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Question 644239: A group of women decided to contribute equal amounts toward obtaining a speaker for a book review. If there were 10 more women, each would have paid $2 less. However, if there were 5 less women, each would have paid $2 more. How many women were in the group and how much was the speaker paid?
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Interesting problem, but straight forward for you "to be" math wizards!
Let n = the number of women in the group, and
let d = $ amount of each one's donation, and
let A = the total $ amount donated to retain a speaker
Then
(1) A = n*d
Givens:
a) If we had 10 more women they would have each donated $2 less or
(2) A = (n+10)*(d-2)
b) If there were 5 less women in the group, they each need to donate $2 more or
(3) A = (n-5)*(d+2)
Our algebra problem is to solve (1), (2) and (3) simultaneously to find the numerical values of n and A.
Can you do it w/o my help? Try it.
Here's my solution:
Equate (2) to (3) and get
(4) 4n = 15d - 10
Equate (1) to (2) and get
(5) -2n + 10d = 20
Equate (1) to (3) and get
(6) 2n - 5d = 10
Now add (5) and (6) and get
(7) 5d = 30 or d = 6
Now let d = 6 in (4) to yield
(8) 4n = 15*6 - 10 or n = 20
Now put n and d in (1) to find
(9) A = 120
We must always (if possible) check our answer.
Check (2);
Is (A = (n+10)*(n-2))?
Is (120 = 30*4)?
Is (120 = 120)? Yes
Check (3);
Is (A = (n-5)*(d+2))?
Is (120 = 15*8)?
Is (120 = 120)? Yes
Answer: There are 20 women in the group that donated a total of $120 to retain a speaker for their book review.
Comment: In any problem with n unknowns requires n equations to obtain a solution. In this problem we need n and d, two unknowns, so we must come up with two independent (not the same equation) equations. These two equations are (2) and (3). I've experienced problems that require as many as 100 unknowns to be solved in "real time". Maybe someday you will also. Yes it requires a very high-speed computer.
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