Question Number 170986 by Tawa11 last updated on 05/Jun/22
Commented by mr W last updated on 05/Jun/22
$${too}\:{easy}\:... \\ $$$${A}\:{covers}\:\frac{\mathrm{15}}{\mathrm{15}+\mathrm{10}}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${B}\:{covers}\:\frac{\mathrm{10}}{\mathrm{15}+\mathrm{10}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$
Commented by Tawa11 last updated on 06/Jun/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by thfchristopher last updated on 05/Jun/22
$$\mathrm{Let}\:{t}\:\mathrm{be}\:\mathrm{the}\:\mathrm{time}\:\mathrm{lapse}\:\mathrm{for}\:\mathrm{their}\:\mathrm{collision} \\ $$$$\mathrm{Assume}\:\mathrm{they}\:\mathrm{each}\:\mathrm{travel}\:\mathrm{at}\:\mathrm{same}\:\mathrm{speed}\:\mathrm{without}\:\mathrm{acceleration} \\ $$$$\mathrm{15}{t}+\mathrm{10}{t}=\mathrm{300}×\mathrm{1000} \\ $$$$\Rightarrow\mathrm{25}{t}=\mathrm{300000} \\ $$$$\Rightarrow{t}=\mathrm{12000sec} \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{Train}\:\mathrm{A}\:\mathrm{travels}\:\mathrm{15}×\mathrm{12000}=\mathrm{180km} \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{of}\:\mathrm{total}\:\mathrm{distance} \\ $$$$\Rightarrow\mathrm{Train}\:\mathrm{B}\:\mathrm{travels}\:\mathrm{120km}\:=\frac{\mathrm{2}}{\mathrm{5}}\:\mathrm{of}\:\mathrm{total}\:\mathrm{distance} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 06/Jun/22
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$