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I posted this to compare times with individuals that possess intellects lightyears beyond my own. I clocked in at 16 mins for this answer.

I know my time will be overtaken by a time that's less than half of what it was.

But instead of getting too wrapped up into the idea of it happening. I want to be the first to witness how an IQ of 130, 140, 150, or 160 functions/processes information from my own intelligence.

Rules: No paper or notes, but a calculator is accepted.

Post the exact time min wise of how long it took.

And obviously give the answer.

http://i.stack.imgur.com/BPJBh.jpg

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    $\begingroup$ I'll admit, I would've taken forever without writing it down $\endgroup$ – Areeb Sep 3 '16 at 1:52
  • $\begingroup$ I don't think the pattern tag applies, try replacing it with math or removing it entirely $\endgroup$ – Areeb Sep 3 '16 at 2:02
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    $\begingroup$ Haha i know 7 answers wow. I give that credit torwards my clickbait title lmao $\endgroup$ – bill bsl Sep 3 '16 at 22:44
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    $\begingroup$ @ArkaKarmakar there's a lot of sense to IQ. I find that it correlates extremely well to a person's ability to take IQ tests. $\endgroup$ – ffao Sep 7 '16 at 2:02
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    $\begingroup$ The OP seems to have completely misunderstood what IQ means. $\endgroup$ – vsz Sep 7 '16 at 14:59
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Took one minute to solve.

This question has a reasonably easy solution if you can figure out the pattern in the the provided equations.

Look at the first row equation and the second column equation.
$a+b =8$
$b+c =8$

Notice that there is a common variable in the equations (the intersection of the equations in the picture denotes the common variable.)

Thus we can rewrite the set of equations as:
$a+b=8$
$c+b=8$

Subtracting these two equations can make us realize that $a=c$. Which means the top left box is equal to bottom right box.

The problem can be rewritten as

enter image description here

The values now seem fairly easy to compute.

$a+c=13$
$c-a=6$

By adding the above equations, the variable $c$ can be resolved. By evaluation of $c$ and substituting into either of the above equations, variable $a$ can be calculated. After which variable $b$ can be calculated by substituting variable $a$ in $a+b=8$.

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    $\begingroup$ +1, this is how I solved it. Took me about 45 seconds to do in my head. $\endgroup$ – Rand al'Thor Sep 3 '16 at 12:21
  • $\begingroup$ I failed to solve this because once I got 2c=13+6 I jumped to the conclusion that Heck, there are no solutions by parity. What a trick question. $\endgroup$ – Matsmath Sep 6 '16 at 22:01
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Let a be the top - left square,
Let b be the top - right square,
Let c be the bottom - left square,
Let d be the bottom - right square,

a = 3.5,
b = 4.5,
c = 9.5,
d = 3.5

Method:

Equations:

a + b = 8
c - d = 6
Add to get: a + b + c - d = 14

a + c = 13
b + d = 8
Add to get: a + b + c + d = 21
Subtract equations to get: 2d = 7
d = 7/2 = 3.5
c - 3.5 = 6
c = 9.5
a + 9.5 = 13
a = 3.5
3.5 + b = 8
b = 4.5

Also ~ two minutes.

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  • $\begingroup$ You used virtual paper or you might've done it in your mind first $\endgroup$ – Areeb Sep 3 '16 at 1:57
  • $\begingroup$ Once I saw what to do it was easy. All you need is to get one number from the method, and everything collapses. $\endgroup$ – greenturtle3141 Sep 3 '16 at 1:58
  • $\begingroup$ You had a great explanation regardless $\endgroup$ – Areeb Sep 3 '16 at 2:00
  • $\begingroup$ Great explanation. $\endgroup$ – LeppyR64 Sep 3 '16 at 2:17
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    $\begingroup$ @TylerJohnson I don't think it's fair to judge someone's intelligence on how fast they solve math problems $\endgroup$ – Areeb Sep 3 '16 at 3:29
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3.5 + 4.5 = 8
9.5 - 3.5 = 6

This took me two minutes to solve. There are four variables and four equations. Basic algebra rules the day.

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    $\begingroup$ 3.5 + 3.5 != 8 so it doesn't work $\endgroup$ – Areeb Sep 3 '16 at 1:53
  • $\begingroup$ Transcription error. Should be fixed now. $\endgroup$ – LeppyR64 Sep 3 '16 at 1:56
  • $\begingroup$ Eyyy it works now and just before all the other answers! $\endgroup$ – Areeb Sep 3 '16 at 1:56
  • $\begingroup$ That's absolutely insane. What's your estimated iq? i would definitely love to know. $\endgroup$ – bill bsl Sep 3 '16 at 1:56
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    $\begingroup$ I did it in my head using three equations noting the two opposite corners would have the same value. It was very easy to get the 9.5 that way. $\endgroup$ – Will Sep 3 '16 at 1:59
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Let squares be $a$, $b$, $c$, $d$. We have $a$ $+$ $b$ $+$ $c$ $+$ $d$ $=$ $21$, $a$ $+$ $b$ $+$ $c$ $-$ $d$ $=$ $14$, so $2d$ $=$ $7$, $d$ $=$ $3.5$. Got this bit in 5 seconds. Full square: 

3.5 + 4.5 = 8
 +     +
9.5 - 3.5 = 6
 =     =
 13    8

$20$ seconds in total - roughly

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  • $\begingroup$ 20 f@×king seconds. Ok i have to know the number to your iq $\endgroup$ – bill bsl Sep 3 '16 at 2:44
  • $\begingroup$ it's more of a concept :) $\endgroup$ – JMP Sep 3 '16 at 2:45
  • $\begingroup$ So your saying you solving that problem had nothing to do with intelligence, but more with the way you solved it? $\endgroup$ – bill bsl Sep 3 '16 at 2:48
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    $\begingroup$ it's a standard method - the sum of the sum of the rows = the sum of the sum of the columns in a usual array - in this case there is a minus, so we can apply simultaneous equations $\endgroup$ – JMP Sep 3 '16 at 2:55
  • $\begingroup$ Experience with these types of puzzles (and perhaps just puzzles/math in general) definitely would help. $\endgroup$ – greenturtle3141 Sep 3 '16 at 6:16
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Here is how to solve in your mind (I did it in this way in 2-3 minutes to solve and confirm):

if you sum all equations you will be summing up the squares two times but the square with opposite signs (the forth square) cancels and you left with summing up squares with + signs two times. So 2x(first+second+third squares) = 8 + 8 + 6 + 13 = 35, hence first+second+third = 17.5, you know first + second = 8 therefore third = 9.5. Then it follows.

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Pretty much the same as everybody else, with a slightly different twist:

Add the columns: \begin{align}a + c = 13\\+\qquad b + d = ~~8\\\hline\llap{\text{Add to get:}\quad}a + b + c + d = 21\\\llap{\text{Subtract:}\qquad}-\qquad a+b=~~8\\\hline\llap{\text{to get:}~~\qquad\qquad\qquad}c+d=13\end{align} So we have $c+d=13$ and $c-d=6$, so $c$ must be the average of $13$ and $6$.  We can calculate this as \begin{align}c+d&=13\\+\qquad c-d&=~~6\\\hline\llap{\text{Add to get:}\quad\qquad}2c~\phantom{+d}&=19\\[2ex]\llap{\text{So,}\quad\qquad\qquad}c~\phantom{+d}&=19\rlap{/2}\\[2ex]\llap{\implies\qquad\qquad}c&=~9\rlap{.5}\end{align} but I didn't need to do that.  I saved a few seconds by realizing that, since $c$ is equally distant from $13$ and $6$, it must be halfway between $13$ and $6$, i.e., their average.

Then (the same as everybody else), it's a simple matter to solve for the other three:

$c+d=13\implies 9.5+d=13\implies\boxed{d}=13-9.5=\boxed{3.5}$
$a+c=13\implies a+9.5=13\implies\boxed{a}=13-9.5=\boxed{3.5}$
$b+d=~8~\implies\,b+3.5=~8~\implies\boxed{b}=~~8-3.5=\boxed{4.5}$
Interestingly, I didn't notice that $a=d$ until after I had solved for their values.

Results: $\qquad a=3.5\qquad b=4.5\qquad c=9.5\qquad d=3.5$

So the filled-in grid is:
$$\begin{array}{c}\Large{3.5}&+&\Large{4.5}&=&\Large{8}\\+& &+\\\Large{9.5}&-&\Large{3.5}&=&\Large{6}\\||& &||\\\Large{13}&&\Large{8}\end{array}\hskip1.1in$$

I didn't exactly time myself, but I'm pretty sure that I did this in under a minute.  In deference to rand al'thor, I'll stipulate that I took at least 45 seconds.

P.S. For convenience, I copied some of the equations from greenturtle3141's answer.

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This took me a minute using brute force.

I instinctively started with the bottom row first because it seemed to have a narrower range of reasonable answers. After plugging 8,9,10 in the bottom left corner it was pretty obvious that a whole number wasn't going to work - my answers for the top row were bracketing 8. Once I reached that conclusion the problem was basically solved.

I can see from other answers that there is a more theoretical way to solve the problem but I figured it would take me longer to work out the theory than to solve it with brute force.

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