Question Number 56503 by otchereabdullai@gmail.com last updated on 17/Mar/19
$$\mathrm{A}\:\mathrm{boat}\:\mathrm{travels}\:\mathrm{30km}\:\mathrm{upstream}\:\mathrm{and}\: \\ $$$$\mathrm{44km}\:\mathrm{downstream}\:\mathrm{in}\:\mathrm{10}\:\mathrm{hours}.\: \\ $$$$\mathrm{in}\:\mathrm{13}\:\mathrm{hours}\:\mathrm{it}\:\mathrm{can}\:\mathrm{travel}\:\mathrm{40km}\:\mathrm{upstream} \\ $$$$\mathrm{and}\:\mathrm{55km}\:\mathrm{downstream}.\:\mathrm{Determine}\:\mathrm{the} \\ $$$$\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{stream}\:\:\mathrm{and}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{boat}\:\mathrm{in}\:\mathrm{still}\:\mathrm{water}.\:\left(\mathrm{in}\:\mathrm{km}/\mathrm{hr}\right) \\ $$
Answered by MJS last updated on 17/Mar/19
$$\frac{\mathrm{30}}{\left({v}_{{b}} −{v}_{{s}} \right)}+\frac{\mathrm{44}}{\left({v}_{{b}} +{v}_{{s}} \right)}=\mathrm{10} \\ $$$$\Rightarrow\:\mathrm{5}{v}_{{b}} ^{\mathrm{2}} −\mathrm{5}{v}_{{s}} ^{\mathrm{2}} −\mathrm{37}{v}_{{b}} +\mathrm{7}{v}_{{s}} =\mathrm{0}\:\:\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{40}}{\left({v}_{{b}} −{v}_{{s}} \right)}+\frac{\mathrm{55}}{\left({v}_{{b}} +{v}_{{s}} \right)}=\mathrm{13} \\ $$$$\Rightarrow\:\mathrm{13}{v}_{{b}} ^{\mathrm{2}} −\mathrm{13}{v}_{{s}} ^{\mathrm{2}} −\mathrm{95}{v}_{{b}} +\mathrm{15}{v}_{{s}} =\mathrm{0}\:\:\left(\mathrm{2}\right) \\ $$$$−\frac{\mathrm{13}}{\mathrm{5}}\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{6}}{\mathrm{5}}{v}_{{b}} −\frac{\mathrm{16}}{\mathrm{5}}{v}_{{s}} =\mathrm{0}\:\Rightarrow\:{v}_{{s}} =\frac{\mathrm{3}}{\mathrm{8}}{v}_{{b}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right) \\ $$$$\frac{\mathrm{275}}{\mathrm{64}}{v}_{{b}} ^{\mathrm{2}} −\frac{\mathrm{275}}{\mathrm{8}}{v}_{{b}} =\mathrm{0} \\ $$$${v}_{{b}} >\mathrm{0}\:\Rightarrow\:{v}_{{b}} =\mathrm{8}{km}/{hr}\:\Rightarrow\:{v}_{{s}} =\mathrm{3}{km}/{hr} \\ $$
Commented by otchereabdullai@gmail.com last updated on 18/Mar/19
$$\mathrm{thanks}\:\mathrm{prof} \\ $$