Question

O suma de bani a fost data ca premiu pentru 3 elevi x,y,z in parti direct proportionale cu numerele 2, 3, 5. Alta data, aceeasi suma de bani a fost impartita ca premiu acelorasi elevi x,y,z in parti invers proportionale cu 2,3,5. S.a constatat ca dupa cele doua premieri,al treilea elev,z, a incasat cu 132 lei mai mult decat al doilea elev,y. Calculati suma totala de bani care a fost oferita in cadrul celor doua premieri

Answer 1

[tex]\{x,y,z\}dp\{2,3,5\}\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow x=2k,y=3k,z=5k\\ a=suma\ de\ bani\\ x+y+z=a\\ 2k+3k+5k=a\\ 10k=a\Rightarrow k=\frac{a}{10}\\ \{x,y,z\}ip\{2,3,5\}\Rightarrow \frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=m\Rightarrow x=\frac{m}{2},y=\frac{m}{3},z=\frac{m}{5}\\ x+y+z=a\\ \frac{m}{2}+\frac{m}{3}+\frac{m}{5}=a|\cdot 30\\ 15m+10m+6m=30a\\ 31m=30a\Rightarrow m=\frac{30a}{31}\\ 5k+\frac{m}{5}=132+3k+\frac{m}{3}\\ [/tex][tex]Inlocuim:5\cdot \frac{a}{10}+\frac{\frac{30a}{31}}{5}=132+3\cdot \frac{a}{10}+\frac{\frac{30a}{31}}{3}\\ \frac{a}{2}+\frac{6a}{31}=132+\frac{3a}{10}+\frac{10a}{31}|\cdot 620\\ 310a+120a=81840+186a+200a\\ 430a=81840+386a\\ 44a=81840 \Rightarrow a=1860\ lei[/tex]