Question Number 40012 by rahul 19 last updated on 15/Jul/18
Answered by ajfour last updated on 15/Jul/18
$$\left({mg}\right)_{{max}} =\frac{\mathrm{3}{Mg}}{\mathrm{5}}+\frac{\mathrm{0}.\mathrm{6}\left(\mathrm{4}{Mg}\right)}{\mathrm{5}} \\ $$$$\:\:\:\:\Rightarrow\:{m}_{{max}} =\mathrm{30}{kg}+\mathrm{24}{kg}\:=\:\mathrm{54}{kg} \\ $$$$\left({mg}\right)_{{min}} =\frac{\mathrm{3}{Mg}}{\mathrm{5}}−\frac{\mathrm{0}.\mathrm{6}\left(\mathrm{4}{Mg}\right)}{\mathrm{5}} \\ $$$$\:\:\:\Rightarrow\:{m}_{{min}} =\:\mathrm{30}{kg}−\mathrm{24}{kg}\:=\mathrm{6}{kg}\:. \\ $$
Commented by rahul 19 last updated on 15/Jul/18
Thank you sir ����