# Manipulate percentages like a Pro

We learned in the previous post to add a percentage with a multiplication, let’s now learn the different manipulations of percentage.

The first of these manipulation is to “take a percentage” (\x\%\) of
a certain quantity $$A$$:
$$x\%$$ of the quantity A is obtained in dividing x by 100, and then multiply A by the result.
But a division by 100 is very easy to process:
- if the number $$x$$ is a two digits integer, for instance $$x=25\%$$, then $$x÷100$$ is a zero followed by a decimal point, and then the two digits: $$25÷100=0.25$$.
So that take $$25\%$$ of a quantity is multiply it by $$0.25$$.
- if the number $$x$$ is a one digits integer, for instance $$x=5\%$$, then $$x÷100$$ is a zero followed by a decimal point and an extra $$0$$, and then the digit: $$5÷100=0.05$$. So that take $$5\%$$ of a quantity is multiply it by $$0.05$$.
- if the number $$x$$ is a more than two digits integer, for instance $$x=325\%$$,then $$x÷100$$ is obtained by inserting a decimal point just before the two last digits: $$325÷100=3.25$$. So that take $$325\%$$ of a quantity is multiply it by $$3.25$$.

The second manipulation is to “add a percentage” $$x\%$$, that is a multiplication by $$(1+(x÷100))$$
), as we saw in the post “Add a percentage is a multiplication”. Thus, when you have divided $$x$$ by $$100$$ as explained above, just add $$1$$ to it, before you multiply the quantity by the result. For instance, add $$15\%$$ to a quantity is multiplying it by $$1.15$$.

But what about the operation “subtract a percentage”? As it seems to be the reverse operation from adding a percentage, it should be a division, no?
No, is is also a multiplication, and it is NOT the reverse operation than adding a percentage.
Namely, it is taking the percentage of the quantity and then subtracting the result to the quantity. Or else multiply the quantity by $$(1-x÷100)$$ to subtract $$x\%$$, because of the process of extracting a percentage, and because of the distributivity of the multiplication on the subtraction.
For instance, if you have a $$40\%$$ coupon on a good priced
$$$150$$, then all you have to pay is $$(1-0.40) \times$$$$$150$$=$$0.60 \times$$$$$150=$$$$$90$$.

And let’s finish that post by showing how the operations “add $$x\%$$” and then “subtract $$x\%$$”, and reversely, behave on the given quantity $$A$$.

First, adding $$x\%$$, and then subtracting $$x\%$$, is consecutilely
multiplying by $$(1+x/100)$$, and then by $$(1-x/100)$$, that is
multiplying by the product $$(1-x/100)(1+x/100)$$, that is not
equal to $$1$$, but a little less: $$(1-(x/100)^2)$$. You have subtracted $$(x^2/100)\%$$ to the initial quantity.
For instance, if you add $$20\%$$ to $$100$$, and then subtract $$20\%$$ to the result, you obtain $$0.80 \times 1.20 \times 100 = 96$$, that is $$4\%$$ less than $$100$$. And $$20^{2}/100=400/100=4$$

But what if subtracting $$x\%$$ first, and then adding $$x\%$$?
The result is the same because of the commutativity of the multiplication.
Indeed, it is subsequently multiplying by $$(1-x/100)$$, and then
by $$(1+x/100)$$, that is multiplying by the product $$(1+x/100)(1-x/100)$$, that is equal to $$(1-(x/100)^2)$$as well.

Thus the operations “add $$x\%$$” and “subtract $$x\%$$” to a given
quantity $$A$$ commute for a given percentage $$x\%$$.
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