Question

Impartiti numarul 63 in trei numere invers proportionale cu numerele 3,5, si 6.

Answer 1

            
$a+b+c=63 \\ 3a=5b=6c=k \\ \\ a= \frac{k}{3} \;\;\;\;\;\; b= \frac{k}{5} \;\;\;\;\;\; c= \frac{k}{6} \\ \\ a+b+c=63 \\ \\ \frac{k}{3}+ \frac{k}{5}+\frac{k}{6}=63 \\ \\ \frac{10k+6k+5k}{30} =63 \\ \\ 21k=63*30 \\ \\ k= \frac{63*30}{21} =3*30=90 \\ \\ a= \frac{k}{3}=\frac{90}{3}=\boxed{30} \;\;\;\;\;\; b= \frac{k}{5}=\frac{90}{5}=\boxed{18} \;\;\;\;\;\; c= \frac{k}{6}=\frac{90}{6}=\boxed{15}$