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#1
04-24-2017, 11:43 AM
 Craig44 Junior Member Join Date: Apr 2017 Posts: 2
Logic puzzle (please help)

Does anyone know the answer to this one? If so please explain why
#2
04-24-2017, 01:15 PM
 uigrad Senior Member Join Date: Jan 2014 Posts: 110

I think the fifth choice is the solution. I can't completely describe the thought process of the test maker, but I came up with a consistent pattern between them. But it's complicated to explain, and probably not exactly exactly what the test maker intended:

Basically, each row and each column is an addition problem, if you define the tiles as two digit numbers using the following rules:

Rule 1: The black dots are the "tens" digit, and white dots determine the "ones" digit
Rule 2: The digits are 0-7, corresponding with binary 000 to 111
Rule 3: If the black dots are in the top row, then it's a positive number. If they are in the bottom row, it's a negative number.

So, we have:

24 52 76
-66 24 -42
-42 76 ??

Within a single row, we have valid addition problems:

24 + 52 = 76
-66 + 24 = -42
-42 + 76 = ??

Within a single column, we have valid addition problems:

24 + (-66) = -42
52 + 24 = 76
76 + (-42) = ??

To make the third row correct, we need the question marks to be 34. Same for the column, we need 34. So, 34 is the solution.

So, we want black dots on top (positive number), and we want 011 for the black dots and 100 for the white dots. That is the 5th choice!!
#3
04-24-2017, 01:24 PM
 aceinnatailsuit Junior Member Join Date: Oct 2016 Posts: 7

The way this puzzle works is that the pattern of dots in the third box is always the combination of the dots in the first two. In the first example, we see that the first box has a black dot in the centre on top and a white dot on the left on the bottom, and the second one has black dots on the left and right on top and a white dot in the centre on the bottom; the third box has black dots in all three spaces on top and white dots in the left and centre on the bottom. The second example is a little tricker, since the orientation of the dots is inverted between the first two - one has white on top and black on the bottom, and the other has the reverse. In this case, any dot in a shared position is 'cancelled' from the side that has more (think of it like simplifying fractions, or dealing with negative numbers). In our second example, we see that the first box has white dots on the left and centre on top and black dots on the left and centre on the bottom, and the second box has a black dot in the centre on top and a white dot on the left on the bottom; the third box (our 'answer') has a white dot in the centre on top and a black dot on the left on the bottom - the black and white dots from the second box 'cancelled out' dots from the first box.

Now, on to the question. The first box has a white dot in the centre on top, and a black dot on the left on the bottom. The second box has black dots in all three spaces on top, and white dots on the left and centre on the bottom. For our black dots, we have them in all three spaces on top, and on the left on the bottom, giving us two black dots on the right and centre on top. For our white dots, we have one white dot in the centre on top, and white dots on the left and centre on the bottom, giving us one white dot on the left on the bottom.

We want a box with black dots on the right and centre on top, and a white dot on the left on the bottom. The box one from the right matches our answer.
#4
04-24-2017, 02:25 PM
 Craig44 Junior Member Join Date: Apr 2017 Posts: 2

Thanks you so much!

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