need to get 2 formulas from this info
Let = number of members currently in the club
Let = cost per person
given:
The current cost per person is:
also:
This is 2 equations and 2 unknowns
so it's solvable
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
We'll set up "x" as the number of members and "y" as the cost. From this we can set up 2 equations like this:
y=200,000/x
y-2000= 200,000/(x+5)
Make sense so Far?
OK, now plug the first equation into the second like this:
200,000/(x)-2000=200,000/(x+5)
Get rid of the fractions by multiplying by (x)(x+5)to get:
200,000(x+5)-2000(x)(x+5)=200,000x Expand this out to be:
200,000x + 1,000,000 - 2000x^2 - 10,000x = 200,000x
Subtract 200,000x from both sides to get:
-2000x^2 - 10,000x + 1,000,000 = 0
Take out -2000 to be:
-2000(x^2+5x-500)=0 Factor this to get:
-2000(x-20)(x+25)=0
Now we know that :
x=-25
x=20
20 is the only reasonable answer so the number of members is:
20 members.
You can check this out by dividing $200,000 by 20 members which is $10000 each has to pay. Adding 5 more members: divide $200,000 by 25 and we get $8000 each member pays. So it works out.
Got it?
Good problem.
RJ
Check out a book I wrote:
www.math-unlock.com
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