Question

impartiti  numarul  106  in parti  direct  proportionale  cu  numerele  0,(3);2;1,2;

Answer 1

a+b+c=106

a=3/9k
a=k/3
b=2k
c=12k/10=> c=6k/5

k/3+2k+6k/5=106 numitor comun 15
5k+30k+18k=1590
53k=1590
k=30

a=30/3
a=10

b=2k
b=2*30
b=60

c=6*30/5
c=180/5
c=36