Is there a way to derive the percentage to add, from the percentage to subtract?

Question

Here is an example:

24.30 + 66.6% = 40.50
40.50 - 60% = 24.30
or (24.30 + 66.6%) - 60% = 24.30

I know if I add $66.6\%$ to $24.30$, I get $40.50$ and if I subtract $60\%$ from $40.50$ I get the other value back (values are being rounded here!)

$$(X + Y\%) - Z\% = X$$

My variable is $60\%$ ($Z\%$). Is there a way to derive $Y\%$ from that, so I don't have to calculate it ($Y\%$) by hand each time ?

Thanks!

Answer 1

Your notation is a bit deceptive.

What you obviously mean instead of Adding 60% to X is Adding 60% **of X** to X which means the same as Multiply X by 1,60

In your concrete example you want to add Y% of X to X, and afterwars you want to substract 60% of this result from this result.

You have to find Y such that $$ X\cdot Y \cdot \left(1-0.60\right) = X\cdot Y \cdot 0.40 = X $$ which means that Y has to be the inverse of $0.40$ which is equal to 2.50

The result means, that you have to add $2.50-1.00=1.50 = 150$% of X to X in the first step.

Answer 2

Let's start with the fact the percentage are not magic numbers, no need to express it with %.. they are just numbers. $60\%$ means $\frac{60}{100} = 0.6$

Saying "add 60 % only means that, taken any number $x$, the resulting number will be

$$x + 60\% \ x = x + 0.6x = x(1 + 0.6) = 1.6x = z$$

In the general case, an increment of $b$ means $x(1+b) = z$

Now you want to find a number such that

$$z - az = z(1 - a) = x$$ Since $x = \frac{z}{1+b}$, you can write $$z(1-a) = \frac{z}{1+b} \Rightarrow a = \frac{b}{1+b}$$

for example if $b = 40\% = 0.4$, and $x = 40.25$, we have

$40.25 + 40\% (40.25) = 56.36$; From the formula above we get $a = \frac{0.4}{1.4} = 0.2857 = 28.57\%$

And in fact $56.36 - 28.57\% (56.36) = 40.25$