Fie a, b, c sumele primite de cei trei muncitori, si S suma totala.
Avem;
[tex]\dfrac{a}{\dfrac16}=\dfrac{b}{\dfrac15}=\dfrac{c}{\dfrac13}=\dfrac{a+b+c}{\dfrac16+\dfrac15+\dfrac13}=\dfrac{S}{\dfrac{7}{10}}\Rightarrow a=\dfrac{5S}{21};\ b=\dfrac{2S}{7};\ c=\dfrac{10S}{21}[/tex]
Daca s-ar fi distribuit ca in varianta a doua am fi avut:
[tex]\dfrac{a'}{\dfrac{1}{12}}=\dfrac{b'}{\dfrac{1}{10}}=\dfrac{c'}{\dfrac{1}{15}}=\dfrac{S}{\dfrac14}\Rightarrow a'=\dfrac13S; \ b'=\dfrac25S;\ c'=\dfrac{4}{15}S[/tex]
unde a', b' c' sunt sumele pe care le-ar fi primit muncitorii in cazul al doilea.
Se observa ca : a<a', b<b', c>c'
Deci muncitorul al treilea este singurul care a primit mai multi bani, fata de varianta a doua.
Deci c=c'+462⇒10S/21=4S/15+462⇒
S=2205 leia=2205·5/21=
525 lei; b=2205·2/7=
630 lei; c=2205·10/21=
1050 lei
