# How to add up two percentages

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

• $\frac{a}{b}$ is not a "percentage," but instead a fraction. – David G. Stork Jun 6 at 22:05

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 $$A=\frac ab=\frac cd$$.

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.

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.

• $q$ cannot be zero as defined in the problem. As with $k$. – Kraig Jun 6 at 22:07
• @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. – TonyK Jun 6 at 22:10
• And if that doesn't convince you, try putting $q=2k$, and then $k=2q$. – TonyK Jun 6 at 22:10