Question Number 195661 by otchereabdullai@gmail.com last updated on 06/Aug/23
Commented by otchereabdullai@gmail.com last updated on 07/Aug/23
$${thanks}\:{prof} \\ $$
Commented by mr W last updated on 07/Aug/23
$${even}\:{when}\:{the}\:{whole}\:{road}\:{is}\:{good},\:{he} \\ $$$${needs}\:\frac{\mathrm{23}}{\mathrm{52}}×\mathrm{60}=\mathrm{26}.\mathrm{5}\:{minutes}.\:{that} \\ $$$${means}\:{he}\:{can}\:{never}\:{drive}\:\mathrm{23}\:{km}\:{in} \\ $$$${only}\:\mathrm{25}\:{minutes}. \\ $$$$\Rightarrow{question}\:{is}\:{wrong}! \\ $$
Answered by som(math1967) last updated on 07/Aug/23
$$\:{let}\:{good}\:{road}={xkm} \\ $$$$\therefore\:{bad}\:{road}=\mathrm{23}−{x} \\ $$$$\:\frac{{x}}{\mathrm{52}}\:+\frac{\left(\mathrm{23}−{x}\right)}{\mathrm{28}}=\frac{\mathrm{25}}{\mathrm{60}} \\ $$$$\Rightarrow\:\frac{\mathrm{7}{x}+\mathrm{13}\left(\mathrm{23}−{x}\right)}{\mathrm{4}×\mathrm{13}×\mathrm{7}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\Rightarrow−\mathrm{6}{x}+\mathrm{13}×\mathrm{23}=\frac{\mathrm{5}×\mathrm{13}×\mathrm{7}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{18}{x}=\mathrm{13}×\mathrm{23}×\mathrm{3}−\mathrm{35}×\mathrm{13} \\ $$$$\Rightarrow{x}=\frac{\mathrm{13}\left(\mathrm{69}−\mathrm{35}\right)}{\mathrm{18}}=\frac{\mathrm{13}×\mathrm{17}}{\mathrm{9}}=\mathrm{24}.\mathrm{5}{km} \\ $$$$\:\mathrm{24}.\mathrm{5}>\mathrm{23}\:? \\ $$$$\:\:{length}\:{of}\:{good}\:{road}>{total}\:{length} \\ $$$$? \\ $$$$\: \\ $$
Commented by otchereabdullai@gmail.com last updated on 07/Aug/23
$${thanks}\:{sir} \\ $$