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I have two fraction:

  • first one: $\frac{a}{b} $, where $0<\frac{a}{b}<1$
  • second one: $\frac{c}{d}$,where $0<\frac{c}{d}<1$

I apply this two percentages on two positive numbers: $q>0,k>0$ :

  • $\frac{a}{b} \cdot q = \alpha $
  • $\frac{c}{d} \cdot k = \beta$

I have to find an expression A depending only on $a,b,c,d$ such that $A(a,b,c,d) \cdot (q+k) = \alpha+\beta$

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    $\begingroup$ $\frac{a}{b}$ is not a "percentage," but instead a fraction. $\endgroup$ – David G. Stork Jun 6 at 22:05
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You have $\displaystyle A(q+k)=\alpha+\beta=\frac abq+\frac cdk\iff A=\frac q{q+k}\left(\frac ab-\frac cd\right)+\frac cd$

So to have $A$ depend only on $a,b,c,d$ we need either $(\frac q{q+k}=cst)$ or $(\frac ab=\frac cd)$.

Since $q,k$ are considered free variables, the first solution is to be discarded.

If $ad-bc\neq 0$ then there is no solution, else $A=\frac ab=\frac cd$.

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Just (try to) do it.

$A(a,b,c,d)(q+k) = \alpha + \beta$ so

$A(a,b,c,d) = \frac {\alpha + \beta}{q+k}= \frac {\frac ab q + \frac cd k}{q+k}=$

$\frac ab\frac {q + \frac {cb}{ad}k}{q+k}$

Such a number is dependent upon the values of $q$ and $k$. Simply try plugging in different values for $q$ and $k$ and you will get entirely different answers.

This can not be done.

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It can't be done.

Putting $q=0$ gives $A(a,b,c,d)\cdot k=\beta=\frac{c}{d}\cdot k$, therefore $A(a,b,c,d)=\frac{c}{d}$.

Similarly, putting $k=0$ gives $A(a,b,c,d)=\frac{a}{b}$, a contradiction.

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  • $\begingroup$ $q$ cannot be zero as defined in the problem. As with $k$. $\endgroup$ – Kraig Jun 6 at 22:07
  • $\begingroup$ @Kraig: Oh, you're right. Sorry! But the principle remains: make $q$ very small, and you can force $A(a,b,c,d)$ to be as close to $\frac{c}{d}$ as you like; make $k$ very small, and you can force $A(a,b,c,d)$ to be as close to $\frac{a}{b}$ as you like. $\endgroup$ – TonyK Jun 6 at 22:10
  • $\begingroup$ And if that doesn't convince you, try putting $q=2k$, and then $k=2q$. $\endgroup$ – TonyK Jun 6 at 22:10

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