Fie x = prețul inițial.
[tex]\it 25\%=\dfrac{\ \ 25^{(25}}{100}=\dfrac{1}{4}\\ \\ \\ I)\ Dup\breve a\ prima\ majorare, \ pre\c{\it t}ul\ devine:\ x+\dfrac{1}{4}x=x(1+\dfrac{1}{4})=\dfrac{5}{4}x\\ \\ \\ II)\ Dup\breve a\ a\ doua\ a\ majorare, \ pre\c{\it t}ul\ devine:\ \dfrac{^{4)}5}{\ 4}x+\dfrac{1}{4}\cdot\dfrac{5}{4}x=\dfrac{25}{16}x[/tex]
[tex]\it III)\ Dup\breve a\ reducerea\ cu\ p\%,\ rezult\breve a:\\ \\ \dfrac{25}{16}x-\dfrac{p}{100}\cdot\ \dfrac{\ \ 25^{(25}}{16}x=x\Big|_{:x} \Rightarrow \dfrac{25}{16}-\dfrac{p}{64}=1\Big|_{\cdot64} \Rightarrow 100-p=64 \Rightarrow p=36\\ \\ \\ R\breve aspuns\ corect:\ \bf A.\ \it 36\%[/tex]
