Question

Both of lalima and romen clean their garden.if lalima works for 4 days and romen works for 3 days then 2/3 part of the work is completed.Again if lalima works for 3 days and romen works for 6 days then 11/12 part of the work is completed .let us form the simultaneous equations and write the number of days required to complete the work separate by lalima and romen by calculating the solutions?​

Answer 1

Answer:

$\huge\ \sf{\red {[[«\: คꈤ \mathfrak Sฬєя \: » ]]}}$

$\sf \: Let \:the \:whole \: work \:done \:by \:Lalima \:be \: x \: and \: Romen \:be \:y$

$\sf \: work \: done \:by \:lalima \:in \:1 \:day \:= 1÷x$

$\ = \dfrac{1}{x}$

$\sf \: work \:done \:by \:lalima \: in \:4 \:days \:= 1/x ×4$

$\ = \dfrac{4}{x}$

$\sf \: work \:done \:by \: Romen \: in \:1 \:day \:= 1÷y$

$\ = \dfrac{1}{y}$

$\sf \: work \:done \:by \:the \:Romen \:in \: 3 \: days \:=1/y×3$

$\ = \dfrac{3}{y}$

$\sf \: Both \:worked \:and \:completed \:2/3 \:part \: of \:work$

$\ {\dfrac{4}{x}+\dfrac{3}{y} = \dfrac{2}{3}}$ --------- equation 1

$\sf \: again,$

$\sf \: work \: done \: by \:lalima \: in \: 3 \:days \:= 1/x×3$

$\ = \dfrac{3}{x}$

$\sf \: work \: done \: by \: the \: Romen \:in \:3 \: days \:=1/y×6$

$\ = \dfrac{6}{y}$

$\sf \: Both \:worked \:and \:completed \:11/12 \: part \: of \: work$

$\ {\dfrac{3}{x}+\dfrac{6}{y} = \dfrac{11}{,12}}$ $\sf- ------ equation \: 2$

From equation 1 and 2

$\ {\dfrac{4}{x}+\dfrac{3}{y} = \dfrac{2}{3}}$

$\ {\dfrac{3}{x}+\dfrac{6}{y} = \dfrac{11}{,12}}$

$\sf \: - \: \: - \: \: - {by \: subtracting}$

___________

$\ 0x - {\dfrac{15}{y} =-\dfrac{5}{3}}$

$\ {\dfrac{1}{y} =\dfrac{5}{3×15}}$

$\ {\dfrac{1}{y} =\dfrac{1}{9}}$

$\sf \: or \: y=9$

$\sf \: putting \:the \: value \: of \:9 \: in \:equation \: 1 ,we \:get$

$\ {\dfrac{4}{x}+\dfrac{3}{9} = \dfrac{2}{3}}$

$\ {\dfrac{4}{x}= \dfrac{2}{3} -\dfrac{3}{9}}$

$\ {\dfrac{4}{x} = \dfrac{1}{3}}$

$\ {\dfrac{1}{x} = \dfrac{1}{12}}$

$\sf \: x=12$

$\sf \: Therefore,\:Lalima\:required\: 12\:days\:for\:complete\:the\:work\:and\:Romen\:required\:9\:days \:to\:complete\:$

Answer 2

Answer:

$\huge\ \sf{\red {[[«\: คꈤ \mathfrak Sฬєя \: » ]]}}$

$\sf \: Let \:the \:whole \: work \:done \:by \:Lalima \:be \: x \: and \: Romen \:be \:y$

$\sf \: work \: done \:by \:lalima \:in \:1 \:day \:= 1÷x$

$= \dfrac{1}{x}}$

$\sf \: work \:done \:by \:lalima \: in \:4 \:days \:= 1/x ×4$

$\sf \: =4/x$

$\sf \: work \:done \:by \: Romen \: in \:1 \:day \:= 1÷y$

$\sf \: = 1/y$

$\sf \: work \:done \:by \:the \:Romen \:in \: 3 \: days \:=1/y×3$

$\sf \: =3/y$

$\sf \: Both \:worked \:and \:completed \:2/3 \:part \: of \:work$

$\sf \: 4/x+3/y=2/3 - eq1$

$\sf \: again,$

$\sf \: work \: done \: by \:lalima \: in \: 3 \:days \:= 1/x×3$

$\sf \: =3/x$

$\sf \: work \: done \: by \: the \: Romen \:in \:3 \: days \:=1/y×6$

$\sf \: =6/y$

$\sf \: Both \:worked \:and \:completed \:11/12 \: part \: of \: work$

$\sf \: 3/x+6/y=11/12 -eq2$

$\sf \: 4/x+3/y=2/3 equation,1 [by \:method \: of \: elimination]$

$\sf \: 3/x+6/y=11/12 equation.2$

$\sf \: - \: \: - \: \: - {by subtracting}$

___________

$\sf \: 0x - 15y= -5/3 \: cancelling- \:from \:both \:sides$

$\sf \: or y=9$

$\sf \: putting \:the value \: of \:9 \: in \:equation \: 1 ,we \:get$

$\sf \: 4/x+3/9=2/3$

$\sf \: 4/x=2/3-1/3$

$\sf \: 4/x=1/3$

$\sf \: 1/x=1/12$

$\sf \: x=12$

$\sf \: Therefore,\:Lalima\:required\: 12\:days\:for\:complete\:the\:work\:and\:Romen\:required\:9\:days \:to\:complete\:$