Menu Close

Question-195661




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
$${thanks}\:{prof} \\ $$
Commented by mr W last updated on 07/Aug/23
even when the whole road is good, he  needs ((23)/(52))×60=26.5 minutes. that  means he can never drive 23 km in  only 25 minutes.  ⇒question is wrong!
$${even}\:{when}\:{the}\:{whole}\:{road}\:{is}\:{good},\:{he} \\ $$$${needs}\:\frac{\mathrm{23}}{\mathrm{52}}×\mathrm{60}=\mathrm{26}.\mathrm{5}\:{minutes}.\:{that} \\ $$$${means}\:{he}\:{can}\:{never}\:{drive}\:\mathrm{23}\:{km}\:{in} \\ $$$${only}\:\mathrm{25}\:{minutes}. \\ $$$$\Rightarrow{question}\:{is}\:{wrong}! \\ $$
Answered by som(math1967) last updated on 07/Aug/23
 let good road=xkm  ∴ bad road=23−x   (x/(52)) +(((23−x))/(28))=((25)/(60))  ⇒ ((7x+13(23−x))/(4×13×7))=(5/(12))  ⇒−6x+13×23=((5×13×7)/3)  ⇒ 18x=13×23×3−35×13  ⇒x=((13(69−35))/(18))=((13×17)/9)=24.5km   24.5>23 ?    length of good road>total length  ?
$$\:{let}\:{good}\:{road}={xkm} \\ $$$$\therefore\:{bad}\:{road}=\mathrm{23}−{x} \\ $$$$\:\frac{{x}}{\mathrm{52}}\:+\frac{\left(\mathrm{23}−{x}\right)}{\mathrm{28}}=\frac{\mathrm{25}}{\mathrm{60}} \\ $$$$\Rightarrow\:\frac{\mathrm{7}{x}+\mathrm{13}\left(\mathrm{23}−{x}\right)}{\mathrm{4}×\mathrm{13}×\mathrm{7}}=\frac{\mathrm{5}}{\mathrm{12}} \\ $$$$\Rightarrow−\mathrm{6}{x}+\mathrm{13}×\mathrm{23}=\frac{\mathrm{5}×\mathrm{13}×\mathrm{7}}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{18}{x}=\mathrm{13}×\mathrm{23}×\mathrm{3}−\mathrm{35}×\mathrm{13} \\ $$$$\Rightarrow{x}=\frac{\mathrm{13}\left(\mathrm{69}−\mathrm{35}\right)}{\mathrm{18}}=\frac{\mathrm{13}×\mathrm{17}}{\mathrm{9}}=\mathrm{24}.\mathrm{5}{km} \\ $$$$\:\mathrm{24}.\mathrm{5}>\mathrm{23}\:? \\ $$$$\:\:{length}\:{of}\:{good}\:{road}>{total}\:{length} \\ $$$$? \\ $$$$\: \\ $$
Commented by otchereabdullai@gmail.com last updated on 07/Aug/23
thanks sir
$${thanks}\:{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *