Question Number 66865 by John Kaloki Musau last updated on 20/Aug/19
$${A}\:{and}\:{B}\:{are}\:{two}\:{towns}\:\mathrm{360}{km}\:{apart}. \\ $$$${An}\:{express}\:{bus}\:{departs}\:{from}\:{A}\:{at} \\ $$$$\mathrm{8}{a}.{m}\:{and}\:{maintains}\:{an}\:{average} \\ $$$${speed}\:{of}\:\mathrm{90}{km}/{h}\:{between}\:{A}\:{and}\:{B}. \\ $$$${Another}\:{bus}\:{starts}\:{from}\:{B}\:{also}\:{at} \\ $$$$\mathrm{8}{a}.{m}\:{and}\:{moves}\:{towards}\:{A}\:{making} \\ $$$${four}\:{stops}\:{at}\:{four}\:{equally}\:{spaced} \\ $$$${points}\:{between}\:{B}\:{and}\:{A}.\:{Each}\:{stop} \\ $$$${is}\:{of}\:{duration}\:\mathrm{5}\:{minutes}\:{and}\:{the} \\ $$$${average}\:{speed}\:{between}\:{any}\:{two}\:{stops} \\ $$$${is}\:\mathrm{60}{km}/{h}.\:{Calculate}\:{the}\:{distance} \\ $$$${between}\:{the}\:{two}\:{buses}\:{at}\:\mathrm{10}{p}.{m}. \\ $$
Commented by John Kaloki Musau last updated on 20/Aug/19
$$\boldsymbol{{The}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\mathrm{65}\boldsymbol{{km}}. \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{do}}\:\boldsymbol{{it}}. \\ $$
Answered by John Kaloki Musau last updated on 20/Aug/19
$$\boldsymbol{{distance}}\:\boldsymbol{{travelled}}\:\boldsymbol{{by}}\:\boldsymbol{{bus}}\:\mathrm{1} \\ $$$$\boldsymbol{{before}}\:\mathrm{10}\boldsymbol{{a}}.\boldsymbol{{m}}=\mathrm{180}\boldsymbol{{km}} \\ $$$$\boldsymbol{{distance}}\:\boldsymbol{{travelled}}\:\boldsymbol{{by}}\:\boldsymbol{{bus}}\:\mathrm{2} \\ $$$$\boldsymbol{{before}}\:\mathrm{10}\boldsymbol{{a}}.\boldsymbol{{m}}=\mathrm{120}\boldsymbol{{km}} \\ $$$$\mathrm{120}\boldsymbol{{km}}\:\boldsymbol{{occus}}\:\boldsymbol{{btn}}\:\boldsymbol{{departure}} \\ $$$$\boldsymbol{{point}}\:\boldsymbol{{and}}\:\boldsymbol{{first}}\:\boldsymbol{{stop}} \\ $$$$\mathrm{1}\boldsymbol{{st}}\:\boldsymbol{{stop}}=\mathrm{5}\boldsymbol{{min}} \\ $$$$\mathrm{10}\boldsymbol{{a}}.\boldsymbol{{m}}−\mathrm{8}\boldsymbol{{a}}.\boldsymbol{{m}}−\mathrm{5}\boldsymbol{{min}}=\mathrm{115}\boldsymbol{{min}} \\ $$$$\boldsymbol{{D}}=\left(\frac{\mathrm{115}}{\mathrm{60}}{hrs}×\mathrm{60}{km}/{h}\right){km}\:\:\:=\mathrm{115}{km} \\ $$$$\boldsymbol{{distance}}\:\boldsymbol{{btn}}\:\boldsymbol{{buses}}\:\boldsymbol{{at}}\:\mathrm{10}\boldsymbol{{a}}.\boldsymbol{{m}}=\left\{\mathrm{360}−\left(\mathrm{115}+\mathrm{180}\right)\right\}\boldsymbol{{km}} \\ $$$$=\mathrm{65}\boldsymbol{{km}} \\ $$