[tex]\it Fie\ \ p\ -\ pre\c{\it t}ul\ \ ini\c{\it t}ial\\ \\ \\ 20\%=\dfrac{\ \ 20^{(20}}{100}=\dfrac{1}{5}[/tex]
După prima majorare, prețul devine:
[tex]\it p+\dfrac{1}{5}\cdot p=p\cdot\Big(1+\dfrac{1}{5}\Big)=p\cdot\dfrac{6}{5}[/tex]
După a doua majorare, prețul devine:
[tex]\it p\cdot\dfrac{6}{5}+\dfrac{1}{100}\cdot p\cdot\dfrac{6}{5}=p\cdot\dfrac{6}{5}\Big(1+\dfrac{1}{100}\Big)=p\cdot\dfrac{6}{5}\cdot\dfrac{101}{100}=p\cdot\dfrac{606}{500}[/tex]
Dar, ultimul preț este cunoscut, deci:
[tex]\it p\cdot\dfrac{606}{500}=6060|_{:606} \Rightarrow p\cdot\dfrac{1}{500}=10|_{\cdot500} \Rightarrow p=5000\ \ell ei[/tex]
