Question Number 47302 by JDlix last updated on 08/Nov/18
$${A}\:{train}\:{which}\:{travels}\:{at}\:{a}\:{uniform}\:{speed}\:{due}\:{to}\:{mechanical}\: \\ $$$${fault}\:{after}\:{traveling}\:{for}\:{an}\:{hour}\:{goes}\:{at}\:\mathrm{3}/\mathrm{5}\:{th}\:{of}\:{the}\:{original}\: \\ $$$${speed}\:{and}\:{reaches}\:{the}\:{destination}\:\mathrm{2}\:{hours}\:{late}.{If}\:{the}\:{fault} \\ $$$${occured}\:{after}\:{traveling}\:{another}\:\mathrm{50}\:{miles}\:{the}\:{train}\:{would}\:{have} \\ $$$${reached}\:\mathrm{40}\:{minutes}\:{earlier}.\:{What}\:{is}\:{the}\:{distance}\:{between}\:{the}\: \\ $$$${two}\:{stations}\:? \\ $$
Answered by math1967 last updated on 08/Nov/18
$${let}\:{distance}\:{between}\:{two}\:{station}\:{ymiles} \\ $$$${original}\:{speed}\:{of}\:{train}={xmiles}/{hr} \\ $$$${for}\:\mathrm{1}{hr}\:{train}\:{covers}\:{xmiles} \\ $$$${remaining}\:{dist}.=\left({y}−{x}\right){miles} \\ $$$$\therefore\:\mathrm{1}+\frac{{y}−{x}}{\frac{\mathrm{3}{x}}{\mathrm{5}}}=\frac{{y}}{{x}}+\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{5}{y}−\mathrm{5}{x}}{\mathrm{3}{x}}\:−\frac{{y}}{{x}}=\mathrm{2}−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{2}{y}−\mathrm{5}{x}}{\mathrm{3}{x}}=\mathrm{1}\Rightarrow\mathrm{2}{y}=\mathrm{8}{x}\Rightarrow{y}=\mathrm{4}{x} \\ $$$${from}\:\mathrm{2}{nd}\:{condition} \\ $$$$\frac{\mathrm{50}}{\frac{\mathrm{3}{x}}{\mathrm{5}}}\:−\frac{\mathrm{50}}{{x}}=\frac{\mathrm{40}}{\mathrm{60}}\Rightarrow\frac{\mathrm{250}−\mathrm{150}}{\mathrm{3}{x}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\therefore{x}=\mathrm{50}\:\therefore{y}=\mathrm{4}×\mathrm{50}=\mathrm{200} \\ $$$${dist}.\:{between}\:{two}\:{stn}.=\mathrm{200}{milesAns}. \\ $$