There is a relation between the shapes in the top equation on the left side. If we apply such an equation to the bottom shapes, which one from the right choices can fill the question mark?
Source: Iranian Ph.D. entrance exam 2017
2
-
$\begingroup$ (You've now posted at least 5 questions in a row from the same exam. First of all, thank you for posting attributions on these! Secondly, thanks for contributing some interesting questions. However, I do want to note that many exams of this type expressly prohibit their reproduction elsewhere; are you sure you have permission to post these questions here? I'll also note that, permission issues aside, we don't really want you, or anyone, largely reproducing an exam one question at a time on P.SE. At some point It looks like rep farming, begging the question of why you're doing it.) $\endgroup$– Rubio ♦Feb 17, 2018 at 11:28
-
$\begingroup$ @Rubio You may be right about the permission issue. However, in Iran, they are available on many websites and answers with explanations also exist. I myself didn't know that I can find the answers, so I posted the questions I couldn't solve easily. Now, I will probably stop it. However, as I said, I just aimed to post those difficult ones to learn a method for solving them. $\endgroup$– AhmadFeb 17, 2018 at 12:26
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
On the first row,
the RHS is the exclusive-or of the first image on the LHS rotated clockwise through a right angle, and the second image on the LHS rotated by 180 degrees. That is: if we rotate those images in those ways and overlay them, the RHS has ink just where exactly one of the LH images has ink.
If we do the same on the second row (it seems like we need to shift the LHS images a little, which makes me uneasy, but never mind) we get
option 3.
answered Feb 15, 2018 at 19:46
-
$\begingroup$ Correct! by "rotated clockwise through a right angle" do you mean rotating 90 degrees on right? $\endgroup$– AhmadFeb 16, 2018 at 5:08
-
1$\begingroup$ Yes. 90 degrees = a right angle = a quarter turn = pi/2 radians; clockwise = the direction such that the top moves right and the bottom moves left. $\endgroup$– Gareth McCaughan ♦Feb 16, 2018 at 12:55