Question Number 202079 by necx122 last updated on 19/Dec/23
$${A}\:{man}\:{travelled}\:{from}\:{town}\:{A}\:{to}\:{B},\:{a} \\ $$$${distance}\:{of}\:\mathrm{360}{km}.\:{He}\:{left}\:{A}\:{one}\:{hour} \\ $$$${later}\:{than}\:{he}\:{had}\:{planned}\:{so}\:{he}\:{decided} \\ $$$${to}\:{drive}\:{at}\:\mathrm{5}{km}/{h}\:{faster}\:{than}\:{his} \\ $$$${normal}\:{speed},\:{in}\:{order}\:{to}\:{reach}\:{B}\:{on} \\ $$$${schedule}.\:{If}\:{he}\:{arrived}\:{B}\:{at}\:{exactly}\:{the} \\ $$$${scheduled}\:{time},\:{find}\:{the}\:{normal}\:{speed}. \\ $$
Answered by AST last updated on 19/Dec/23
$${t}=\frac{\mathrm{360}}{{x}+\mathrm{5}}=\frac{\mathrm{360}}{{x}}−\mathrm{1}\Rightarrow\frac{−\mathrm{360}×\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}{x}}=−\mathrm{1}\Rightarrow{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1800}=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{40} \\ $$
Commented by necx122 last updated on 20/Dec/23
Sir AST, you've also been very helpful on this platform since you joined. This is very clear. Thank you so much sir.
Commented by necx122 last updated on 20/Dec/23
textbook says the answer is 80.
Commented by AST last updated on 20/Dec/23
$$\frac{\mathrm{360}}{\mathrm{80}}−\frac{\mathrm{360}}{\mathrm{85}}\neq+\mathrm{1} \\ $$
Commented by necx122 last updated on 20/Dec/23
true proof. Thanks