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Question-184218




Question Number 184218 by katana last updated on 04/Jan/23
Answered by Rasheed.Sindhi last updated on 04/Jan/23
Suppose car catches the bus after x  hours.  •When the car catces the bus,the bus  has travelled for  2(1/2)+x  hours,(while  the car has travelled for x hours.)  •The bus travelled the distance in  ((5+2x)/2) hours is equal to the distance  travelled by car in x hours:  ((5+2x)/2)∙60=x∙100  150+60x=100x⇒x=((150)/(40))=3(3/4)hoursT  The distance travelled by bus from  Mombasa:  3(3/4)×60=((15)/4)×60=225 km  a(i)The distance of bus from Nairobi:            =500−225=275 km  a(ii) The car travelled 225 km to  catch up the bus.  (c)If the car would have continued  his first average speed it would have   completed its journey in 5 hours.  •In case car stays for 25 minutes  (=(5/(12)) hours) let the remaining journey  is completed in  y hours  x+(5/(12))+y=5  3(3/4)+(5/(12))+y=5  y=5−((25)/6)=(5/6) hours  In this duration((5/6) hours) the car  has to travel 275 km  v=((225)/(5/6))=225×(6/5)=270 km/hour  (c)  270 km/hour
$$\mathrm{Suppose}\:\mathrm{car}\:\mathrm{catches}\:\mathrm{the}\:\mathrm{bus}\:\mathrm{after}\:\mathrm{x} \\ $$$$\mathrm{hours}. \\ $$$$\bullet\mathrm{When}\:\mathrm{the}\:\mathrm{car}\:\mathrm{catces}\:\mathrm{the}\:\mathrm{bus},\mathrm{the}\:\mathrm{bus} \\ $$$$\mathrm{has}\:\mathrm{travelled}\:\mathrm{for}\:\:\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{x}\:\:\mathrm{hours},\left(\mathrm{while}\right. \\ $$$$\left.\mathrm{the}\:\mathrm{car}\:\mathrm{has}\:\mathrm{travelled}\:\mathrm{for}\:\mathrm{x}\:\mathrm{hours}.\right) \\ $$$$\bullet\mathrm{The}\:\mathrm{bus}\:\mathrm{travelled}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{in} \\ $$$$\frac{\mathrm{5}+\mathrm{2x}}{\mathrm{2}}\:\mathrm{hours}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{travelled}\:\mathrm{by}\:\mathrm{car}\:\mathrm{in}\:\mathrm{x}\:\mathrm{hours}: \\ $$$$\frac{\mathrm{5}+\mathrm{2x}}{\mathrm{2}}\centerdot\mathrm{60}=\mathrm{x}\centerdot\mathrm{100} \\ $$$$\mathrm{150}+\mathrm{60x}=\mathrm{100x}\Rightarrow\mathrm{x}=\frac{\mathrm{150}}{\mathrm{40}}=\mathrm{3}\frac{\mathrm{3}}{\mathrm{4}}\mathrm{hoursT} \\ $$$$\mathrm{The}\:\mathrm{distance}\:\mathrm{travelled}\:\mathrm{by}\:\mathrm{bus}\:\mathrm{from} \\ $$$$\mathrm{Mombasa}: \\ $$$$\mathrm{3}\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{60}=\frac{\mathrm{15}}{\mathrm{4}}×\mathrm{60}=\mathrm{225}\:\mathrm{km} \\ $$$$\mathrm{a}\left(\mathrm{i}\right)\mathrm{The}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{bus}\:\mathrm{from}\:\mathrm{Nairobi}: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{500}−\mathrm{225}=\mathrm{275}\:\mathrm{km} \\ $$$$\mathrm{a}\left(\mathrm{ii}\right)\:\mathrm{The}\:\mathrm{car}\:\mathrm{travelled}\:\mathrm{225}\:\mathrm{km}\:\mathrm{to} \\ $$$$\mathrm{catch}\:\mathrm{up}\:\mathrm{the}\:\mathrm{bus}. \\ $$$$\left(\mathrm{c}\right)\mathrm{If}\:\mathrm{the}\:\mathrm{car}\:\mathrm{would}\:\mathrm{have}\:\mathrm{continued} \\ $$$$\mathrm{his}\:\mathrm{first}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{it}\:\mathrm{would}\:\mathrm{have} \\ $$$$\:\mathrm{completed}\:\mathrm{its}\:\mathrm{journey}\:\mathrm{in}\:\mathrm{5}\:\mathrm{hours}. \\ $$$$\bullet\mathrm{In}\:\mathrm{case}\:\mathrm{car}\:\mathrm{stays}\:\mathrm{for}\:\mathrm{25}\:\mathrm{minutes} \\ $$$$\left(=\frac{\mathrm{5}}{\mathrm{12}}\:\mathrm{hours}\right)\:\mathrm{let}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{journey} \\ $$$$\mathrm{is}\:\mathrm{completed}\:\mathrm{in}\:\:\mathrm{y}\:\mathrm{hours} \\ $$$$\mathrm{x}+\frac{\mathrm{5}}{\mathrm{12}}+\mathrm{y}=\mathrm{5} \\ $$$$\mathrm{3}\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{12}}+\mathrm{y}=\mathrm{5} \\ $$$$\mathrm{y}=\mathrm{5}−\frac{\mathrm{25}}{\mathrm{6}}=\frac{\mathrm{5}}{\mathrm{6}}\:\mathrm{hours} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{duration}\left(\frac{\mathrm{5}}{\mathrm{6}}\:\mathrm{hours}\right)\:\mathrm{the}\:\mathrm{car} \\ $$$$\mathrm{has}\:\mathrm{to}\:\mathrm{travel}\:\mathrm{275}\:\mathrm{km} \\ $$$$\mathrm{v}=\frac{\mathrm{225}}{\mathrm{5}/\mathrm{6}}=\mathrm{225}×\frac{\mathrm{6}}{\mathrm{5}}=\mathrm{270}\:\mathrm{km}/\mathrm{hour} \\ $$$$\left(\mathrm{c}\right)\:\:\mathrm{270}\:\mathrm{km}/\mathrm{hour} \\ $$
Commented by mr W last updated on 05/Jan/23
Happy New Year to you too!  Your answer is surely correct.   I just like the fast car which can   catch up the bus. only few cars in  the world can drive so fast.
$${Happy}\:{New}\:{Year}\:{to}\:{you}\:{too}! \\ $$$${Your}\:{answer}\:{is}\:{surely}\:{correct}.\: \\ $$$${I}\:{just}\:{like}\:{the}\:{fast}\:{car}\:{which}\:{can}\: \\ $$$${catch}\:{up}\:{the}\:{bus}.\:{only}\:{few}\:{cars}\:{in} \\ $$$${the}\:{world}\:{can}\:{drive}\:{so}\:{fast}. \\ $$
Commented by mr W last updated on 04/Jan/23
Commented by Rasheed.Sindhi last updated on 04/Jan/23
Sir Mr W!              determinant (((  determinant ((( HAPPY_(NEW_(YEAR ! ) ) ^  )))_  ^  )))
$$\boldsymbol{\mathrm{Sir}}\:\boldsymbol{\mathrm{Mr}}\:\boldsymbol{\mathrm{W}}! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\:\begin{array}{|c|}{\:\boldsymbol{\mathrm{HAPPY}}_{\boldsymbol{\mathrm{NEW}}_{\boldsymbol{\mathrm{YEAR}}\:!\:} } ^{\:} }\\\hline\end{array}_{\:} ^{\:} }\\\hline\end{array}\: \\ $$
Commented by Rasheed.Sindhi last updated on 04/Jan/23
Are the above answers correct?
$$\mathrm{Are}\:\mathrm{the}\:\mathrm{above}\:\mathrm{answers}\:\mathrm{correct}? \\ $$

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