Calculating overall Profit/Loss I was reading a text book and came across the following interesting short-cut:

If two items are sold, each at $X$, one at a gain of $P\%$ and the other at a loss of $P\%$, then overall loss percentage  $= P^2/100 \%$.

Can anyone please explain the underlying logic on the basis of which this shortcut works?
Thanks in advance!

A gain of P% that results in an item cost of $X$ means that the original price, $C_1$, was such that
$$\left(1 + \frac{P}{100}\right)C_1 = X.$$
A loss of P% that results in an item cost of $X$ means that the original price, $C_2$, was such that
$$\left(1 - \frac{P}{100}\right)C_2 = X.$$
In other words, your costs were
$$C_1 + C_2 = \frac {X}{1 + \frac{P}{100}} + \frac{X}{1-\frac{P}{100}} = \frac{100X}{100+P} + \frac{100x}{100-P}.$$
On the other hand, you received $X+X = 2X$. So what is the total profit/loss? It's equal to the amount received minus the amount spent:
$$\begin{align*}
\text{Profit} &= \text{Revenue} - \text{Cost}\\
&= 2X - \frac{100X}{100+P} + \frac{100X}{100-P} \\
&= 2X- \frac{100X(100-P)+100X(100+P)}{(100+P)(100-P)} \\
&= 2X - 100X\left(\frac{100-P+100+P}{(100+P)(100-P)}\right)\\
&= 2x - 100X\left(\frac{200}{(100+P)(100-P)}\right)\\
&= 2x\left( 1 - \frac{10000}{(100+P)(100-P)}\right)\\
&= 2X\left(1 - \frac{10000}{10000 - P^2}\right)\\
&= 2X\left(\frac{10000-P^2 - 10000}{1000-P^2}\right)\\
&= 2X\left(-\frac{P^2}{10000}\right)\\
&= 2X\left( - \frac{(P^2/100)}{100}\right).
\end{align*}$$
So your total loss is $(P^2/100)$%. 
