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Mad Ade has found himself at a party. There are 120 people at the party and 3 fifths of them are women. Also, 2 thirds of the 120 people are married. How many unmarried women could Mad Ade possibly meet at the party?
Answer
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What am I missing? I figure there are 72 women at the party (3/5 of 120) and 80 married people (2/3 of 120) which I assume means 40 women are married. That leaves me with 32 single women - did I misunderstand something or do I need to actually get out the calculator?
You did it the wrong way around. 80 people are married, 40 people are unmarried. Assume as many women as possible are unmarried. If there were only 40 women at the party, its possible they are the fourty unmarried people. If there are 72 women assume the 40 are still unmarried women, and 32 are married...
ok, so are gay marriages allowed in britain? in the united states (except vermont, i think), the answer would be 32.
May 23, 2002
The puzzle doesn't say anywhere that the people at the party that are married are married to other people at the party.
the information given is not enough for a definitive answer. it could be anything between 32 & 40 as we are not told how many, even as a % of the married are women or vica verca. maybe this should have been under a subject heading other than math, such as probability or logic?
I believe the answer is fine. The final question simply says "How many unmarried women could Mad Ade possibly meet at the party?". Well, we know that there are 40 unmarried and we know that 72 are women, therefore it is a possibility that all 40 of the unmarried people are women therefore '40' is the amount of unmarried women Mad Ade could 'possibly' meet at the party?".
Basically the first clue is just a distraction.
No, the first clue's not a distraction. What's being asked, essentially, is the total possible single women at the party, given the two restrictions. I figured both 32 and 40 also. I went with 32, figuring the clues implied that the marrieds were there as couples. Since that's not stated, the 40 is correct. As for the possibility of there being 80 married men at a party with 40 unmarried women, I'd venture to guess that those 80 may not all be married for much longer!
Well, I'll go with 40 being the correct answer. This is not much of a math problem, more of a logic or trick teaser.
I don't see a trick. All I see is mental arithmetic. How many unmarried women COULD he meet implies what is the maximum number. I suppose it might also imply a range from a minimum to a maximum.
Anyway I interpreted it as the maximum number so I assumed that all of the men were married (doesn't really matter if they are married to some women at the party as long as it's not all 48 of them). There are 80 married people so that leaves 32 women who must be married out of the 72 women present. maximum of 40 married.
Since there are only 72 women an 80 married people it could be that all of the women are married in which case the answer is between 0 and 40. I'm happy with 40.
Anyway I interpreted it as the maximum number so I assumed that all of the men were married (doesn't really matter if they are married to some women at the party as long as it's not all 48 of them). There are 80 married people so that leaves 32 women who must be married out of the 72 women present. maximum of 40 married.
Since there are only 72 women an 80 married people it could be that all of the women are married in which case the answer is between 0 and 40. I'm happy with 40.
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