Question related to percents of increase and decrease … we know what happens if you increase an original quantity by adding x% of it, then subtracting that same x%: you’ll wind up with the quantity you started with. That is to say, the value of the original quantity will stay the same.
Now let’s pose the same question with regard to multiplication and division. Suppose you take an original quantity and increase it by x%. Then you turn around and decrease the new quantity by x%?
Will the new amount be: a) the same? b) greater? c) less? or d) might it depend on the value of x? What if you reverse the order … you first decrease the quantity by x%, and then you turn around and increase that new quantity by x%. Will the amount you wind up with be the same or different than what you got the other way around?
Many students intuitively think the final result will be the same as the original quantity. That makes this concept fun to teach; math lessons always are more fun when the results go against intuition.
We’ve already said that the amount of the increase and decrease is x%.
Now to keep the math as simple as possible, we’ll just use 1.00 as the original quantity. That way all values that we wind up with will automatically be in percent form.
When we increase our original quantity, 1.00, by x percent, we multiply it by 1.00 + x% = 1.00 + x/100 = 100/100 + x/100 = (100 + x)/100.
Similarly, when we decrease the original quantity, 1.00, by x percent, we multiply it by 1.00 – x% = 1.00 – x/100 = 100/100 – x/100 =(100 – x)/100.
So, then what are the steps performed in the problem? In the first way described, we 1st) increase the original quantity by x%, then 2nd) starting with the increased value, decrease it by x%. Let’s do that now.
1st) 1.00 [times] (100 + x)/100 = (100 + x)/100
2nd) (100 + x)/100 [times] (100 – x)/100 = (100 + x)(100 – x)/100^2 = (100^2 – x^2) / 10,000 = (10,000 – x^2) / 10,000.
We can draw a number of conclusions from this final algebraic expression, boldfaced.
1) It represents the quantity we are left with after increasing, then decreasing a quantity of 1.00, by x%.
2) Since this was done in a general way, this expression serves as a formula to predict the new value for any situation where an original quantity gets increased or decreased by the same percent.
Example: if $2,000 gets increased by 3%, then decreased by 3%, the amount left would be given by this:
2000 [times] (10,000 – 3^2) / 10,000 = 2000(10,000 – 9)/10,000 =
2000(9991/10,000) = 1998.2
3) Since this expression by which the original value gets multiplied does not changed, no matter the original value, no matter the percent, it behooves us to deeply understand this expression:
(10,000 – x^2) / 10,000
4) First, let’s notice that this expression does not change whether we increase the original value first, then decrease — or if we switch the order by decreasing the original value first, then increasing the result. We can see this is true because the numerator of the expression: (10,000 – x^2) arises as the product of (100 + x) and (100 – x). The commutative property guarantees that the order of the factors does not change their product.
So the point here is that it doesn’t matter whether we first increase the original quantity, then decrease it, or if we first decrease the original quantity, then increase it; we’ll get the same result each time.
5) Now let’s look at the relationship between the value of the percent, x, and the outcome, the change in the original quantity.
From the expression we see that as x^2 gets larger, the numerator,
10,000 – x^2, will get smaller. And of course, the value of x^2 increases as the value of x increases.
So in a transitive way of thinking, as the percent, x, that we increase and decrease by gets larger, the ultimate decrease of the original value gets larger. Or, slightly differently, as the percent of change we put the value through increases, the original quantity will end up decreasing more at the end.
Here’s a table that shows the relationship between the x-value, the percent of increase and decrease, and the percent of the original value you’re left with after the increase and decrease are carried out.
This is the first time I’ve explored this topic. So feel free to share your thoughts, insights on it.
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