Răspuns :
Fie x,y,z, sumele de bani primite decei trei muncitori
0,5=1/2; 0.(6)=6/9=2/3; 0,4=4/10=2/5
parti invers prop cu 1/2 2/3 si 2/5
parti d.p cu 2 3/2 si 5/2
x/2=y/(3/2)=z/(5/2)=(x+y+z)/(2+3/2+5/2)=(x+y+z)/6=2350/6=1175/3
x=1175/3 *2=2350/3
y=1175/3*(3/2)=1175/2
z=1175/3 *5/2=5875/6
verificare
2350/3+1175/2+5875/6=(4700+3525+5875)/6=14100/6=2350
adevarat , bine rezolvat
numere nenaturale
posibil ca sumasa fie alta, nu 2350, ci un nr care se impartea la 6,. gen 2450, sau sa se fi omis/gresit niste paranteze la tehnoredactarea problemei
a,b,c sumele primite de muncitori, a+b+c=2350
0,5a=0,(6)b=0,4c, a=0,(6)b:0,5,=(2b/3):(1/2)=(2b/3)x2=4b/3, c=0,(6)b:0,4=(2b/3):(2/5) =(2b/3)x(5/2)=5b/3, inlocuim a si c in suma
4b/3+b+5b/3=2350, numitor comun 3, 4b+3b+5b=7050, b=7050/12=587,5 lei
a=(4x587,5)/3=2350/3 lei=783,(3) lei, c=(5x587,5)/3=2937.5/3=979,1(6) lei