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#46
October 5th, 2005, 05:25 PM
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Visual aid for the solution provided by elijathegold
Step) Boat bank Side people 1)L L MMM CCC R 2)R L MMM C R CC 3)L L MMM CC R C 4)R L MMM R CCC 5)L L MMM C R CC 6)R L M C R MM CC 7)L L MM CC R M C 8)R L CC R MMM C 9)L L CC R MMM C 10)R L R MMM CCC Hope that helps someone 
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#47
October 5th, 2005, 05:52 PM
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Quote:
Erm... Not really... 
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#48
October 5th, 2005, 07:51 PM
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Here's a classic I'm sure some of you know: ********* Three friends walk into a hotel, and agree to share a room. They split the $30 cost by each paying $10. After they are on their way to the room, the cashier realizes it's discount day and the room is only $25. He gives the bellhop $5, in ones, to return to the members of the room. The bellhop, not knowing how to split up the money, gives $1 to each of them, and keeps $2 for himself. We know that each roommember paid $9, totalling $27, and the bellhop had the other $2. Where did the extra dollar go? *********
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#49
October 6th, 2005, 01:30 AM
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<Comment>On Friday I'd know it'd have to be that day because Saturday is already ruled out and therefore Friday will be ruled out.</Comment> How can he rule out Saturday because the present day is Friday and there are chances to be hanged today (Friday) or the next day(Saturday). Like this for all other cases. Am I on the right track gecko36........ Rajesh Sivaraman.
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#50
October 6th, 2005, 04:01 AM
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There is no extra dollar Cashier $25 Bellhop $2 Guests $3 This totals $30. But the prisonner is killing me too
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And he picked it all up... in his pick-up.  Friends of Shemzilla
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#51
October 6th, 2005, 08:22 AM
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etxra dollard ,lol. i remember that one from years ago
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#52
October 6th, 2005, 10:57 AM
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Here'z my riddle...
A 40Kg weight (one they use in balances) was broken into four pieces, such that you can make measurements from 1 to 40kg. If so how much did each piece weigh.
Hint: Suppose you have 3 and 4kg weights you can measure 1kg with the weights in opposite pans. Cheers!!! Rajesh Sivaraman.
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#53
October 6th, 2005, 03:43 PM
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now thats intressting. im not sure if i understand what you saying with the 3 and 4 kg weght. but this is what im thinking let me know if im on the right track. 1 of the weights has to be 1kg none of them weight 40kg , 39kg im sure this is not the right track. let me know
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#54
October 7th, 2005, 01:02 AM
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If two of them are of 3 and 4 Kgs then by placing them on two different sides of the weighing scale you will be able to measure 1 kg by keeping the material on the 3 kg side of the pan. ie; when the scales become equal if one kg material comes on the 3kg side of the pan. Hope this clears your doubt. Like that if the weights are of 2, 5 and 6 kgs then you can measure 9 kgs by 6 + 5 - 2. Cheers!!! Rajesh Sivaraman.
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#55
October 8th, 2005, 10:42 AM
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First a hint... "you won't be able to deduce what day you'll die" translates into, to say the least, "you won't die on Saturday because that's the last possible day to die" Take a calandar and start with Saturday. It's kind of like probability. There, I've said too much. Right track but here's his logic in probably a more understandable format. Saturday can't be the day because if he survives until Saturday (i.e. not being killed on subsequent days) through process of elimination that'd be the day. Friday was eliminated because Saturday was eliminated and Friday was the only day left. (it doesn't hurt to grant those two!) Thursday, like Friday, includes two eliminated days and he uses the process of elimination to eliminated Thursday. Logic builds in that fashion until Sunday and he concludes that it's impossible Well, with that hint I'll leave another I guess. Seeing as how I'm so kind...  Each day's logical test is seperate. Saturday and Friday have been eliminated. However if you build on Friday's logic for Thursday the probability changes so the logic must change. Not the logic itself (you can use Friday's logic like he did just fine) but the application of the logic (you can't just cancel the next days).
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#56
October 10th, 2005, 01:45 AM
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So gecko36 the solutions which I gave is right or wrong...  Rajesh Sivaraman.
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#57
October 10th, 2005, 02:29 PM
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you asked how can he rule Sat. out. That's okay to do that. IF he survived until Sat. he'd live because it's the last day. Therefore on Fri. he could rule out Fri. because if Sat. is always ruled out, Fri. would be the last day. On Thurs. Fri. can't be ruled out because Fri. was only ruled out because Sat was ruled out. The problem is people are inclined to think ahead of themselves. (things to pick from) nPr* (total items) Sat is out! (0/0) 0 = nPr 1 Fri probability is 1 = 1 nPr 1 Thurs probability is 2 = 2 nPr 1 Wed. probability is 3 = 3 nPr 1 Tues. probability is 4 = 4 nPr 1 Mon. probability is 5 = 5 nPr 1 Sun probability is 6 = 6 nPr 1 Or just do n! where n is the days to pick from. The numbers aren't important, the fact that Thurs. thru Sun. are >1 then it's like programming  IF day > 1 Then die
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