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Question Number 135420 by liberty last updated on 13/Mar/21
Algebra  Pipe A can fill a tank in two hours and pipe B can fill it in half the time it takes pipe C to empty it. When all 3 are opened, it takes 1.5 hours to fill the pool how much time is required for pipe 'C to empty the tank?
$${Algebra} \\ $$Pipe A can fill a tank in two hours and pipe B can fill it in half the time it takes pipe C to empty it. When all 3 are opened, it takes 1.5 hours to fill the pool how much time is required for pipe 'C to empty the tank?
Answered by john_santu last updated on 13/Mar/21
(1)pipe A ⇒V_1 = (W/2)  (2) pipe C ⇒ V_2  = (W/a)  (3) pipe B ⇒V_3  = (W/(a/2))  (4) When all 3 pipe are opened  ⇒ (W/(V_1 +V_2 +V_3 )) = (3/2)  ⇒(W/((W/2)+(W/a)+((2W)/a))) = (3/2)  ⇒ (1/2)+(3/a) = (2/3) ; (3/a) = (1/6)  we get a = 18 hour
$$\left(\mathrm{1}\right){pipe}\:{A}\:\Rightarrow{V}_{\mathrm{1}} =\:\frac{{W}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:{pipe}\:{C}\:\Rightarrow\:{V}_{\mathrm{2}} \:=\:\frac{{W}}{{a}} \\ $$$$\left(\mathrm{3}\right)\:{pipe}\:{B}\:\Rightarrow{V}_{\mathrm{3}} \:=\:\frac{{W}}{{a}/\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:{When}\:{all}\:\mathrm{3}\:{pipe}\:{are}\:{opened} \\ $$$$\Rightarrow\:\frac{{W}}{{V}_{\mathrm{1}} +{V}_{\mathrm{2}} +{V}_{\mathrm{3}} }\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{W}}{\frac{{W}}{\mathrm{2}}+\frac{{W}}{{a}}+\frac{\mathrm{2}{W}}{{a}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{{a}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:;\:\frac{\mathrm{3}}{{a}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${we}\:{get}\:{a}\:=\:\mathrm{18}\:{hour} \\ $$$$ \\ $$

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