In a chess tournament in which there were no draws, every pa

Question

In a chess tournament in which there were no draws, every participant played every other exactly once, and 4 of the 16 participants were women. If women won \(3/10\) of all matches played, what fraction of the matches they played against men did they win?

A. \(3/5\)
B. \(5/8\)
C. \(13/20\)
D. \(2/3\)
E. \(3/4\)

Show Answer

Official Answer

B

A.  3/5 
B.  5/8 
C.  13/20 
D.  2/3 
E.  3/4

paragw · Answer

Here's my solution

findingmyself · Answer

I am getting 3/4 as the answer.

Total matches: 16*15/2=120
Matches won= 3/10 *120=36

Matches between 4 women and 12 men, each woman plays with 12 men, total match =48
Matches between 4 women among themselves =4*3/2=6

What fraction of the matches they played against men did they win?
For me, matches playes against men is 48
Matches won overall is 36, which includes Matches won against men and matches won against women

paragw : Can you explain why you subtracted 6 from 36

SuchTask · Answer

The question asks for fraction of matches women won vs. men.

So you need to subtract the number of wins women had vs. themselves (6) from the total matches they need to have won (36) to get wins vs. men (30).

Total wins = 36
Wins vs. women = 6
Wins vs. men = 30
Matches vs. men = 48

Fraction of matches won vs. men = 30/48 = 5/8

paragw · Answer

As you've mentioned in your solution that "Matches between 4 women among themselves =4*3/2=6".
In these matches, it doesn't matter who wins, all these matches are already counted in the total number of victories of Women (because all are Women).
And total number matches won by Women is 36.
Therefore, the matches Women played against men and won will be 36 - 6 = 30.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=89632 - There are 4 women and 12 men. - Total # matches = number of ways of selecting 2 from 16 people = 16C2 = 120 - Total # of matches which only feature men = number of ways of selecting 2 men from 12 men = 12C2 = 66 - Total # of matches which feature only women = number of ways of selecting 2 women from 4 women = 4C2 = 6 - Total # of matches between a man and a woman = 120 - (66+6) OR (4 x 12) = 48 Women won 3/10 of the total matches i.e., 3/10 of 120 = 36 matches. - 6 of these wins are from the women-only matches - None of these 36 wins could come from the 66 men-men matches - So, the rest of the female wins come from men-women matches. The number is 36 - 6 = 30. GMAT-Club-Forum-5f1k0jae.png

egmat · Answer

Hi findingmyself,

Your setup is spot on. Total matches  120 , women's total wins  36 , women-vs-men matches  48 , and women-vs-women matches  6 . The only slip is in the last division.

Here's the key question:  what exactly does the  36  count? It's  every  match a woman won - and a woman can win in two different arenas:

- matches against  men 
- matches against  other women

Those women-vs-women matches matter because there are  no draws  - so in each of the  6  matches between two women, a woman  has  to win. That means  6  of your  36  wins were earned against women, not men.

So when you do  36/48 , you're crediting all  36  wins as if they came against men - but  6  of them didn't. To get wins against  men only , strip out the  6  automatic women-vs-women wins first:

- Wins against men =  36 - 6 = 30 
- Fraction =  30/48 = 5/8

That's why paragw subtracted the  6  - it removes the wins that belong to a different bucket. The denominator ( 48 ) is purely women-vs-men matches, so the numerator must be purely women-vs-men wins too. Matching the numerator's bucket to the denominator's bucket is the whole game here.

Quick gut-check to lock it in:  suppose women had won only  10  matches total. Six of those are guaranteed against women, so wins against men would be just  10 - 6 = 4 , not  10 . The  6  women-vs-women wins are fixed no matter what - always peel them off before comparing to the  48 .

Answer: B

In a chess tournament in which there were no draws, every pa

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