Question

Both Lalima and Ramen cleaned the gardens of the house. Lalima 4 days and Ramen cleaning 3 days simultaneously
So 2/3 part work is done. If redness is cleaning for 3 days and ramen for 6 days simultaneously, then 11/12 parts work
Complete . By making simultaneous equations and solving lalima and ramen separately, work alone
In how many days Both .
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Answer 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

___________

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

$\ ⟹{\dfrac{1}{y} =\dfrac{5}{3×15}}$

$\ ⟹{\dfrac{1}{y} =\dfrac{1}{9}}$

$\sf \: ⟹\: y=9$

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

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

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

$\ ⟹{\dfrac{4}{x} = \dfrac{1}{3}}$

$\ ⟹{\dfrac{1}{x} = \dfrac{1}{12}}$

$\sf \: ⟹x=12$

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