
#1
Nov2108, 04:16 PM

P: 116





#2
Nov2208, 01:06 AM

P: 1,520

Spoiler
1. 2 The 2nd square in a row shares all of the 3rd square's lines, not accounting for translation. This leaves 2 as the only choice. 2. 5 This is about having the same number of black squares inside a square and outside. Their numbers are already the same, which means the answer is 5. The other 2 are obvious... 



#3
Nov2208, 04:31 PM

P: 116

2 and 4?




#4
Nov2208, 05:27 PM

P: 657

Four IQ test problems
The only one I haven't gotten so far is the 1st one...
Spoiler
I dunno if I like that reasoning since it would appear to ignore the 1st square in each row, making the 1st square's content arbitrary. In other words, you could place an irregular pentagon circumscribed by an oval into any square in the 1st column, and your solution would appear to be unaffected. I don't like my answer (as it's sort of similar), but here's what I've come up with: The number of distinct white regions in each square is: 4 2 1 8 4 2 3 2 ? Now, this means from the top row to the middle row, each of the numbers of white regions double. And from the middle to the bottom row, they're log(root 2)'d. So the number of white regions in a given column corresponds to: A, 2*A, log base 2(2*A). Hence, the only viable option is choice #1, because it's the only one with a single white region. DaveE 



#5
Nov2808, 08:21 PM

P: 323

Please correct me if one or more of my answers are wrong. I am not completely sure about 1 and 3.
Problem 1:
Spoiler
I believe that the answer is 5.
Problem 2:
Spoiler
The correct answer is 5. Assign each of the squares inside the box a value of 1 and each of the outside ones a value of +1. The rest should be easy
Problem 3:
Spoiler
I believe that the answer is 2. One reason would be symmetry. 1, 3, and 5 would work because each of them would cause for there to be an equal number of white and red squares, but then it would not be symmetrical.
Problem 4:
Spoiler
I think that the correct answer is 4. Each of the shapes shifts one unit to the right and wraps around to the next line. With the exception of the circles, the shapes cycle through the colors blue, red, and green. In that order. So the next triangle is green. There is a second way to look at this, rather than looking that rows one looks at diagonals. Each way leads to the same answer though.




#6
Nov2908, 12:12 AM

P: 657

DaveE 



#7
Nov2908, 02:28 AM

P: 655

Spoiler
Imagine the shape folded into a cube. 2 would have the cube cleanly divided half and half into red and white parts.




#8
Nov2908, 03:34 PM

P: 323

Hm... 



#9
Nov2908, 03:35 PM

P: 323

Does anyone happen to know what the correct answers are?




#10
Jan1209, 09:03 PM

P: 18

pretty simple:
Spoiler
1. 2
2. 5 3. 2 4. 4 



#11
Jan1909, 01:52 PM

P: 2

Am sure about 3 and 4, they are pretty simple but dunno about 1, 2.
3 2nd 44th 



#12
Jan1909, 02:37 PM

P: 411

2 3 4 I thought were pretty obvious.
#1 is kicking my *** though. I call defeat. here's looking at me, kid: FAIL 



#13
Feb2609, 11:12 AM

P: 52

I got:
Spoiler
1. 3
number of straight lines equals number of curved lines 2. 5 or 4 5 with the rule number of black squares inside equals number outside or 4 with the rule number of squares increases by 2 for each row/column starting from the top/left 3. 2 4. 4 



#14
Feb2609, 01:47 PM

P: 657

[edit]
Spoiler
I did just notice an interesting pattern with the number of intersections, though. The number of curved line segments in each square is equal to the number of line *intersections* (not corners). Which would similarly make the answer #3. Huh. I like that answer, actually.
DaveE 



#15
Feb2609, 03:12 PM

P: 52

Sorry yeah I should have been more explicit, I meant:
Spoiler
The totals when you add all 9 boxes together, and that e.g the two circles in the first box would be counted as two curved lines.
I think I prefer your explanation for why it would be 3 better though. I'm not sure I really like the answer to the second question being 5 either. EDIT
Spoiler
In answer four would the touch of thecircle and oval count as an intersection and would the touch be considered a double touch/intersection?




#16
Feb2709, 12:21 PM

P: 657

DaveE 



#17
Apr1110, 07:58 PM

P: 88

I think posting to an old thread is not prohibited, correct me(and delete this post:) ) if I am wrong.
My question to the OP, leonard. I just want to know the name of the program in the pictures above. I tried to contact him but he doesn't receive PMs. 



#18
Jun2510, 03:15 AM

P: 88

Hey, people, can anyone help me ??
I want to know the name of the program in the OP's post. thanks 


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