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: 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 $(P^2/100)$%. 
A: Given that 


*

*Item 1 was bought at a cost I call $C_{1}$, sold at $X$ with a relative gain $p=\frac{X-C_{1}}{C_{1}
}>0$, with $p=\frac{P}{100}$, where $P$ is in percentage. 

*Item 2 was bought at a cost I call $C_{2}$, sold at $X$ with a relative loss $p=\frac{C_{2}-X}{C_{2}}>0$. 

*And items 1+2 were sold at $2X$, 
then the overall relative loss $q>0$ is given by $$q=\frac{C_{1}+C_{2}-2X}{C_{1}+C_{2}}.$$
From 1 and 2 we get respectivelly
$$C_{1} =\frac{X}{p+1},\qquad C_{2} =-\frac{X}{p-1}.$$
So $q$ can be rewritten as
$$q =\dfrac{\dfrac{X}{p+1}-\dfrac{X}{p-1}-2X}{\dfrac{X}{p+1}-\dfrac{X}{p-1}}=\dfrac{\dfrac{1}{p+1}-\dfrac{1}{p-1}-2}{\dfrac{1}{p+1}-\dfrac{1}{p-1}}=\dfrac{-2p^{2}}{-2}=p^{2}=\left( \dfrac{P}{100}\right) ^{2}.$$
The overall loss in percentage is 
$$100q=\frac{P^{2}}{100},$$
which proves the assertion.
Numerical example:


*

*Item 1: $C_{1}=100$, sold at $X=108$. The relative gain is $p=\frac{108-100}{%
100}=\frac{8}{100}=0.08$, $P=8.$

*Item 2: $C_{2}=\frac{108}{0.92}\approx 117.39$, sold at $X=108$. The relative loss is  $\frac{108/0.92-108}{108/0.92}=0.08=p$.

*Item 1 + Item 2: $C_{1}+C_{2}=100+108/0.92\approx 217.39.$ 


The overall loss is
$$q=\frac{100+108/0.92-2\cdot 108}{100+108/0.92}=0.0064=0.08^{2}=\left( \frac{8}{100}\right) ^{2},$$ 
and $$100q=100\left( \frac{8}{100}\right) ^{2}=0.64.$$
A: To work it out, the original costs of the items $C_1$ and $C_2$ are such that $\dfrac{X-C_1}{C_1}= \dfrac{P}{100}$ and $\dfrac{X-C_2}{C_2}= -\dfrac{P}{100}$ so $C_1=\dfrac{100X}{100+P}$ and $C_2=\dfrac{100X}{100-P}$.  Add the two costs together and you get $C_1+C_2=\dfrac{20000X}{10000-P^2}$
The overall profit percentage (it is a loss since it is negative) is 
$$ 100 \left(\dfrac{2X-(C_1+C_2)}{C_1+C_2}\right) = 100 \left(\dfrac{2X-\frac{20000X}{10000-P^2}}{\frac{20000X}{10000-P^2}}\right)  =  \dfrac{-P^2}{100} $$    
