Answer:
[tex]The\ number\ of\ wagons\ on\ Station\ A\ at\ the\ beginning\ = 90\\ The\ number\ of\ wagons\ on\ Station\ B\ at\ the\ beginning\ =45[/tex]
Step-by-step explanation:
[tex]We\ are\ given:\\Total\ no.\ of\ wagons\ in\ both\ the\ stations=135\ wagons\\Now,\\Let\ both\ the\ stations\ be\ denoted\ as\ Station\ A\ and\ Station\ B.\\\\Let\ the\ number\ of\ wagons\ on\ Station\ A\ at\ the\ beginning\ be\ x\\Let\ the\ number\ of\ wagons\ on\ Station\ B\ at\ the\ beginning\ be\ y\\Hence,\\We\ are\ also\ given\ that,\\Number\ of\ wagons\ that\ removed\ from\ Station\ A=45\\Number\ of\ wagons\ that\ were\ added\ to\ Station\ A=36\\Similarly,\\[/tex]
[tex]Number\ of\ wagons\ that\ removed\ from\ Station\ B=36\\Number\ of\ wagons\ that\ were\ added\ to\ Station\ B=45\\Hence,\\The\ final\ number\ of\ wagons\ in\ Station\ A=x-45+36=x-9\\The\ final\ number\ of\ wagons\ in\ Station\ B=x+45-36=y+9\\Now,\\As\ we\ already\ know,\\No.\ of\ wagons\ in\ Station\ A\ at\ the\ beginning\\ +No.\ of\ wagons\ in\ Station\ B\ at\ the\ beginning=135\\Hence,\\x+y=135\\y=(135-x)\\Hence,\\The\ final\ number\ of\ wagons\ in\ Station\ B=y+9=(135-x)+9=(144-x)[/tex]
[tex]We\ are\ also\ given\ that,\\Final\ number\ of\ wagons\ in\ Station\ A=\frac{3}{2}* Final\ number\ of\ wagons\\ in\ Station\ B\\Hence,\\(x-9)=\frac{3}{2}*(144-x)\\By\ simplifying:\\2(x-9)=3(144-x)\\2x-18=432-3x\\3x+2x=432+18\\5x=450\\x=\frac{450}{5}=90\\Hence,\\The\ number\ of\ wagons\ on\ Station\ A\ at\ the\ beginning\ = x=90\\ The\ number\ of\ wagons\ on\ Station\ B\ at\ the\ beginning\ =y=(135-x)=(135-90)=45[/tex]
