Logic Puzzles

Aqua wrote: Time for a hard one. How can it be that there is a empty square?

Warning: Spoiler!

So I did some trig for this one, but the answer lies in the angles of the triangles and how those angles affect their lateral and vertical areas. After adding up the areas of each shape, I got 32.5. If the top triangle was a true triangle, it would've had an area of 33. That .5 difference, as goes the difference between a triangle and rectangle, will be multiplied by 2 when considered in the whole of the shape. (Hopefully I'm explaining myself well enough ). Anyways, so noticing this, I measured the angles of the triangles out which came to Red= 20 degrees/ Blue= 21 degrees. The values don't matter as much as the difference itself. So, since the shorter triangle takes up less lateral space, and the taller triangle takes up more vertical space, when the shorter is placed next to the rectangular shapes and the taller triangle is placed above them, that .5 difference becomes available, which is multiplied by two. Therefore, there is an extra 1x1 cube available within the formation.
Again, hopefully I explained what I found well enough. Feel free to call me an idiot .

Izoshua wrote: another one I like:
You have 10 boxes of chocolates and a digital weighing machine. Each box has 20 chocolates in it that weigh 20 grams each. One of the boxes, however, is full of chocolates that are defective and only weigh 19 grams each. How can you use the scale only once to find which box has the defective chocolates in it?

Izoshua wrote:

Aqua wrote: Time for a hard one. How can it be that there is a empty square?

Warning: Spoiler!

So I did some trig for this one, but the answer lies in the angles of the triangles and how those angles affect their lateral and vertical areas. After adding up the areas of each shape, I got 32.5. If the top triangle was a true triangle, it would've had an area of 33. That .5 difference, as goes the difference between a triangle and rectangle, will be multiplied by 2 when considered in the whole of the shape. (Hopefully I'm explaining myself well enough ). Anyways, so noticing this, I measured the angles of the triangles out which came to Red= 20 degrees/ Blue= 21 degrees. The values don't matter as much as the difference itself. So, since the shorter triangle takes up less lateral space, and the taller triangle takes up more vertical space, when the shorter is placed next to the rectangular shapes and the taller triangle is placed above them, that .5 difference becomes available, which is multiplied by two. Therefore, there is an extra 1x1 cube available within the formation.
Again, hopefully I explained what I found well enough. Feel free to call me an idiot .

Izoshua wrote:

Akkarin wrote: Answer?

Warning: Spoiler!

The only thing I can think of is timing the speed the odd ball out falls at, though I'm hesitant, because I'm not sure whether the difference would be great enough to measure without tools (since you only have 9 balls and a scale, no stopwatch).

Nope. I did consider it to be something like that as well, but the answer does involve using the scale three times.

You have 10 boxes of chocolates and a digital weighing machine. Each box has 20 chocolates in it that weigh 20 grams each. One of the boxes, however, is full of chocolates that are defective and only weigh 19 grams each. How can you use the scale only once to find which box has the defective chocolates in it?

Akkarin wrote:

You have 10 boxes of chocolates and a digital weighing machine. Each box has 20 chocolates in it that weigh 20 grams each. One of the boxes, however, is full of chocolates that are defective and only weigh 19 grams each. How can you use the scale only once to find which box has the defective chocolates in it?

Answer?

Warning: Spoiler!

I did it the other way round to Edan, I was going to put them all on the scale and then take them off one at a time to see when the scale went down by 380 instead of 400. If it's the case that there would be too much weight on the scale to make the measurement an alternative would be to remove all but one chocolate from each box and use the same method.

Edan wrote:

Akkarin wrote:

You have 10 boxes of chocolates and a digital weighing machine. Each box has 20 chocolates in it that weigh 20 grams each. One of the boxes, however, is full of chocolates that are defective and only weigh 19 grams each. How can you use the scale only once to find which box has the defective chocolates in it?

Answer?

Warning: Spoiler!

I did it the other way round to Edan, I was going to put them all on the scale and then take them off one at a time to see when the scale went down by 380 instead of 400. If it's the case that there would be too much weight on the scale to make the measurement an alternative would be to remove all but one chocolate from each box and use the same method.

You know a better and much tastier way and would avoid using the scales at all is just to open each box and eat the chocolates until you find a duff one