Question

impartiti nr 108 in parti invers proportionale cu 2, 3 si cu 6

Answer 1

Răspuns:

a/(1/2)=b/(1/3)=c/(1/6)=k

2a=3b=6c=k

a=k/2

b=k/3

c=k/6

a+b+c=108

³⁾k/2+²⁾k/3+k/6=⁶⁾108

3k+2k+k=648

k=648:6=>k=108

a=108:2=>a=54

b=108:3=>b=36

c=108:6=>c=18