Question

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

Answer 1

a+b+c =106
a= 3/9 k
b= 2k
c=12/10 k
3/9k +2k +12/10 k =106
1/3 k +2k + 6/5 k =106
5k+30 k+ 18k/15=106
53k/15=106
k= 106:53/15
k=106×15/53
k=30
a=3/9×301/3×30= 30/3= 10
b=2×30 =60
c=12/10×30= 6/5×30= 6×6= 36

Answer 2

Consideram x , y , z
x / 0,3 + y / 2 + z / 1 , 2
= 106 / 3 , 5(3)
k = > 30
30 x 0 , ( 3 ) = 10 ( x )
30 x 2 = 60 ( y )
30 x 1 , 2 = 36 ( z )