Question

Pretun unui obiect se mareste de doua ori succesiv cu cate 20% .
a) Aflati pretul initial al obiectului stiind ca pretul acestuia dupa cele doua mariri este de 432.
b) Cu ce procent ar fi trebuit facuta o singura marire de pret astfel incat obiectul sa ajunga la pretul dupa cele doua mariri ?
-Pe a) l-am facut, iti da pretul inital 300. Da nu stiu la b)
Va rog ,ajutati-ma !

Answer 1

x- pret initial

x+20x/100=A
A+20A/100=432 /*100
100A+20A=43200
120A=43200
A=360

x+20x/100=360  /*100
100x+20x=36000
120x=36000
x=300 pretul initial

b)

300+b*300/100=432  amplificam cu 100
30000+300b=43200
300b=43200-30000
300b=13200
b=44, deci cu 44%

Answer 2

$a)x\to pretul \ initial \\\\ 20\% \ din \ x= \frac{\not20}{\not100} *x = \frac{1}{5}*x \to \frac{x}{5} -prima \ marire \\\\ x^{(5}+ \frac{x}{5}=\frac{5x+x}{5}= \frac{6x}{5}-noul \ pret \\\\ 20\% \ din \frac{6x}{5}= \frac{\not20}{\not100}*\frac{6x}{5}= \frac{1}{5}*\frac{6x}{5}\to \frac{6x}{25}-a \ doua \ marire \\\\ \frac{6x}{5}^{(5}+\frac{6x}{25}=\frac{30x+6x}{25} \to \frac{36x}{25}-noul \ nou \ pret \\\\ \frac{36x}{25}= 432 \\\\ x= \frac{25*432}{36} \\\\ x=25*12 \\\\ \boxed{x=300}$

$b)p\% \ din \ 432=300 \\\\ \frac{p}{100}*432=300 \\\\ \frac{p}{100}=\frac{300}{432} \\\\ p=\frac{100*\not300}{\not432} \\\\ p=\frac{100*100}{144} \\\\ p=\frac{10000}{144} \\\\ p=\frac{625}{9} \\\\\ \boxed{p=69,4\%}$