Brain Teasers

[plater]

From an old mathamatics book

(1) A clock strikes six in five seconds, How long does it take to strike twelve ?? 11 Seconds

(2) If five spiders can catch five flies in five minutes, How many spiders are required to catch a hundred flies in a hundred minutes ?? Five Spiders

(3) The vicar saved the game for his side and scored 40 runs,thereby raising his average for the season from 27 to 28, How many runs would he have required to bring his average up to 30 ?? 66 Runs

(4) The gambler had lost all his money, he had a gold chain of seven links which his opponent agreed to value at £1 per link.

The play proceeded at £1 per game and the debts were settled after each play, The gambler lost each time as the opponent expected, What was the least number of links that had to be cut ??.Only one the third

[markwales]

2 = 5
4 = 6?

[plater]

2 correct
4 incorrect

[vegyjones]

1 = 21605 seconds

2 = 5

[plater]

1 incorrect :D

[vegyjones]

4 = 5

[markwales]

1 = 10 ?? :doh

[vegyjones]

Is 1 a trick answer and

= 10

[vegyjones]

Bloody mark ;fire

[plater]

Only question 2 solved come on easy peasy (i've got the answers :wink )

[markwales]

1 = 11 seconds

[Workshy]

1. five seconds, the minute hand hits twelve as the hour hand hits six. ???

[plater]

1 = 11 seconds correct Mark

[vegyjones]

1 = 11 seconds correct Mark
the reason being???

[plater]

I've only got the answers Vegy, not the solutions I'm afraid :geek

[vegyjones]

3 = 64

[plater]

Getting Warmer

[markwales]

Because the time before the first strike isn't counted....or at least that's the way I looked at it.

It's strikes once each second, strike 1 being on ZERO seconds (the clock now starts), strike 2 being on 1, strike 3 being on 2...and so on.

[Workshy]

1. five seconds, the minute hand hits twelve as the hour hand hits six. ???

My answer is correct until Mark proves his, ner! :D

[markwales]

My answer is correct until Mark proves his, ner! :D

See above ....Ner Ner no returns :D:D:D

[vegyjones]

Because the time before the first strike isn't counted....or at least that's the way I looked at it.

It's strikes once each second, strike 1 being on ZERO seconds (the clock now starts), strike 2 being on 1, strike 3 being on 2...and so on.
Okay, I see! :D

[vegyjones]

Getting Warmer
3 = 67

[Workshy]

See above ....Ner Ner no returns :D:D:D

You know what mate, I still (stubborn as it might sound) believe my answer is correct too!! :)

[plater]

3 strikes and yer out :laugh

[vegyjones]

3 = 66

[plater]

Because the time before the first strike isn't counted....or at least that's the way I looked at it.


That what I thought as well Mark

[plater]

Correct Mr Vegy, how did you arrive at the answer ??

[Roberto]

4) The answer must be "3"?

[vegyjones]

Well,

I did it,like this, but it pe=robably not the correct mathematiclal method

40 is 12 past 28,

so I added on another 12 to make 29 and another 12 to make 30
then add on an extra one for each increase in average,

so 26 altogether! :D

[plater]

4 Nope not 3, I'm still trying to get my head round the answer though :doh

[vegyjones]

Is 4 none!

[plater]

Well,

I did it,like this, but it pe=robably not the correct mathematiclal method

40 is 12 past 28,

but it's 13 past 27 :wink

[plater]

None is incorrect

[Roberto]

4 Nope not 3
Very strange!
After G loses the first one, clearly one must be broken off.
After G loses the second one, clearly another must be broken off.
After G loses the third one, clearly a third one must be broken off.
After G loses the fourth one, he can give the remaining linked-up chain of 4 to his opponent and take the 3 loose ones back.
After G loses the fifth, sixth and seventh ones, he simply passes loose ones over to his opponent one at a time.
So my answer is "3".
What have I missed?! :)

[vegyjones]

But it's not the answer in his book! :D

4 = 2

[Roberto]

But it's not the answer in his book! :D

4 = 2
Yes, that's right, of course!!
How stupid of us!
The second time, the link broken is one from the middle of the chain, not one of the end, leaving a piece of 2-units and a piece of 3-units which can be "used as change" in the way I mentioned above.
Very clever! :)

[plater]

Still trying to work the answer out myself, I think it's a play on words Roberto, when, If anyone gets it correctly perhaps they can explain why :yikes:

[Roberto]

The second time, you must break the third one along in the remaining chain of 6 links! No word-play involved, just a very clever puzzle. :)

[plater]

Very strange!
After G loses the first one, clearly one must be broken off.


But which one ???????????

[vegyjones]

sort it out Plater, me and roberto have solved it

It's 2!

Cut 1 off after losing Game 1
Cuts 2 off together after losing 2nd

so has peces of 1 2 and 4 in length,
after losing game 3 he gives him the single one back

after 4 he exchasnges his 4 for the other 3 pieces,

then gives him 1
then gives him 2 piece gets one back
then gives him the last piece! :D

[plater]

BTW 2 is the wrong answer

[markwales]

BTW 2 is the wrong answer

:laugh:laugh:laugh:laugh:laugh:laugh:laugh:laugh

[vegyjones]

BTW 2 is the wrong answer
;fire

[vegyjones]

4 must = 1

[Roberto]

BTW 2 is the wrong answer ::swear

[Roberto]

4 must = 1 ... if only because we've tried everything else! :doh

[plater]

Correct Vegmeister, why ??

[vegyjones]

I could make up any old crap as you don't know either? :D

[Roberto]

Maybe it IS one ...
You start with a straight-line chain of 7 links and break the third one. This leaves you with pieces of 1, 2 and 4 in length. I think you can cover every number from there?
But this doesn't work if it starts out as a circular chain, only as a straight-line one, I think? So a little bit "tricky", no? :censored:

[vegyjones]

Maybe it IS one ...
You start with a straight-line chain of 7 links and break the third one. This leaves you with pieces of 1, 2 and 4 in length.
That still needs 2 cuts!

[plater]

Thank gawd for that, the third link is the only one to be cut :wiggle:

[Roberto]

That still needs 2 cuts!No ... only one cut ... if the chain starts out as "linear" rather than "circular". It's a bit of a trick question, but we got there in the end.

[plater]

As I said Roberto, is it a play on words, " settled after each play", would you not have 3 links after one play if you won ??

[vegyjones]

Right...

If I have 7 segments,

cutting the 3rd segment will leave me with pieces of 3 and 4 segments

Another cut is needed to make the 3 segment into 1 and 2 segments.

Someone explain as I don't get it??

[Roberto]

If I have 7 segments, cutting the 3rd segment will leave me with pieces of 3 and 4 segmentsNot so. Cut the third link and remove it completely. You have 2 (which were on one side of it) and the loose one you cut (damaged now of course) and four more (which were on the other side of it). Try it with loops of paper or something!

[plater]

of course, when you cut that link it falls off so to speak, I'll get me coat

[vegyjones]

Ah I see...

I was thinking lof it like if you cut at one end,
it would still be attached at the other side!

For example separating bits of scalextric track! :D

[Roberto]

Great brain-teasers, must be a good book you've got there ... many thanks. See you ...