Question Number 202086 by necx122 last updated on 20/Dec/23
$${There}\:{are}\:{two}\:{possible}\:{routes}\:{from} \\ $$$${Zindhi}\:{to}\:{Katifa}.\:{One}\:{route}\:{is}\:{through} \\ $$$${Zindhi}/{Chadler}\:{expressway}\:{which}\:{is} \\ $$$$\mathrm{100}{km}\:{and}\:{the}\:{other}\:{is}\:{through}\:{Adfeti}\:{and} \\ $$$${Ngonu}\:{covering}\:{a}\:{distance}\:{of}\:\mathrm{80}{km}.\:{A} \\ $$$${motorist}\:{going}\:{through}\:{the}\:{expressway} \\ $$$${can}\:{travel}\:\mathrm{10}{km}/{h}\:{faster}\:{than}\:{the}\:{one} \\ $$$${going}\:{through}\:{Adfeti}\:{and}\:{Ngonu},\:{and} \\ $$$${arrive}\:{Katifa}\:\mathrm{5}\:{minutes}\:{earlier}\:{as}\:{well}. \\ $$$${What}\:{is}\:{the}\:{time}\:{spent}\:{on}\:{the}\:{journey} \\ $$$${to}\:{Katifa}\:{by}\:{the}\:{motorist}\:{travelling} \\ $$$${through}\:{the}\:{expressway}. \\ $$
Answered by mr W last updated on 20/Dec/23
$$\frac{\mathrm{100}}{{t}}=\frac{\mathrm{80}}{{t}+\frac{\mathrm{1}}{\mathrm{12}}}+\mathrm{10} \\ $$$$\mathrm{12}{t}^{\mathrm{2}} −\mathrm{23}{t}−\mathrm{10}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{23}+\sqrt{\mathrm{1009}}}{\mathrm{24}}\approx\mathrm{2}.\mathrm{28}\:{h}=\mathrm{2}{h}\mathrm{16}{m} \\ $$
Commented by necx122 last updated on 20/Dec/23
This too is clear. Thank you so much sir.