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 $100q=100\left( \frac{8}{100}\right) ^{2}=0.64.$