"Reverse" percentage difference

Question

Using the percentage difference formula, explained here, with the numbers $1$ and $1.5$, I have a result of $40\%$.

So, if i have a $100$ value, how can I find two numbers with a formula that add up to $100$ with the same $40\%$ percentage difference? For example, $60$ and $40$ have a difference of $40\%$ and $60 + 40 = 100$

i.g., for $4$ and $16$: $\frac{|4-16|}{(4+16)/2} * 100 = 120\% $, I would need two values that $x + y = 100$ and $\frac{|x-y|}{(x+y)/2} * 100 = 120\% $

Here is a useful and simple calculator for percentage difference.

This is my first question and I'm not very good at math or English, so I'm sorry for any mistakes.

Answer 1

To solve your question in a general manner, first note there's the required condition

$$x + y = 100 \tag{1}\label{eq1A}$$

Since your equations are symmetric in $x$ and $y$, then WLOG have $x \ge y \implies \lvert x - y \rvert = x - y$. Next, using $120\% = \frac{120}{100} = 1.2$ gives that

$$\begin{equation}\begin{aligned} \frac{x - y}{(x + y)/2} & = 1.2 \\ \frac{x - y}{100/2} & = 1.2 \\ x - y & = 1.2(50) \\ x - y & = 60 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Now, adding \eqref{eq1A} and \eqref{eq2A} results in

$$2x = 160 \implies x = 80 \tag{3}\label{eq3A}$$

Using this in \eqref{eq1A} then gives

$$80 + y = 100 \implies y = 20 \tag{4}\label{eq4A}$$

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Since you already have that $4$ and $16$ give the required difference percentage, here is a bit shorter and easier way to solve the problem. Note that for any $c \neq 0$, we get

$$\begin{equation}\begin{aligned} \frac{\lvert cx - cy \rvert}{(cx + cy)/2} & = \frac{c\lvert x - y \rvert}{c(x + y)/2} \\ & = \frac{\lvert x - y \rvert}{(x + y)/2} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

i.e., multiplying both $x$ and $y$ by $c$ (and, thus, the sum of them) results in the same value difference percentage. Thus, since $4 + 16 = 20$ and $\frac{100}{20} = 5$, we can use $c = 5$ to directly get that the values of $x$ and $y$ are $4 \times 5 = 20$ and $16 \times 5 = 80$.

Answer 2

Now that I actually know what "percentage difference" means (it was a term I had not heard used in that way before), I can help answer the OP's question, and maybe also pose and answer a related question. This might help the OP or anyone else who happens to be working with this sort of thing.

So OP's question was: given a particular percentage difference (call it $P$), and two numbers (call them $b$ for big and $s$ for small) with a known fixed total (call it $T$), can we determine what the two numbers must be?

So we can write: $\frac{b-s}{0.5T} \cdot 100 = P$ which may be rearranged to $b-s = \frac{PT}{200}$

This equation can be solved simultaneously with $b+s = T$ by first adding both to isolate $b$:

$2b = \frac{PT}{200} + T$

$b = \frac T{400}(200+P)$

and then subtracting one from the other to isolate $s$. By analogous steps we end up with:

$s = \frac T{400}(200-P)$

Now we have fairly neat expressions for both the numbers. The OP can use these expressions by substituting for $T$ ($100$ as per his requirement in this question) and $P$ as he sees fit.

But we can also make some interesting observations from the form of the expressions. Note that if we require both numbers to be non-negative, we require $P \leq 200$. If we need both numbers to be positive, we need $P < 200$. Otherwise, the smaller number will necessarily be negative.

Let's go a little further. We've dealt with a fixed total ($T$). A reasonable additional question to pose would be to ask what happens if we're asked to deal with a fixed difference ($D$). In other words, what should $b$ and $s$ be to arrive at a particular $P$ given a particular $D$?

We again start from the definition and write:

$\frac{D}{0.5(b+s)}\cdot 100 = P$ which may be rearranged to: $b+s = \frac{200D}{P}$

This can be solved simultaneously with $b-s = D$ in a similar fashion to what we did earlier to yield:

$b = \frac{D}{2P}(200 + P)$

and

$s = \frac{D}{2P}(200 - P)$

Note that we can deduce the same conclusions about the upper bound of $P$ for both numbers to be non-negative (or positive) from this set of equations.

In summary, for a fixed total $T$ and a particular percentage difference $P$:

Bigger number: $\displaystyle b = \frac T{400}(200+P)$

Smaller number: $\displaystyle s = \frac T{400}(200-P)$

And for a fixed difference $D$ and a particular percentage difference $P$:

Bigger number: $\displaystyle b = \frac{D}{2P}(200 + P)$

Smaller number: $\displaystyle s = \frac{D}{2P}(200 - P)$

I did this mainly for my own amusement, but I hope it helps someone at some point.