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Old 20th February 2019, 04:56 PM   #1
MEequalsIxR
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Not so Simple Aritmetic or is it

Saw this on another forum and the debate on the correct answer was interesting to me. No idea where it was originally was from. I thought I'd bring it here to see what the smart crowd does with it. Two possible answers - 15 or 60 which is correct.

My take on which answer is correct:

I believe 15 is correct based on PEMDAS but the way I was taught in grade school the answer would be 60 but then in those days printing was the new thing.
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Old 20th February 2019, 05:09 PM   #2
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The correct answer is "unknown," because the value of the single-strut symbol is not established. All we can determine is that the double-strut symbol has a value of 2. (We have to treat the double-strut as a single whole symbol because there is no space and no operand between the struts of the double-strut where it appears here.)

We might guess that the single-strut symbol must therefore have a value of 1, but by the same reasoning we could likewise guess that the symbol "3" must have half the value of the symbol "8," or that the symbol "11" must have twice the value as the symbol "1".

I bet 99.99% of people wouldn't be that pedantic about it, though.
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Old 20th February 2019, 05:27 PM   #3
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Sometimes I wonder about who makes these things. What's going through their mind?
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Old 20th February 2019, 05:32 PM   #4
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Originally Posted by Beelzebuddy View Post
Sometimes I wonder about who makes these things. What's going through their mind?
Motor parts.
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Old 20th February 2019, 05:51 PM   #5
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I figured that the answer was

20, by simple algebra. Here's my work:

3x = 30
x = 3x / 3
x = 10

x + 2y = 20
x = 10
2y = 20 - 10
2y = 10
y = 10 / 2
y = 5

5 + 2z = 9
2z = 9 - 5
2z = 4
z = 4 / 2
z = 2

z = x + yz
a = 10 + 5 . 2
a = 20


But then Myriad pointed out that the symbol in the third row was not the same as the symbol in the fourth, and that threw everything out.
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Old 20th February 2019, 06:07 PM   #6
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(x+1/2z) x 10= LX
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Old 20th February 2019, 06:16 PM   #7
MEequalsIxR
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Originally Posted by Myriad View Post
The correct answer is "unknown," because the value of the single-strut symbol is not established. All we can determine is that the double-strut symbol has a value of 2. (We have to treat the double-strut as a single whole symbol because there is no space and no operand between the struts of the double-strut where it appears here.)

We might guess that the single-strut symbol must therefore have a value of 1, but by the same reasoning we could likewise guess that the symbol "3" must have half the value of the symbol "8," or that the symbol "11" must have twice the value as the symbol "1".

I bet 99.99% of people wouldn't be that pedantic about it, though.
Myriad I like the way you think. I have to confess being a little miffed at myself for not seeing that.

The same thread has another similar graphic but there are two undefined symbols. I didn't fall into that trap but did in the first one.
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Old 20th February 2019, 06:21 PM   #8
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If two shocks is 2, then one shock is one. So the final answer is either (5 + 1) * 10 = 60; or 5 + (1 * 10) =15. I'd go with the latter.

I, of course, originally borked the whole thing by noticing neither the change from two shocks to one nor the times sign.
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Old 20th February 2019, 07:00 PM   #9
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Originally Posted by casebro View Post
(x+1/2z) x 10= LX
Not according to PEMDAS protocol.
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Old 20th February 2019, 07:06 PM   #10
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Isn't it another scam to get info from your FB account?
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Old 20th February 2019, 07:43 PM   #11
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This is basically a Monte Hall type problem. In order to get that "algebraic" result, you have to make an assumption, which may or may not be warranted. If your assumption is correct, then you can solve it. But you can't assume the assumption is correct.
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Old 20th February 2019, 07:53 PM   #12
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Originally Posted by pgwenthold View Post
This is basically a Monte Hall type problem. In order to get that "algebraic" result, you have to make an assumption, which may or may not be warranted. If your assumption is correct, then you can solve it. But you can't assume the assumption is correct.

What assumption do you have to make in the Monty Hall problem?
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Old 20th February 2019, 08:12 PM   #13
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Originally Posted by arthwollipot View Post
I figured that the answer was

20, by simple algebra. Here's my work:

3x = 30
x = 3x / 3
x = 10

x + 2y = 20
x = 10
2y = 20 - 10
2y = 10
y = 10 / 2
y = 5

5 + 2z = 9
2z = 9 - 5
2z = 4
z = 4 / 2
z = 2

z = x + yz
a = 10 + 5 . 2
a = 20


But then Myriad pointed out that the symbol in the third row was not the same as the symbol in the fourth, and that threw everything out.
How is this part correct:
5 + 2z = 9
2z = 9 - 5
2z = 4
z = 4 / 2
z = 2
If you use your values you get 1/2 z or one shock in the last calculation.

Seems so simple: 5 + 1 + 10 = 16
Unless they are going with you can't add those items together. One shock isn't half of a double shock?
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Old 20th February 2019, 08:25 PM   #14
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Originally Posted by Skeptic Ginger View Post
How is this part correct:
5 + 2z = 9
2z = 9 - 5
2z = 4
z = 4 / 2
z = 2
If you use your values you get 1/2 z or one shock in the last calculation.

Seems so simple: 5 + 1 + 10 = 16
Unless they are going with you can't add those items together. One shock isn't half of a double shock?
Like me, you've missed that there's a multiplication sign in the last row. So it's either (5+1)*10 = 60, or 5 + (1*10) = 15.
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Old 20th February 2019, 08:45 PM   #15
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All I know is, I'm shocked, tired, and piston ways you can't imagine.
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Old 20th February 2019, 08:50 PM   #16
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The whole PEMDAS concept is fundamentally broken.

https://www.youtube.com/watch?v=y9h1oqv21Vs
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Old 20th February 2019, 08:50 PM   #17
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Originally Posted by Trebuchet View Post
Like me, you've missed that there's a multiplication sign in the last row. So it's either (5+1)*10 = 60, or 5 + (1*10) = 15.
Of course it would be something obvious.
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Old 20th February 2019, 09:18 PM   #18
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Originally Posted by Trebuchet View Post
If two shocks is 2, then one shock is one.
The "two shocks" symbol isn't two "one shock" symbols. It's one "two shocks" symbol.
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Old 20th February 2019, 09:19 PM   #19
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Originally Posted by JoeMorgue View Post
The whole PEMDAS concept is fundamentally broken.

https://www.youtube.com/watch?v=y9h1oqv21Vs
I'd rather not watch a video without some idea of what it's trying to say. Is it a real argument or a joke? PEMDAS is just a convention.
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Old 20th February 2019, 09:34 PM   #20
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Originally Posted by BowlOfRed View Post
I'd rather not watch a video without some idea of what it's trying to say. Is it a real argument or a joke? PEMDAS is just a convention.

It's an error. The video misrepresents PEMDAS as meaning always multiply before dividing and always add before subtracting, rather than M/D and A/S each being equal priorities resolved left to right when not otherwise grouped. And then shows that if you apply those wrong rules, you get wrong results.
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Old 20th February 2019, 09:38 PM   #21
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Originally Posted by BowlOfRed View Post
I'd rather not watch a video without some idea of what it's trying to say. Is it a real argument or a joke? PEMDAS is just a convention.
It's actually a real argument and a joke.


Or, what Myriad said.
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Old 20th February 2019, 09:58 PM   #22
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Originally Posted by alfaniner View Post
All I know is, I'm shocked, tired, and piston ways you can't imagine.
All three in one sentence. Well done!
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Old 20th February 2019, 10:08 PM   #23
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sorry for being dense, but even if one shock is 1/2 of two shocks, how can the answer possibly be 60 ?
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Old 20th February 2019, 10:13 PM   #24
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Originally Posted by Myriad View Post
The correct answer is "unknown," because the value of the single-strut symbol is not established.


Frak.

Didn't notice that.
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Old 20th February 2019, 10:13 PM   #25
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Originally Posted by The Great Zaganza View Post
sorry for being dense, but even if one shock is 1/2 of two shocks, how can the answer possibly be 60 ?

By not realizing that the one-shock isn't a two-shock and then concluding 5x12=60.
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Old 20th February 2019, 11:52 PM   #26
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I am guessing that one is expected to do the operation in entry order, as one might punch it into a calculator, giving 15, to generate the "gotcha" that it's really 60 because PEMDAS.

This of course requires a level of trust in the puzzle maker that the very form of the puzzle puts in doubt.

The obvious answer, I think, is "it depends."
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Old 21st February 2019, 12:42 AM   #27
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The puzzle seems to me (like the second one that MEequalsIxR posted) to be a deliberate trap. I don't know what it's purpose is. Perhaps to make people feel stupid. Why someone would want to do that is beyond me.
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Old 21st February 2019, 02:39 AM   #28
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Originally Posted by arthwollipot View Post
The puzzle seems to me (like the second one that MEequalsIxR posted) to be a deliberate trap. I don't know what it's purpose is. Perhaps to make people feel stupid. Why someone would want to do that is beyond me.
It's a classic instance of https://xkcd.com/169/.

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Old 21st February 2019, 02:49 AM   #29
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The answer is immaterial. They are designed to push clicks to sites that can then be monetized.
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Old 21st February 2019, 03:25 AM   #30
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3a = 30 ∴ a=10

10 + 2b = 20 ∴ b=5

5 + 2c + 2c = 9 ∴ c=1

5 + 1 X 10 = 15
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Old 21st February 2019, 05:06 AM   #31
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Originally Posted by RecoveringYuppy View Post
What assumption do you have to make in the Monty Hall problem?
As the problem is typically stated, you have to assume that Monte is playing fair.
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Old 21st February 2019, 05:16 AM   #32
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Originally Posted by pgwenthold View Post
As the problem is typically stated, you have to assume that Monte is playing fair.
Which, if I recall correctly, Monty himself stated was not the case.

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Old 21st February 2019, 05:22 AM   #33
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Originally Posted by pgwenthold View Post
As the problem is typically stated, you have to assume that Monte is playing fair.
Originally Posted by Dave Rogers View Post
Which, if I recall correctly, Monty himself stated was not the case.

Dave

No, the assumption is that Monty is not playing fairly. He opens a box knowing there's no prize behind it. If he were to open a box at random, there would be some chance that he'd actually reveal the prize (and, thus, there'd be no benefit to switching your guess). Monty must show you a box he knows is wrong in order for the problem to work.
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Old 21st February 2019, 05:26 AM   #34
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Originally Posted by Loss Leader View Post
No, the assumption is that Monty is not playing fairly. He opens a box knowing there's no prize behind it. If he were to open a box at random, there would be some chance that he'd actually reveal the prize (and, thus, there'd be no benefit to switching your guess). Monty must show you a box he knows is wrong in order for the problem to work.
That's fair, it's just not random. Random <> fair.
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Old 21st February 2019, 05:42 AM   #35
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Originally Posted by Loss Leader View Post
No, the assumption is that Monty is not playing fairly. He opens a box knowing there's no prize behind it. If he were to open a box at random, there would be some chance that he'd actually reveal the prize (and, thus, there'd be no benefit to switching your guess). Monty must show you a box he knows is wrong in order for the problem to work.
It's rather more complicated than that, and it depends on your definition of "fair". It's stipulated in the formulation that Monty knows which box contains the prize, and will never open that box; the assumption that is then commonly made is that Monty always opens an empty box, in which case switching boxes can be calculated to give a 2/3 chance of winning. In an interview, though, Monty himself said that he had the choice whether to open an empty box and offer the switch or not, and that he based that choice on his reading of the situation and of the contestant. Knowing that he had that choice blows the standard analysis out of the water; if he only ever offered a switch when he knew the contestant had already selected the right box, then the odds would be zero for switching, whereas if he only offered it when the contestant was wrong, the odds would be zero for sticking. The real life situation, therefore, was that the probability of getting the right box on switching was somewhere between zero and one (i.e. completely unknown), and in order to get a better estimate one would have to analyse Monty's game theory and thought processes.

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Old 21st February 2019, 06:14 AM   #36
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Originally Posted by pgwenthold View Post
As the problem is typically stated, you have to assume that Monte is playing fair.
If he's using three cards, I wouldn't.
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Old 21st February 2019, 06:40 AM   #37
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Originally Posted by The Great Zaganza View Post
sorry for being dense, but even if one shock is 1/2 of two shocks, how can the answer possibly be 60 ?
if you do the adding before the multiplication you get 5 + 1 = 6, then 6 x 10 = 60.
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Old 21st February 2019, 06:43 AM   #38
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In addition to Myriad's correct observation about the shock, I didn't see any direction stating that any of the pictured items represented a consistent value. Hyper pedantic, yes, but showing pictures of car parts is not exactly classical variable nomenclature.
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Old 21st February 2019, 07:22 AM   #39
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Originally Posted by Trebuchet View Post
If two shocks is 2, then one shock is one. So the final answer is either (5 + 1) * 10 = 60; or 5 + (1 * 10) =15. I'd go with the latter.

I, of course, originally borked the whole thing by noticing neither the change from two shocks to one nor the times sign.
I missed the single shock and ended up with 25. Duh.
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Old 21st February 2019, 07:27 AM   #40
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dupe
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