a question.. who knows?

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^ we should use the scale, and just one weighing (one reading).
 
i was going to say to put 5 boxes on each side of the scale (all at one) and the side which weighs less is the one with the fake gold...but that doesn't determine which box contains the fake gold.

EDIT: i guess you could apply the above method, then try moving them (gently and whilst still on the scales) with your bare hands. then the one that's easiest/lightest to move is the one with the fake gold in it?

btw, when putting them on the scales wouldn't you come to know their weight? i say this could be the answer as well. come to think of it, i think that's the answer since you said it doesn't matter if the scale is one sided or two.
 
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^ I mean you should put the weight in the scale once and for all. No progressive addition, and no moving.
PS : Let's say, you can't notice the difference of 100g in the box just by lifting the box with your hands. And the problem is that you should do just one operation, and find the answer from the first time : no repetition.
 
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Ok, I'll give the answer in ... 10 min from ... Now !

you have less than 10 min to post the answer ...
 
ok il try lol
Is it by placing the boxes on a weighing scale but not all at once, but like its the weighing scale that has 2 plates or 2 sides, i am not sure what you call it sorry.

But by placing 1 box on both 2 scales, the side that goes down, means its the one with the fake gold and the one that goes up is the one with real gold.
If you were to place 2 real gold boxes on the weighing scale on each side, they would stay the same, the scales wont go up or down. . .

I guess i tried :-\
 
^ I mean you should put the weight in the scale once and for all. No progressive addition, and no moving.
PS : Let's say, you can't notice the difference of 100g in the box just by lifting the box with your hands. And the problem is that you should do just one operation, and find the answer from the first time : no repetition.

in the answer i gave you're not moving anything off once you put it on the scale. the only thing you are moving is the boxes one by one onto the scales. you don't take them off after that.

when i mentioned this:
i guess you could apply the above method, then try moving them (gently and whilst still on the scales) with your bare hands. then the one that's easiest/lightest to move is the one with the fake gold in it?
i meant sort of like how you move/touch books when they are stacked one on top of the other. you can sort of push them gently, without taking them off the pile...but obviously that isn't he correct answer.

other than that, i reiterate and say it that you put 5 boxes each side and the one that weighs the less is the one with the fake gold.
 
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^ you should use the scale just one time. So your solution is not really good because you're not sure you'll find the fake box from the first time.

Now as it's timeout, i'll give the answer

Question:
We have 10 boxes. Every box contains 10 bricks of gold. Every brick is 1kg. So the weight of every box is 10kg.
But one of the 10 boxes is fake, i.e it contains non-gold bricks, each non-gold brick weighs 0.990 Kg (990g), i.e 10g less than a real gold brick.
We have a scale, but we must use it for just ONE weighing.
How can we know which is the fake gold box by using the scale just ONE time ?

Answer:

We have 10 boxes.
We take :
1 brick from box1
2 bricks from box2
3 bricks from box3
4 bricks from box4
5 bricks from box5
6 bricks from box6
7 bricks from box7
8 bricks from box8
9 bricks from box9
10 bricks from box10

so we have 1+2+3+4+5+6+7+8+9+10 = 55 bricks.

if we had no fake box, then the total weight of these 55 bricks would be 55kg.

But we won't have a whole 55kg, because there is some fake bricks from one box.

we put these all 55 bricks on the scale, and read the weight :

* if the total weight is 10g less (one fake brick), then the fake box is box1, because we get 1 brick from box1.
* if the total weight is 20g less (2 fake bricks), then the fake box is box2, because we get 2 bricks from box2.
* if the total weight is 30g less (3 fake bricks), then the fake box is box3, because we get 3 brick from box3.
* if the total weight is 40g less (2 fake bricks), then the fake box is box4, because we get 4 bricks from box4.
* if the total weight is 50g less (5 fake bricks), then the fake box is box5, because we get 5 bricks from box5.
* if the total weight is 60g less (6 fake bricks), then the fake box is box6, because we get 6 bricks from box6.
* if the total weight is 70g less (7 fake bricks), then the fake box is box7, because we get 7 bricks from box7.
* if the total weight is 80g less (8 fake bricks), then the fake box is box8, because we get 8 bricks from box8.
* if the total weight is 90g less (9 fake bricks), then the fake box is box9, because we get 9 bricks from box9.
* if the total weight is 100g less (10 fake bricks), then the fake box is box10, because we get 10 bricks from box10.
 
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This one sounds familiar. Is it a wheelbarrow?
 
I can't think of a riddle. Does anybody else want to have a go?
 
I talk, but I do not speak my mind I hear words,
but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me

who am i?
 

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