Ok - here we go again. It's a bit of an old one but have fun anyway 
Three people check into a hotel. They pay £30 to the manager and go to their room. The manager suddenly remembers that the room rate is £25 and gives
£5 to the bellboy to return to the people. On the way to the room the bellboy reasons that £5 would be difficult to share among three people so he
pockets £2 and gives £1 to each person. Now each person paid £10 and got back £1. So they paid £9 each, totalling £27. The bellboy has £2, totalling
£29. Where is the missing £1?
£30 - £2(bellboy) = £28, not £27 so they actually spent £9.33 each on the room.
took me a couple of minutes 
[Edited on 8/8/05 by ned]
£27 paid less £2 to the bellboy = £25 to the hotel. 
Manager and secutary have run off with the money so the hotel gets nowt!!
quote:
Originally posted by ned
£30 - £2(bellboy) = £28, not £27 so they actually spent £9.33 each on the room.
took me a couple of minutes
[Edited on 8/8/05 by ned]
Make sure that you add together the right amounts!
Originally, they paid £30, they each received back £1, thus they now have only paid £27. Of this £27, £25 went to the manager for the room and £2 went
to the bellboy.
DJ
nearly got ya - pete's is right though
taken all day for someone to question it lol
quote:
Originally posted by ned
nearly got ya - pete's is right though
taken all day for someone to question it lol
OK - that was far to easy so try this little beauty
The warden of the Arizona State Penitentiary decides to have some fun one day... he decides that he will give three prisoners a chance to go free...
but only if they can prove themselves worthy of being released!
The warden brings out five hats, three of them red and two of them blue. The three prisoners are then blindfolded and one of the hats, at random, is
placed on each of their heads.
He lines up each prisoner in a row, facing a brick wall. The blindfolds are then removed. Prisoner C can see both Prisoners A and B and the color hats
that each of them are wearing. Prisoner B can only see Prisoner A and his hat. Prisoner A, being in the front, cannot see any hats. And no prisoner
can, of course, see the color of his own hat.
============= wall
. A
. B
. C
The problem he puts to the prisoners is simple:
"If you can determine the color of the hat on your own head, I will release you from this prison. If you answer incorrectly, I will add another
20 years to your sentence! However, if you do not know and choose not to answer at all, nothing will come of it."
He first asks Prisoner C, "What is the color of the hat on your head?" Prisoner C thinks for a short while and then replies, "I'm
sorry, I do not know."
He then asks Prisoner B, "What is the color of the hat on your head?" Prisoner B thinks for a short while and then also replies,
"I'm sorry, I do not know."
He finally asks Prisoner A, "What is the color of the hat on your head?" Prisoner A thinks for a short while and then replies, "I
know!" The color of my hat is...."
What color hat is Prisoner A wearing and how did he know?
[Edited on 8/8/05 by DaveFJ]
He can correctly deduce he's wearing a red cap if he's confident enough in both his mate's logic abilities to risk 20 years...
to expand, since C know, you could assume that A and B werent both wearing blue hats (in this case, his would've had to have been red). Since B
didnt know, A must have been wearing red, since if he had had a blue hat on, B would've been able to deduce that his own hat was red, as C didnt
know his one colour. Hence, A must have been wearing red in this case.
Imagine how badly pissed you would be if you were proved wrong cos one of your idiot pals couldn't work it out?!
Anyone I know quick enough to sort that out under pressure also have the sense of humour to get me another 20 years for a laugh 
! I'd keep my
mouth shut and finish my sentence ...
hang on!
if "c" sees a blue on "a" and a red on "b"
he wouldnt know what his colour is as there are still 2 reds and 1 blue unknown to him
"b" would see a blue on "a"
he wouldnt know what his colour is as there are still 2 reds and 1 blue unknown to him
"a" could not know either
but ........
if "c" sees a red on "a" and a blue on "b"
he wouldnt know what his colour is as there are still 2 reds and 1 blue unknown to him
"b" would see a red on "a"
he wouldnt know what his colour is as there are still 2 reds and 1 blue unknown to him
"a" could not know either
............
help!
If C sees either a red and a blue or two reds, he's stuck.
If B sees a red, that means his hat could be either blue or red, so he's jammed.
So A has to be red.
Of course, in real life, chances are about 50-50 that cocksure A would take off his hat only to find that it's blue!!!
Is he standing in front of a mirror?
the key is that, individually, only C has a chance of working it out (2 blues), but if he doesnt know, that tells his mates something.
C sees bb > C is red - B and A would then know they were both blue
C sees rb > C unknown, B then sees blue, and knows he must be red
C sees br > C unknown, B sees red, but could still be red or blue himself
C sees rr > C unknown, B sees red, again he cant know for sure.
Hence, when A hears that neither B or C could work it out, only the last two paths apply, and in both he is wearing a RED hat. 
cheers
i think !
That was a good one....
Thanks
My brain is sore from to much thinking.
so what's the official answer
Is it Thursday or Bobby Charlton?