SOLUTION: In 2005, the members of a club paid a total of $42 in subscription fees. In 2006, the number of members increased by 20 and the subscription fees were subsequently reduced by 10 ce
Question 401405: In 2005, the members of a club paid a total of $42 in subscription fees. In 2006, the number of members increased by 20 and the subscription fees were subsequently reduced by 10 cents per member. The total of the subscription fees for 2006 was $45. How many members were there in 2005?
:) =) Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! In 2005, the members of a club paid a total of $42 in subscription fees.
In 2006, the number of members increased by 20 and the subscription fees were
subsequently reduced by 10 cents per member.
The total of the subscription fees for 2006 was $45. How many members were there in 2005?
:
Let n = original no. of members in 2005
then
(n+20) = no. of members in 2006
:
2005 cost/member - 2006 cost/member = $.10 - = .10
:
multiply eq by n(n+2), results:
42(n+20) - 45n = .10n(n+20)
:
42n + 840 - 45n = .1n^2 + 2n
:
-3n + 840 = .1n^2 + 2n
;
Arrange as a quadratic equation
.1n^2 + 2n +3n - 840 = 0
:
.1n^2 + 5n - 840 = 0
:
Multiply by 10 to get rid of the decimal, results:
n^2 + 50n - 8400 = 0
:
you can use the quadratic formula, but this will factor to:
(n-70) (n+120) = 0
the positive solution
n = 70 members in 2005
:
:
Check this (90 members in 2006), find the cost per member in each year.
42/70 - 45/90 =
.60 - .50 = .10 reduction
