Brain teaser thread - Page 8

Tuberski · October 24, 2002, 17:52

It has to do with sales tax, they can chare you an extra penny in tax, but to the government they don't have to claim that penny.

Doesn't seem like a lot until you consider that McDonald's, for instance, is doing this to EVERY customer in EVERY store in the world that has a sales tax.

Sounds good to me anyway.

Who knows?

Dauphin · October 24, 2002, 20:36

Quote:

Originally posted by Immortal Wombat

I spend much of my Sundays pondering this. It struck me that if you price things just under an amount, people are likely to pay that amount. (eg. £4.99, people give you a fiver) you give a penny change.

This is a damn sight easier than pricing stuff at £5.01, and giving 99p change when they give you £6, or more likely, £4.99 change when they give you a tenner.

It is to do with change, think about what could happen in stores if things were priced in round pounds/dollars.

Oh, it has nothing to do with taxes.

Immortal Wombat · October 24, 2002, 20:40

er... Pretty much obsoletes the smaller denomination of coins. People would get change solely in round £/$, and will be unable (or at least unwilling) to use vending machines.

Dauphin · October 24, 2002, 20:51

You work at a till, or at least you did. Now assume you are an unscrupulous till operator. Someone wants to buy a £5 item, they give you a £5 note....

Immortal Wombat · October 24, 2002, 21:09

...I give them the item, take the money, and

oooh I see...

I think

Dauphin · October 24, 2002, 21:11

To give customer change will require the till to be opened - thus ensuring the recording of a transaction.

Immortal Wombat · October 24, 2002, 21:17

oooh, cunning.

Most till operators are too stupid to figure that out. They just lift it from the till.

Anyway, Paul Hanson's turn then?

Urban Ranger · October 25, 2002, 08:48

Did he flee?

Immortal Wombat · October 25, 2002, 09:48

^bump^

Urban Ranger · October 25, 2002, 10:42

Alright, since that bugger Paul Hanson has fled the scene, I'll fill in with a silly question.

How do you divide a square into seven equal parts?

Paul Hanson · October 25, 2002, 10:50

Oops, sorry. Didn`t realise. Not that I'd have come up with anything remotely taxing anyway, so it's probably beneficial to you all that I left when I did.

Immortal Wombat · October 25, 2002, 11:07

Divide the perimeter by 7, and draw lines to the middle.

100*100 square, cut the perimeter into 57 pixel lengths (ish...) and draw lines to the middle. Each segment has the same area.

Urban Ranger · October 25, 2002, 11:11

I suppose that can work, how about a simpler, more elegant way?

Immortal Wombat · October 25, 2002, 11:14

What's not simple? My way works for every number as well, not just 7

Urban Ranger · October 25, 2002, 11:22

Of course. So does the one I have in mind.

Az · October 25, 2002, 11:57

oh common , UR.

you take one of the sides of the square, and place perpendiculary 6 lines in equal distances from each other, and the other sides of the square.

I would draw it, but it's so plain that it sucks.

Immortal Wombat · October 25, 2002, 12:13

*slaps self*

Urban Ranger · October 25, 2002, 12:15

Quote:

Originally posted by Azazel
oh common , UR.

you take one of the sides of the square, and place perpendiculary 6 lines in equal distances from each other, and the other sides of the square.

I would draw it, but it's so plain that it sucks.

Dang you

Your turn for a question.

Az · October 25, 2002, 12:41

hmmmm.

I can't think of nothing out of the blue, so I'll skip this turn. actually, IW should ask the question, since he answered first.

Az · October 25, 2002, 12:56

ok, I think I managed to scrape something up.

take 3 squares and place them together like this:

Code:

[][]
[]

divide them into 4 shapes, equal in size and form.

Zero-Tau · October 25, 2002, 13:58

Like this:

Code:

[][]()()
[]{}{}()
/\{}
/\/\

Zero-Tau · October 25, 2002, 16:10

Since Azazel seems to have left (and because I know that my answer is correct

), here's the next riddle.

One day Ming was careless enough to leave the forums unattended for a while. When he came back, he found that a lot of new people had arrived. Of these, 63 were simply screaming for a ban.
There were no DLs among the 63.
All 63 fell into at least one, possibly two or three, of these categories: Threadjacker, flamer, spammer.
There was a total of 42 spammers.
There was a total of 20 threadjackers.
There was a total of 30 flamers.
There was 5 people who were both spammers, flamers, and threadjackers.
All people who were flamers and threadjackers were also spammers.
There was twice as many people who were only spammers, as there was people who were only threadjackers.

How many people were only flamers?

Lemmy · October 25, 2002, 16:38

i know it.
but i don't have a riddle, so i'll just shut up

Zero-Tau · October 25, 2002, 17:13

Mmmkay...

Everyone, just go on without me. I'm going to bed. G'night.

Japher · October 25, 2002, 17:19

18 flamers only
30 spammers
15 threadjackets

Japher · October 25, 2002, 17:20

Is that correct?

JohnM2433 · October 25, 2002, 17:25

Exactly 12 people were only flamers.

Japher · October 25, 2002, 17:34

12 can't be right.

That would leave 51 others, from which you would end up with 34 only spammers and 17 only threadjackers. If 5 are threadjackers, spammers, and flammers tha would yeild 22 threadjackers (5+17) total, not 20.

JohnM2433 · October 25, 2002, 17:41

Japher, I'm almost certain my answer is correct. I checked my math very thoroughly. I that found a nice Venn diagram helps to solve this problem. Just so that you can confirm it, though, I got:

18 people are only spammers
9 people are only threadjackers
12 people are only flamers
6 people are both spammers and threadjackers, but not flamers.
13 people are both smammers and flamers, but not threadjackers.
No one is both a threadjacker and a flamer, but not a spammer.
The remaining 5 people are threadjackers, spammers, and flamers.

I suppose that if you interpretted "There was 5 people who were both spammers, flamers, and threadjackers" to mean at least 5 people, rather than exactly 5, more than one correct answer might be possible.

So...

There are two words in the English language that both begin and end with "he". What are they?

If you only get one, feel free to post it by itself and let someone else get the other one.

JohnM2433 · October 25, 2002, 17:54

Quote:

Originally posted by Japher
12 can't be right.

That would leave 51 others, from which you would end up with 34 only spammers and 17 only threadjackers. If 5 are threadjackers, spammers, and flammers tha would yeild 22 threadjackers (5+17) total, not 20.

The non-flamers are not all only spammers or only threadjackers. Indeed, if they were, then the flamers would have to be only flamers as well, since no threadjacker or spammer would be a flamer. That would contradict the given statement that "There was 5 people who were both spammers, flamers, and threadjackers". And it would make no sense to add that figure to the number of people who are only threadjackers, since all 63 have already been accounted for.