Question

Impartiti nr. 7 in parti invers proporționale cu nr. 2; 3; 0.5 si 0. (3)

Answer 1

fie a,b,c si d nr cautate
a*2=b*3=c*0,5=d*0,(3)
a*2=b*3=c/2=d/3, imparti totul cu 6 
a/3=b/2=c/12=d/18=k
a=3k
b=2k
c=12k
d=18k
a+b+c+d=7
35k=7
k=0,2
a=0,6
b=0,4
c=2,4
d=3,6

Answer 2

x * 2 = y * 3 = z * ( 1  / 2 ) = t * ( 1 / 3 ) = k => x = k / 2 ; y = k / 3 ; z = 2k ; t = 3k
x + y+ z + t = 7 ;
Ultima ecuatie devine echivalenta cu k / 2 + k / 3 + 5k = 7 <=> 3k + 2k + 30k = 42 35k = 42 <=> k = 42 / 35 <=> k = 6 / 5 => x = 3 / 5 ; y = 2 / 5 ; z = 12 / 5 si t = 18 / 5 ;

Bafta !