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Considering that

50+20=60

and

60+30=72,

what's 40+30?


Edit (now that this has been answered, I can state): This was earlier posted to alt.usage.english as part of its 2008 Summer Doldrums Competition. (I was the author of that puzzle.) Google has an archive of the discussion at the time.

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4 Answers 4

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The answer is

$58$

Because

$a\star b = a + b - \frac{ab}{100}\\50\star20 = 50+20-5\times2 = 60\\60\star30 = 60+30 - 6\times3 = 72$

So

$30\star40 = 30+40 - 3\times4 = 58$

Use of title and $+$ sign

This is almost the same as the total $\%$ increase when a value is increased by $a\%$ and then by $b\%$ (that would be $a + b +\frac{ab}{100}$) so perhaps 'addition problem' refers to the incorrect sign? The $+$ would then be used to mean that the percents were 'added', but that feels like I'm just grasping at straws.


Note: I've used $\star$ instead of $+$ as was used in the question to make the answer more clear to read - it is confusing to use $+$ as the symbol for both the addition operator and the 'mystery operator'.

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  • $\begingroup$ Yes, excellent, that's the correct answer, and +1, but you haven't explained the use of the "+" (also alluded to, though weakly, in the title). Why is this called addition? $\endgroup$
    – msh210
    May 29, 2016 at 8:59
  • $\begingroup$ I honestly can't think of anything. $\endgroup$
    – KoA
    May 29, 2016 at 9:03
  • $\begingroup$ If you think of it and edit it in, then please ping me in a comment. $\endgroup$
    – msh210
    May 29, 2016 at 9:07
  • $\begingroup$ @msh210 that's the best I've got. $\endgroup$
    – KoA
    May 29, 2016 at 9:21
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KoA already gave the numerical answer, but here's the reasoning:

The numbers are probabilities of events (or whatever), expressed as percentages. The addition operator represents "or" — that is, the probability of either two of the events occurring. Assuming the events are independent, the probability of either of them occurring is $a + b - a \times b$, or as percentages, $\left(\frac a{100} + \frac b{100} - \frac a{100} \times \frac b{100}\right)\times100$, which simplifies to the equation KoA gave.

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  • $\begingroup$ That definitely seems much more plausible than what I said :D $\endgroup$
    – KoA
    May 29, 2016 at 12:03
  • $\begingroup$ Hm, that's a good explanation, +1; and not what I was thinking of. :-) I'm still hoping someone will come up with an answer that's more "additional". $\endgroup$
    – msh210
    May 29, 2016 at 13:50
  • $\begingroup$ The +1 is not because it explains the "addition" but because it's a step toward that explanation. $\endgroup$
    – msh210
    May 29, 2016 at 15:19
  • $\begingroup$ @msh210 man, I was certain about this one... the addition sign is commonly used when expressing this particular operation. I'm curious about what it turns out to be. Got a couple of ideas, but they're far more complicated and contrived. $\endgroup$
    – Sneftel
    May 29, 2016 at 16:01
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You're calling this addition because it's a form of "adding" discounts to sale prices.

A lot of outlet stores have advertised discounts of, say, 20% off everything in the store plus 50% off certain items. This results in a total discount of $1 - (1 - 20\%) (1 - 50\%) = 60\%$, rather than the 70% you'd get by adding things together straight.

If it's 30% off everything and 60% off a certain item, the total discount would 72% instead of 90%.

And 40% off plus 30% off would be 58% off, as some people have mentioned.


In fact, the real numbers in the interval $(0,1)$ form an abelian semigroup (or a monoid if you include 0) under the operation $a + b - ab$, just like addition over the positive (/nonnegative for the monoid) real numbers. So it can rightfully be called "addition" in that way too.

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  • $\begingroup$ Bravo! This is the addition. Now I'm torn as to whether to award you or KoA the checkmark. $\endgroup$
    – msh210
    May 29, 2016 at 19:13
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Or it could be

$60$

with

$a\star b = 6\left(3+\dfrac{a+b}{10}\right)$

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  • $\begingroup$ Yeah, all sorts of arithmetic formulas are possible but, as I commented on the other answers, there's a reason I called it addition. $\endgroup$
    – msh210
    May 29, 2016 at 15:18

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