Question

Impartind numarul 106 in parti direct proportionale cu numerele 0.(3),2 si 1,2.

Answer 1

a+b+c=106;
a/0,(3)=b/2=c/1,2=k ⇒ 3a=b/2=5c/6=k ⇒ a=k/3, b=2k, c=6k/5 ⇒
k/3+2k+6k/5=106 ⇒ 5k+30k+18k=1590, k=30;
a=10, b=60, c=36