My version is No1. You can choose the response you want.
THE RESERVOIR
A woman is sitting in a dingy in a small reservoir. In the boat with her is a hemp sack containing the dead body of her husband. She has weighted the sack down with two large blocks of concrete and the twisted remains of his HiFi home entertainment system and sewn it tightly with wire to ensure he will sink without trace.
Whilst securing the sack she glances over to the reservoir wall and notices a water level marker. With a major struggle she manages to heave the sack of hubby, concrete blocks and HiFi system over board. She settles back in the dingy and smokes a well-earned cigarette. After a while, the reservoir wall catches her eye again; she notices the height of the water line has changed.
Has the water risen or fallen against its original level on the water level marker?
Fallen. But I reserve the right to change my mind if it was a statement system.
No need to change your mind Mike, fallen it is !
Good at Maths ??
read on…
Do the following sums as quickly as you can:-
1 - 1 = ?
4 - 1 = ?
8 - 7 = ?
15 - 12 = ?
Pretty sure you got them all right ?
OK. Now for the next bit. Without ‘messing about’ or ‘thinking too hard’ pick any number between 12 and 5.
Now you are probably wondering what this is all about. You’re not the only one…but be honest…
Chances are, the number you picked is 7.
Scaringly, 7 it was.
Pretty spooky hey ?
Balls in a box
(Many moons ago, Matthew T posted this in one of the early Brian Teaser threads, and it remains one of my favourites to this day)
I have a box 5cm x 5cm x 10cm and spherical balls of diameter 1cm. What is the maximum number of balls can I fit in the box?
(To avoid any doubts, all the dimensions are “mathematically perfect”, eg 5 balls will fit exactly across the internal face of such a box, and a lid would close perfectly on top of 5 balls etc etc - but don’t be misled by these examples !!)
Just passing and saw an easy one!
148
5x5 balls on bottom then alternating 4x4 and 5x5, with 3 layers of 4x4 and 5 of 5x5, the 4x4 sitting in the mid points of the 5x5.
From middle of bottom layer, (which is 1 radius r from base), the ball in next layer centre of 4 forma a regular tetrahedron. Distance between base sides = diameter d = 1
Height = sqrt( 1^2 - 0.5^2 - 0.5^2) = sqrt(0.5) = 0.7071
(If I could work out how to do a sketch I’d show working for the above - description is too involved!)
Layers = 0.5 to middle of bottom layer
- 0.7071 to mid each layer above
+0.5 middle of top layer to top of balls
So in box 5 high, 5-(0.5+0.5) = 4
4/ 0.7071 = 5.657 so 5 layers in between the top and bottom layers, 7 in all.
Largest number will be 4 of 5x5 + 3 of 4x4 rather than other way round
Must have been an exceedingly small reservoir or very fine gauge and very good eyesight!
Quite possibly !
Orrrrr…
A very large dingy, and the two blocks of concrete came in at 10,000 tonnes each and the hifi was a Statement that weighs a tonne and a half etc etc
You might want to retrace your steps IB and wander past that “easy one” again
Oops I said tetrahedron, when I meant square pyramid!
If I’m messed up the maths it escapes me, and will need more time than I have right now to reconsider.
Maybe I counted a layer of 5x5 too many, which would mean only 123 which is less than 5x5x5…
Hmm I see that’s what I did, stupid me - the geometry was right!. So actually 2 layers of 5x5, then alternate 4x4, 5x5, 4x4, 5x5, = 4x 0,7071 + 0.5x2 +1 = 4,828 high, 132 balls.
Well, it’s a promising start, but as with nearly all “easy” teasers there is the risk of the odd mistake
Let’s try to rectify some of the easy mistakes…one step at a time…
The box is 5cm x 5cm x 10cm (not 5cm x 5cm x 5cm)
So I can’t read!
Same calc basis. Option 1 is 13 alternating 5 x 5 and 4 x 4, starting and ending 5x5. That is 12x0.7071 + 2x0.5 = total 9.49 and 271 balls.
The other way round again same condition alternating layers 10 x 5 and 9 x 4 total of five layers like that plus one extra layer of 10 x 5 gives a height of 4×0.7071+2×0.5+ the extra layer = 4.83 high and the total number of balls that way is 272 that one wins!
Yes it was easy once read properly ha ha !
Now that we’ve sorted out the easy bit (ie the reading ) …
…let’s move onto the next bit…
… there seems to be an awful lot of redundant space inside that box with only 272 balls !!!
yes, With more time I can see it fits to stack 5x5 alternating with 5x4, 13 layers in total, exactly 10 high 7x25 +6x20 = 295 balls.
Of course now I realise you haven’t said the balls can’t be cut into smaller parts…
Hi IB,
Just to put your mind at rest…
… the balls can’t be cut into smaller parts, nor can they be ground down to a fine powder, nor can the fine powder be reshaped into little cubes, nor…etc etc
It’s just a rectangular box 5x5x10cm with balls each 1cm diameter none of which are squishy or squashy or anything else. And the dimensions are mathematically perfect.
ie there are no silly little tricks.
I think, when you check the arithmetic (or geometry) your 13 layers of 5x5/5x4/5x5 etc will be too high to fit within the 10cm dimension.
I just checked it. The end layers, as you quite correctly quote are each 0.5cm. However, I calculate the intermediate layers to be 0.866cm (sqrt 0.75)
Perhaps you could review how the 5x5 layers and 5x4 layers lay on top of each other ?
Cheers
Don