Profit & Loss (Solution)

Solution (Question):
\[Cost\;Price = $x\]
\[\%\;Profit = 20\%\]
\begin{align}
Profit & = 20\% \times $x \\
& = $\frac{1}{5}x \\
\end{align}
\begin{align}
Selling\;Price\;before\;Sale & = 120\% \times $x \\
& = $\frac{6}{5}x \\
\end{align}
\[\%\;Loss\;during\;Sale = 20\%\]
\begin{align}
Selling\;Price\;after\;Sale & = 80\% \times $1.2x \\
& = \frac{4}{5} \times $\frac{6}{5}x \\
& = $\frac{24}{25}x \\
\end{align}
\begin{align}
Loss & = $x - $\frac{24}{25}x \\
& = $\frac{1}{25}x \\
\end{align}
\[\%\;Loss = 4\%\]


Solution (Extension):
\[Cost\;Price = $x\]
\[\%\;Profit = y\%\]
\[Profit = y\% \times $x\]
\begin{align}
Selling\;Price\;before\;Sale & = \left(100 + y\right)\% \times $x \\
& = $x \left(1 + \frac{y}{100}\right) \\
\end{align}
\begin{align}
Selling\;Price\;after\;Sale & = \left(100 - y\right)\% \times $x \left(1 + \frac{y}{100}\right) \\
& = $\left(1 - \frac{y}{100}\right) \left(1 + \frac{y}{100}\right)x \\
& = $\left[1^2 - \left(\frac{y}{100}\right)^2\right] x \\
& = $x \left(1 - \frac{y^2}{10000}\right) \\
\end{align}
\begin{align}
Since\;0 < y < 100,\; then\;& 0 < y^2 < 10000. \\ \Rightarrow & 0 < \frac{y^2}{10000} < 1 \\ \Rightarrow & 0 < 1 - \frac{y^2}{10000} < 1 \\ \Rightarrow & Loss \end{align} \begin{align} Loss & = $x - $x \left(1 - \frac{y^2}{10000}\right) \\ & = $\frac{xy^2}{10000} \\ \end{align} \[\%\;Loss = \frac{y^2}{100}\%\]