Going the rounds at school a long time ago were the following, which I have never encountered anywhere since:--
"Why is a mouse when it,s spinning ?" and
"If a hen and a half lays an egg and a half in a day and a half, how long will it take six hens to lay six eggs ?"
We were led to understand that there was a logical explanation of the first, and a formula to answer the second . Can anyone put me out of my misery after all these years ?

Years at KBGS e.g. 1958-1964 (optional) 41/46

Current location (optional) Mexico

Ive no idea about the mouse, but the hen question is straightforward.

How on earth can a hen and a half lay an egg and a half?
Anyway working out theoretically:-

If the hens increase four fold (from 1.5 to 6) and the eggs also increase fourfold (from 1.5 to 6), then the answer is still a day and a half !

Years at KBGS e.g. 1958-1964 (optional) 58-64

Current location (optional) Singapore

Correction. The first should read "Why is a mouse when it,s spinning ?" and the answer was said to be " Because the higher the fewer," and there is supposed to be a rational explanation !

Years at KBGS e.g. 1958-1964 (optional) 41/46

Current location (optional) Mexico

Call me thick but I don't get the mouse thing but I was fortunate enough to pick up a Brownhenway at a local garage sale recently..

Current location (optional) Oliver B.C. (Haworth)

this is even more confusing, what is a brownhenway?

Years at KBGS e.g. 1958-1964 (optional) 1958-61

Current location (optional) Blue Mountains, Australia via Haworth

Years at KBGS e.g. 1958-1964 (optional) 1954-59

Current location (optional) Denholme (garethwhittaker99@hotmail.com)

Quite right Gareth. Somehow, I remember Peter posing the same 'conundrum' some 2/3 years ago, and my brother Arthur posting the very same answer, almost verbatim, so I can only assume he picked it up off Google himself.
Just to keep your 'grey matter' active, try this one:-
'A fox completes the same round trip every evening. If he covers his outward journey at 175 meters/hour, and returns over exactly the same distance at 70 meters/hour, what is his average speed for the entire journey? Blow the dust off those old Algebra books!!

Years at KBGS e.g. 1958-1964 (optional) 1945-1950

Current location (optional) Keighley

Looks like 100m/h. Is that right?

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

Foxed if I know!

Years at KBGS e.g. 1958-1964 (optional) 1952-60

Current location (optional) Lincoln

Yup - 100metres/hour.

Well done, Trevor. Light yourself a big cigar!

Years at KBGS e.g. 1958-1964 (optional) 1945-1950

Current location (optional) Keighley

It's a bloody slow fox.It's going to be in trouble when the Tories get back in power!

Years at KBGS e.g. 1958-1964 (optional) 55-60

Current location (optional) Harrogate

Someone sent me this recently, not too difficult....

Please look at the maths below:

They say only people with an IQ with 120 and over are able to figure this out.

Prove me wrong

If:

2 + 3 = 10

7 + 2 = 63

6 + 5 = 66

8 + 4 = 96

Then:
9 + 7 = ?

Years at KBGS e.g. 1958-1964 (optional) 1952-60

Current location (optional) Lincoln

I work it out to be 144, but then again, I do have an IQ in excess of 120.

Reasoning
2+3=10
2+3=5 (x 2 ie the 1st number of the pair)=10
7+2=63
7+2=9 (x 7 ie the 1st number of the pair)=63
6+5=66
6+5=11 (x 6 ie the 1st number of the pair)=66
8+4=96
8+4=12 (x 8 ie the 1st number of the pair)=96

Beaky would have been proud of me, probably not. I still don't know what a Brownhenway is though.

Years at KBGS e.g. 1958-1964 (optional) 1958-61

Current location (optional) Blue Mountains, Australia via Haworth

OK you get 10/10 for that one John. Of course using that same system 1 + 1 = 2 as it always has!

Years at KBGS e.g. 1958-1964 (optional) 58-64

Current location (optional) Wirral

Have you changed your name or have you got a cold.

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

I always had trouble with maths. Did you know that I'm not on my own, because 4 out of every 3 people have trouble with fractions.....

Years at KBGS e.g. 1958-1964 (optional) 1960-1964

Current location (optional) Bradford

If it helps, that's fraction 4/3 of the population Steve.

Years at KBGS e.g. 1958-1964 (optional) 1958 - 65

Current location (optional) Dudley, West Midlands

who worked at the accountants Rhodes & Co in the 1980,stel 01535 272095

Years at KBGS e.g. 1958-1964 (optional) 1960 -1965

Current location (optional) 7 Chapel row, Wilsden, Bradford W Yorks BD15 0EQ

Thank you ,David, for reminding me of Arthur´s explanation of a couple of years ago. As I read Gareth´s response I realised I had heard it before------it must be the tequila !

Years at KBGS e.g. 1958-1964 (optional) 41/46

Terry your maths problem is neat. Ever the pedant especially in things mathematics which prides itself in precision of language I have to insist that in a way the question is unfair, or rather incorrectly posed. The use of conventional notation + (plus) and = (equals) is misleading and of course deliberately so to fog the solution. What we are required to really do is define the function that maps a given pair ( x, y) to a single element(z).
I am not sure of the correct functional notation but it probably will look something like this:
If f ( x, y) -> z
And given that f(2,3) -> 10 and f (4,5) ->36 etc ……… define f

The answer being that for any pair ( x,y) the function f maps the pair(x,y) to z where z = x( x+y)

Sounds like one of Zeno's Paradoxes,Arthur.

Years at KBGS e.g. 1958-1964 (optional) 55-60

Current location (optional) Harrogate

What is the smallest number of people you would need so that there is a greater than 50:50 change of two having the same birthday ?

I was asked this the other day, as my daughter and I shared the same birthday last week.

It's not too difficult really, but I thought I'd pass it on for your perusal.

Years at KBGS e.g. 1958-1964 (optional) 58-64

Current location (optional) Wirral

23 ...... or possibly 22 if you count 29th February (not sure about that though).

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

I can remember the time when giving an answer without an explanation as to how you had arrived at it would have got you a clip round the lug'ole. It may well be right(it might also be a wild guess!)

Years at KBGS e.g. 1958-1964 (optional) 55-60;

Current location (optional) Harrogate

I'd better steer clear o' thee - or at least not tek mi lug'oles wi' me.
OK.
You need at least 50% probability of 2 people with the same b'day - so less than 50% probability of all having different b'days.
First person certainly has a birth date - prob 100%.
Second person has a different birth date - prob (364/365 x 100)% = 99.7% approx.
Third person has a birth date different from both these - prob (363/365 x 364/365 x 100)% = 99.2% approx.
Etc. etc.
When you get to the 23rd person this equals slightly less than 50%. Q.E.D.

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

Yes I got 23 as well Shaun , so it must be right !

Years at KBGS e.g. 1958-1964 (optional) 58-64

Current location (optional) Wirral

Now I remember why we used to call you Gammy Scott

Years at KBGS e.g. 1958-1964 (optional) 1958-61

Current location (optional) Blue Mountains, Australia via Haworth

Can you find integers x, y and z such that x cubed plus y cubed = z cubed, or can you show that this can't be done. First correct answer gets a pint of Landlord.

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

Come home Fermat,all is forgiven.

Years at KBGS e.g. 1958-1964 (optional) 55-60;

Current location (optional) Harrogate

I was just hoping someone might help me win the million dollars.

Years at KBGS e.g. 1958-1964 (optional) 58-65

Current location (optional) leeds

Now, let me see; a million dollars or a pint of Landlord - difficult choice!

Years at KBGS e.g. 1958-1964 (optional) 55-60;

Current location (optional) Harrogate