Lost Wages, Nevada
Three men and their wives recently vacationed in Las Vegas. Each person gambled by themselves ... they agreed to stop whenever each couple's gain or loss reached two hundred dollars. All the men lost money, but as a couple, each won exactly $200. Bernard lost $504 more than Harry. Betty won $2,376 more than Vera. Oswald was glad his wife won. Belinda laughed at her husband for losing at what she called "these easy games".
Can you correctly identify the couples? ... and tell how much money each person won or lost?
Can you correctly identify the couples? ... and tell how much money each person won or lost?
posted by Zaux at
9:20 PM
19 Comments:
Bernard Belinda
Harry Betty
Oswald Vera
nope ... not right
also added the question as to how much each person won or lost?
Oswald -2900 & Betty +3100
Harry - 20 & Belinda + 220
Bernard - 524 & Vera + 724
?
This one is hard ;.;
All I've succeeded in doing is having eliminated these combinations:
Bernard and Vera
Oswald and Belinda
Harry and Betty
and
Bernard and Betty
Oswald and Belinda
Harry and Vera
Ultimately, Oswald is not with Belinda. But now I'm stuck -_-
How about:
Oswald lost $2377
Betty won $2577
Henry lost $1
Vera won $201
Bernard lost $505
Belinda won %705
no correct answer yet
Johnny ...
it's Harry, not Henry
but you got the couples right, however the money is wrong
From what I can tell, there could be an infinite number of money options that could meet these criteria ...
Hi Ekatahuna,
I agree. Your answer of:
Oswald-2900 Betty+3100
Harry-20 Belinda+220
Bernard-524 Vera+724
Works. It fits all criteria. From the math I've done(which is a lot ;.;) any combination not involving Oswald and Belinda has numbers that works.
Not to mention, that is not the only possible combination of numbers for that combo.
Oswald-2899 Betty+3101
Harry-19 Belinda+221
Bernard-523 Vera+725
That works too.
Hate to say this Zaux, but I think you made a mistake somewhere :(
Wow. Just reread the "answer" I posted. woops.
Epic fail, I apologize.
However, I stand firm on there are a number of different solutions.
Oswald-2901 Betty+3101
Harry-21 Belinda+221
Bernard-525 Vera+725
There we go.
Again, my fail.
I look forward to seeing the answer tomorrow ... I'm off to force down a few pints of Guinness, to be sure, to be sure :-)
This is a rarity ... a puzzle not solved by the ToM wizards.
Published Solution:
Oswald (lost $2401) is married to Betty (won $2601)
Bernard (lost $529) is married to Belinda (won $729)
Harry (lost $25) is married to Vera (won $225)
It looks like Johnny worked it out like I did. So I checked his solution, and his worked also.
You can pick out any lost for Harry and his wife wins 200 more than he lost.
Let's pick -1000 for Harry
Oswald lost......-3376 Betty won.....3576 Sum 200
Harry lost ......-1000 Vera won......1200 Sum 200
Bernard lost.....-1504 Belinda won...1704 Sum 200
There seems to something in the ToM missing, that would tie this up a little tighter.
Hi Ragknot ...
the problem did state that the couples agreed to stop when each couple had lost $200 ... that would indicate the lowest figures which satisfy the conditions ... not just any amounts which work ... right? ...
or am I looking at it incorrectly?
Zaux,
You are right.
I knew something was missing.
But I was wrong about it missing from the Tom.
I can't think of a way to tell when a couple might have first reached $200.
I would expect it would be the minimum that satisfied the equations I constructed , but those are the same that Johnny submitted.
My equations are as follows
Harry $ x
Vera $ 200-x
Bernard $ x-504
Belinda $ 200 - (x-504)
Betty $ 200-x+2376
Oswald $200- (200-x+2376)
If you let x be -25 then you get the "published" answers. If the logic is minimum then x should be -1 (if each guy must have a loss)
If x is -25 then $6510 must have been won-lost.
If x is -1 then only $6366 was won-lost.
Something is wrong with this.
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