Alan, Bruce, Colin and David each began a game of poker with $7 made up solely of 20 cent pieces. At the end of the game, they each added up their coins. They found that Alan and David together had as many coins as Bruce and Colin together. Moreover, Alan had 2 coins more than Colin and David together, while Bruce and David together had 8 coins more than the other two together. What was each player's profit or loss from the game? Thanks for help.
Let Alan =A, Bruce=B, Colin=C, David =D
Because each player began with $7 in the form of 35 coins, the first clue can be expressed as:
1 - A+B+C+D=140. The remaining clues can be expressed similarly:
2- A+D=B+C
3- A=C+D+2
4- B+D=A+C+8
From 3 above we get =C+D=A-2, which when substituted into 1 yields:
5- 2A + B=142, From 2 above we get: D=B+C-A, which when substituted into 4 above, yields:
6- 2B-2A=8, solving 5 and 6 above together yields:A=46, B=50.
Substituting A into 3 above yields:
7 - C+D =44, and substituting A and B into 4 above yields:
8 -D - C =4, solving 7 and 8 together we get:
C=20 and D=24
Expressed in dollars, we get:
A=$9.20, B=$10, C=$4 and D=$4.80 as the money each player had left after the game.
P.S. Somebody should check my "Algebra!."
+129840 The had a total of $28/.20 = 140 coins...so...
A + B + C + D = 140 ( 1)
A + D = B + C → A + D - B - C = 0 (2)
A - C - D = 2 → A - 2 = C + D (3)
B + D - 8 = A + C (4)
Add (1) and (2) → 2A + 2D = 140 → A + D = 70 → D = 70 - A (5)
Put (5) into (3) → A - 2 = C + ( 70 - A) → 2A - 72 = C (6)
Put (5) and (6) into (4)
B + [70 - A] - 8 = A + [2A - 72] → B = 4A - 134
So we have
A + B + C + D = 140
A + [4A - 134] + [2A - 72] + [70 - A] = 140
6A = 276 → A = 46 → $9.20 = $2.20 profit
D = 70 - 46 = 24 → $4.80 = $2.20 loss
C = 2(46) - 72 = 20 → $4.00 → $3.00 loss
B = 4(46) - 134 = 50 → $10.00 → $3.00 profit
Notice that this is a "zero sum" game with regard to profit and losses...just as we would expect !!!!
