SOLUTION: there are 3 males A,B,C and 3 females X,Y,W. they played 18 games of golf altogether. 1. A scored 94 2. X scored 106 3. Y scored 102. 4. Z scored 100. 5. B and C scored 96 an

Question 900953: there are 3 males A,B,C and 3 females X,Y,W. they played 18 games of
golf altogether.
1. A scored 94
2. X scored 106
3. Y scored 102.
4. Z scored 100.
5. B and C scored 96 and 98 and don't no who's score what??
6. A's wife beats C's wife.
7. there are two couples whose sum of scores is same.
Determine who's wife is who and scores of B and C.

Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
there are 3 males A,B,C and 3 females X,Y,W. they played 18 games of
golf altogether.
1. A scored 94
2. X scored 106
3. Y scored 102.
4. Z scored 100.
5. B and C scored 96 and 98 and don't no who's score what??
6. A's wife beats C's wife.
7. there are two couples whose sum of scores is same.
Determine who's wife is who and scores of B and C.

From Clue #6, we know that A's wife is not Z and C's wife is not X.
From Clue #7, we see that the choices for pairs that have the same sum are:
{100,96}and{102,94} OR {102,98}and{106,94}. Both of these contain A (94).
We know that A scored 94 (from Clue #1). Therefore A's wife could be Y, who scored 102, or she could be X, who scored 106. Use a grid like a tic-tac-toe board to set this up for exclusions (I don't know how to do that on this forum).

Let's look at each scenario, the pair {100,96}and{102(Y),94(A)} first.
If A's wife is Y, then one of the other pairs is {100,96}.
We already know that C's wife is not X. Therefore, with this pair, B's wife must be X. If that is so, then C's wife must be Z.
This gives us the pairs {94(A),102(Y)}, {98(B),106(X)}, and {96(C),100(Z)}.
Does this fit with all the clues? If so, we don't need to look at the other scenario.
6. A's wife beats C's wife. Yes, A's wife scored 102 and C's wife scored 100.
7. there are two couples whose sum of scores is same. Yes, A+Y=196 and C+Z=196.

(You can look at the other possibility, just for fun, to see whether there is more than one solution.)
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