Question Number 127006 by liberty last updated on 26/Dec/20
$${Two}\:{pipes}\:{A}\:{and}\:{B}\:{together}\:{can}\:{fill}\: \\ $$$${a}\:{cistern}\:{in}\:\mathrm{5}\:{hours}.\:{Had}\:{they}\:{been}\:{opened} \\ $$$${separately},\:{then}\:{B}\:{would}\:{have}\:{taken}\: \\ $$$$\mathrm{6}\:{hours}\:{more}\:{than}\:{A}\:{to}\:{fill}\:{the}\:{cistern}. \\ $$$${How}\:{much}\:{time}\:{will}\:{be}\:{taken}\:{by}\:{A}\:{to}\:{fill} \\ $$$${the}\:{cistern}\:{separately}?\: \\ $$
Answered by bramlexs22 last updated on 26/Dec/20
$${Let}\:{pipe}\:{A}\:{take}\:{x}\:{hours}\:{and}\:{pipe}\:{B} \\ $$$${take}\:\left({x}+\mathrm{6}\right)\:{hours}\:{to}\:{fill}\:{the}\:{cistern} \\ $$$${each}\:{working}\:{alone}. \\ $$$$ \\ $$$${if}\:{both}\:{are}\:{open}\:{together}\::\:\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}+\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}{x}+\mathrm{6}}{{x}^{\mathrm{2}} +\mathrm{6}{x}}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:;\:{x}^{\mathrm{2}} +\mathrm{6}{x}=\mathrm{10}{x}+\mathrm{30} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{30}=\mathrm{0}\:;\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{34}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{x}\:=\:\mathrm{2}+\sqrt{\mathrm{34}}\:\approx\:\mathrm{7}.\mathrm{83}\:{hours} \\ $$$$ \\ $$