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Good day to you all. I have a little problem I have been banging my head on for a while.

I have come to think that it is impossible, but I hope you can save me.

I have a fraction, $ \frac{num_i}{den_i} $., which takes different values over time.

The objective is to calculate $ \frac{average(num_i)}{average(den_i)} $.

I have at my disposal $ average (\frac{num_i}{den_i}) $ and $average(num_i) $

Is there any way to do this, or do I need to get $average(den_i)$ also?

Thanks a lot for your help.

Edit with an example

Let's take $\frac {1}{2},\frac{1}{3},\frac{1}{4}$

The information at my disposal is:

  • The average of numerators is 1.
  • The average of fractions is 0.36

Is it possible with the information I have to retrieve the average of denominators?

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closed as unclear what you're asking by Harish Chandra Rajpoot, user99914, Aloizio Macedo, N. F. Taussig, Ali Caglayan Oct 30 '15 at 10:10

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What do you mean by averages here? Each time it seems to me like you are taking the average of a single number. $\endgroup$ – davidlowryduda Oct 30 '15 at 8:09
  • $\begingroup$ I have added indexes for clarity. The fraction can take different values. For example, let's take $ \frac {1}{2}, \frac {1}{3}, \frac{1}{4} $ I know that the average of numerators is 1. I know that the average of fractions is 0.36 Is there any way from this to get that the average of denominators is 3? $\endgroup$ – Maxime Oct 30 '15 at 8:10
  • $\begingroup$ Are we supposed to assume that the fractions are in simplest form? $\endgroup$ – N. F. Taussig Oct 30 '15 at 10:04
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No, consider the example:

$A=\{{1\over2},{1\over3},{1\over6}\}$

$B=\{{1\over2},{1\over4},{1\over4}\}$

We have the numerator average $1$ and fraction average $1\over3$ for both cases yet average for denomiators are not the same.

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  • $\begingroup$ OK, thanks for the example, it helps the with the grieving process :( $\endgroup$ – Maxime Oct 30 '15 at 8:20
  • $\begingroup$ I am sorry but math only talks about truth and bad things can happen all the time. $\endgroup$ – cr001 Oct 30 '15 at 8:22
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This is impossible.

First case, consider the series $\frac{1}{2},\frac{2}{6}$.
Then, $avg(num)=\frac{3}{2}, avg(den)=4, avg(num/den)=\frac{5}{12}$ and $avg(num)/avg(den)=\frac{3}{8}$.

Second case, consider the series $\frac{2}{4},\frac{1}{3}$.
Then, $avg(num)=\frac{3}{2}, avg(den)=\frac{7}{2}, avg(num/den)=\frac{5}{12}$ and $avg(num)/avg(den)=\frac{3}{7}$.

In both cases your two givens are the same, but the answer is different. Therefore you cannot find the answer from the givens.

Edit: I see you got another answer that was both faster and better. Well done.

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