Question

Impartiti nr 106 in parti direct proportionale cu nr 0,(3); 2; 1,2.

Answer 1

a+b+c=106
3a=$\frac{b}{2} = \frac{5c}{6} =k$
a=$\frac{k}{3}$
b=2k
c=$\frac{6k}{5}$
[tex] \frac{k}{3} +2k+ \frac{6k}{5} =106 [/tex]
5k+30k+18k=1590
53k=1590
k=30
a=10
b=60
c=36