help w/ "the card game": who was the biggest loser

K

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Please help me out guys!!

Alyssa, Mulunda, and Corina decided to play a game of cards. They agree on the following procedure:

  • When a player loses a game, she will double the amount of money that each of the other players already has.
First Alyssa loses a hand and doubles the amount of money that Mulund and Corina each have.

Then Mulunda loses a hand and doubles the amount of money that Alyssa and Corina each have.

Lastly, Xorina loses a hand and doubles the amount of money that Alyssa and Mulunda each have.

The three players then decide to quit, and they find that each player now has $8. Who is the biggest loser?

Thanks guys!
 
what i got by myself

here is what i did but somehow i think im wrong.

for the equations i got this

A=2(M+C)
M=2(A+C)
C=2(A+M)

but when i do some substitution i get negative numbers....is that even possible??

like this sub A for the second equation

M= 2A+2C and A=2M+2C-------> you get this M=2(2m+2c)+2c-----> M= 4m+6c

then you get this M-4M=6C......then -3m=6c then m=-2c.....is this even possible??? i think i have my first three equations wrong can someone plz help??

thanks guys.
 
What was the reasoning by which you developed your equations?

Thank you.

Eliz.
 
Alyssa, Mulunda, and Corina decided to play a game of cards. They agree on the following procedure:

When a player loses a game, she will double the amount of money that each of the other players already has.


First Alyssa loses a hand and doubles the amount of money that Mulund and Corina each have.

Then Mulunda loses a hand and doubles the amount of money that Alyssa and Corina each have.

Lastly, Corina loses a hand and doubles the amount of money that Alyssa and Mulunda each have.

The three players then decide to quit, and they find that each player now has $8. Who is the biggest loser?

==================

Rather than using "equations", I used "reasoning" as follows.

Consider the end of the game:

C=8, M=8, A=8

Since C loses the last game and doubles the amount of money each of the others had, then before the last hand of the game, M had to be 4 and A had to be 4. Thus, C a had to supply the additional money, 4 + 4 = 8. So, just before the last hand:

C=16, M=4, A=4

Continuing in this manner, since M loses the second hand and doubles the others' money, C had to be 8 and A had to be 2. So C had to provide 8+2 = 10. Therefore, before the second hand:

C=8, M=14, A=2

A loses the first hand and doubles the others' money, so A had to provide 4+7 =11. Therefore before the first hand we had:

C=4, M=7, A=13

Since they all end up with 8, A is the ONLY loser ($5) and each of the others wins ($4 for C and $1 for M).

Hope that helps...

Steve
 
Nice solution Steve!
IF you need to do it by setting up equations:
Code: 
start               a                     m                   c
game1           a - m - c                2m                  2c
game2         2a - 2m - 2c          -a + 3m - c              4c
game3         4a - 4m - 4c         -2a + 6m - 2c         -a - m + 7c
So now you need to solve the following system:
4a - 4m - 4c = 8
-2a + 6m - 2c = 8
-a - m + 7c = 8

....why, o why, use three exotic first names; why not simply: A, B and C ... :?:
 
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