Aplici relatia:c%= m,dx100/m,s unde m,s = m,d +m,apa
Solutia 1; se stie ms1 si c1 deci putem afla md1
25/100=md1/20---> md1=5g
Solutia 2; are acelasi md deci md2=md1=5g
are ms3= ms1+mapa--->20+10
c2%= 5x100/30 calculeaza !!!!! Te verifici la calc, deducand ca apa adaugata dilueaza sol si, atunci, c2 trebuie sa fie mai mic decat c1
