6
$\begingroup$

I was solving the sample problems for my school's IQ society and there are some I don't get. Since all I get is a final score, I wanted to ask puzzling if someone might be able to make things clearer. For the problem that I've answered I'm writing down the answer and explanation below. I'm not sure if it's right though.

1 4 5 8

1

I believe the answer is C. For each column, move from top to bottom on each side (left/right) of the rectangle. If there are two occurrences of the black dot, it is cancelled out, so there's a blank in the bottom rectangle. If there's only one black dot, it is added to the bottom rectangle. So the right column is basically the same as the middle column apart from left/right being switched, hence the resulting image is the same.

$\endgroup$
1
  • 2
    $\begingroup$ Welcome to Puzzling.SE! If the puzzles aren't original, make sure you add a reference to the original work. Also be careful that the questions don't relate to an open competition, otherwise that could potentially provide other participants with the answers $\endgroup$
    – Dmihawk
    Commented Nov 25, 2018 at 18:24

4 Answers 4

5
$\begingroup$

Problem 1:

The answer is C. The last element in each row is made by drawing a point if only the 2 elements in the same position have no point, then rotating it 180 degrees.

Problem 4:

The answer is F. Same explanation in Problem 1, but it's rotated 90 degrees anticlockwise.

Problem 5:

The answer is A. The last element in each row is made firstly by rotating the center element in the row by 180 degrees, then subtracting the first and the middle. Blank - black = still blank.

$\endgroup$
3
  • 1
    $\begingroup$ I would have never figured that for myself. I guess when you say row you mean both rows and columns? (because it holds for columns too). Also, in problem 5, when you say "the first and the last" I think you meant "the first and the middle". $\endgroup$
    – abl
    Commented Nov 27, 2018 at 21:16
  • $\begingroup$ A good argument for 1C! And seeing as how the same principle can be applied to 4, I'm quite sure this is the solution. $\endgroup$
    – Chiquitain
    Commented Nov 28, 2018 at 10:54
  • $\begingroup$ Regarding the edit, another user suggested the answer, but had written it more definitively. I thought the idea had some merit so allowed the edit with a change of presentation. $\endgroup$
    – hexomino
    Commented Mar 10, 2021 at 18:17
6
$\begingroup$

I think for 1st puzzle it is C and for 5th puzzle correct answer is A. I think in these two diagrams rotation is being performed on the diagrams of second columns. I am trying to get solutions for other diagrams.

$\endgroup$
3
$\begingroup$

Problem 1:

I don't totally get this problem, but I suspect the answer might be (b). Note that for each of the first two rows, there are exactly three dots in the top of the squares and one dot in the bottom. And for each of the first two columns, there are exactly two dots on the left and two dots on the right. If this pattern has to be maintained in the third row and third column, the answer is (b).

$\endgroup$
0
0
$\begingroup$

A possible solution for 8 could be...

B

Because...

In each row you overlap the first two items with OR. After you make the picture negative (so white squares gets black and black squares gets white) and rotate it 90° * row

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.