# Finding average of denominator knowing average of numerator and average of fraction Good day to you

Finding average of denominator knowing average of numerator and average of fraction
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, $\frac{nu{m}_{i}}{de{n}_{i}}$., which takes different values over time.
The objective is to calculate $\frac{average\left(nu{m}_{i}\right)}{average\left(de{n}_{i}\right)}$
I have at my disposal $average\left(\frac{nu{m}_{i}}{de{n}_{i}}\right)$ and $average\left(nu{m}_{i}\right)$
Is there any way to do this, or do I need to get $average\left(de{n}_{i}\right)$ 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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robegarj
No, consider the example:
$A=\left\{\frac{1}{2},\frac{1}{3},\frac{1}{6}\right\}$
$B=\left\{\frac{1}{2},\frac{1}{4},\frac{1}{4}\right\}$
We have the numerator average $1$ and fraction average $\frac{1}{3}$ for both cases yet average for denomiators are not the same.

kixEffinsoj
This is impossible.
First case, consider the series $\frac{1}{2},\frac{2}{6}$.
Then, $avg\left(num\right)=\frac{3}{2},avg\left(den\right)=4,avg\left(num/den\right)=\frac{5}{12}$ and $avg\left(num\right)/avg\left(den\right)=\frac{3}{8}$
Second case, consider the series $\frac{2}{4},\frac{1}{3}$
Then, $avg\left(num\right)=\frac{3}{2},avg\left(den\right)=\frac{7}{2},avg\left(num/den\right)=\frac{5}{12}$ and $avg\left(num\right)/avg\left(den\right)=\frac{3}{7}$
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.