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 \(x\%\).
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 \(x\%\).