Question

impartiti nr 117 in parti egale invers proportionale cu 3 ,4 si 6

Answer 1

x/(1/3)=k  3x=k  x=k/3
y/(1/4)=k  4y=k  y=k/4
z/(1/6)=k  6z=k  z=k/6
x+y+z=117
k/3+k/4+k/6=117
4k+3k+2k=1404
9k=1404
k=156
x=k/3=156/3=52
y=k/4=156/4=39
z=k/6=156/6=26