Question

Impartiti numarul 7 in parti invers proportionale cu numerele 2;3;0,5 si 0,(3)

Answer 1

Fie a, b, c, d partile  invers proportionale cu nr 2,3, 0,5 si 0, (3)
adica a*2=b*3=c*0,5=d*3/9=k
a=k/2
b=k/3, 
c=k/0,5=10k/5=2k
d=9k/3=3k
si a+b+c+d=7
inlocuind am
k/2+k/3+2k+3k=7
numitorul comun e 6
3k+2k+30k=42
35k=42
5k=6 ⇒ k=6/5
a= k/2=6/5*1/2=6/10=3/5
b=k/3=6/15=2/5
c=2k=6*2/5=12/5
d=3k=6*3/5=18/5