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The "average" of a series is x. A new data point comes in with the value x + 1.

But the new average is less than x.

How is this possible?

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  • $\begingroup$ Why is average in quotation marks? $\endgroup$ – Tweakimp Dec 12 '17 at 11:00
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This feels like there are probably a number of solutions all of which depend on the original statement being wilfully misleading, but how about this one?

The "average" is some sort of rolling/weighted average, where datapoints' weights depend on how long ago they are. As the new datum comes in, old weights change. Simplest example: consider a simple moving average where we take the average of the last (say) 10 points. If the oldest one in the current average is $x+10$ and a new one $x+1$ comes along, the average will decrease.

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  • $\begingroup$ Bingo! Exactly what I had in mind. $\endgroup$ – Ajay Brahmakshatriya Dec 12 '17 at 10:31
  • $\begingroup$ I am new to puzzling.stackexchange and this is my first question. Can you tell me why is it being downvoted so much? Is it just the quality of the puzzle or the way it is worded? I am asking so I can keep it in mind next time. $\endgroup$ – Ajay Brahmakshatriya Dec 13 '17 at 3:59
  • $\begingroup$ @Ajay This puzzle seems to be more of a "guess what I'm thinking" puzzle, that depends on information that isn't explicitly stated in the puzzle. That means that there isn't really one unambiguous answer, which is what we aim for on Puzzling. For most puzzles, once I see an answer, I think, "Oh, yes, that's obviously the correct answer." For this one, it was not at all obvious that Gareth's answer was definitively the correct answer. $\endgroup$ – GentlePurpleRain Dec 14 '17 at 18:23
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I failed math in school but could the series be

modular 2?

So the average (x) = 1 (or 1.#)

and x + 1 in a mod 2 system results in 0 (or 0.#)?

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  • $\begingroup$ I get what you are trying to suggest, but it would be wrong to say that the new data point added is x+1. $\endgroup$ – Ajay Brahmakshatriya Dec 12 '17 at 10:30
  • $\begingroup$ @AjayBrahmakshatriya I accept that this isn't the intended answer but are you certain that "it would be wrong to say that the new data point added is x+1"? I just checked in code and 2 mod (2 + 1) = 1 so in a "2 mod" 'system' 2 added to 1 does result in 1 which is less than 2 (but like I said, I failed math in school). $\endgroup$ – Brent Hackers Dec 12 '17 at 10:32
  • $\begingroup$ I meant that the new point being added to the series is (x+1)%2. Mathematically you are correct. I just meant that the wording of my question doesn't go with this. $\endgroup$ – Ajay Brahmakshatriya Dec 12 '17 at 10:36
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There are 3 kind of average : mean, modus and median.

The first average is mean, but the second average is modus or median.

Example 1,2,2,31. Average (Mean) = 9.
You add 10 to the data the data became 1,2,2,10,31
the average(median) become 2.
2 < 9

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  • $\begingroup$ You are right about the fact that there are multiple kind of averages. But here I am looking only at mean. But different kind of averages is on the right path. $\endgroup$ – Ajay Brahmakshatriya Dec 12 '17 at 10:08
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Just to be clear, the mathematical definition of a serie does not clearly admit a mean. Probably you refer to a set of numbers, rather than a serie or even a sequence.

Assuming that, a possible solution is to be in the ring of integers modulo n. For example, in n=2. Here, the set {1} has an average of x=1. By adding x+1=2=0, we have {1,0}, and therefore we are deceasing the average to 0.5

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There is my answer, this looks like not possible mathematically :

two number $a,b$ of average $x$. so $(a+b)/2 = x$
if whe add $x+1$ , the new average $y$ is $(a+b+(x+1))/3$
$a+b+x+1 = 3y$
with $a+b = 2x$ we have $2x + x + 1 = 3y $
$3x + 1 = 3y$
$x + 1/3 = y$
so $x < y$

that's seems not possible so i tried with $n$ number $(n > 0)$

at the end i have something like $y = x + 1/(n+1)$ so $x < y$ again, this probleme looks like unsolvable for me, but maybe need lateral thinking.

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  • $\begingroup$ @JamalSenjaya I just prove that it's not possible mathematically $\endgroup$ – Skyvask Dec 12 '17 at 9:07
  • $\begingroup$ There are 3 type of average : Modus, Mean, Median. You can explore another kind of average. $\endgroup$ – Jamal Senjaya Dec 12 '17 at 9:16
  • $\begingroup$ This doesn't prove anything; it's a mess of random symbols. Furthermore, it's quite obvious that the question is not possible with an arithmetic mean. $\endgroup$ – Wen1now Dec 12 '17 at 10:28
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With a small amount of lateral thinking:

If we have a series with a negative average, adding a datapoint 1 above the average makes the new average greater, i.e. "less average"

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  • $\begingroup$ "But the new average is less than x" I thought about negative average too, but the new average will always be greater than the previous one, even if it's negative... -1, -11 : -6 / -1, -5, -11 : -5.66667 / -6 < -5.666667 $\endgroup$ – Saeïdryl Dec 12 '17 at 10:15
  • $\begingroup$ Yes, but 5.66... is less negative (i.e. greater) than 6. In other words, the magnitude is less. $\endgroup$ – Lolgast Dec 12 '17 at 10:17
  • $\begingroup$ What is the series is modular? Where the numbers 'wrap around' so a series where mod(3) 3 + 1 = 1 $\endgroup$ – Brent Hackers Dec 12 '17 at 10:20
  • $\begingroup$ @Lolgast I understand your point but "less than" could not be anything else than "<", if "less than" means something else in this riddle, that's misuse of words. $\endgroup$ – Saeïdryl Dec 12 '17 at 10:30
  • $\begingroup$ @Saeïdryl That's why I explicitly wrote "with a small amount of lateral thinking" at the very top of my question. $\endgroup$ – Lolgast Dec 12 '17 at 10:32

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