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Question Number 184218 by katana last updated on 04/Jan/23

Answered by Rasheed.Sindhi last updated on 04/Jan/23

$$\mathrm{Suppose}\mathrm{car}\mathrm{catches}\mathrm{the}\mathrm{bus}\mathrm{after}\mathrm{x}$$$$\mathrm{hours}.$$$$\bullet \mathrm{When}\mathrm{the}\mathrm{car}\mathrm{catces}\mathrm{the}\mathrm{bus},\mathrm{the}\mathrm{bus}$$$$\mathrm{has}\mathrm{travelled}\mathrm{for}2\frac{1}{2}+\mathrm{x}\mathrm{hours},(\mathrm{while}$$$$\mathrm{the}\mathrm{car}\mathrm{has}\mathrm{travelled}\mathrm{for}\mathrm{x}\mathrm{hours}.)$$$$\bullet \mathrm{The}\mathrm{bus}\mathrm{travelled}\mathrm{the}\mathrm{distance}\mathrm{in}$$$$\frac{5+2\mathrm{x}}{2}\mathrm{hours}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{distance}$$$$\mathrm{travelled}\mathrm{by}\mathrm{car}\mathrm{in}\mathrm{x}\mathrm{hours}:$$$$\frac{5+2\mathrm{x}}{2}\cdot 60=\mathrm{x}\cdot 100$$$$150+60\mathrm{x}=100\mathrm{x}\Rightarrow \mathrm{x}=\frac{150}{40}=3\frac{3}{4}\mathrm{hoursT}$$$$\mathrm{The}\mathrm{distance}\mathrm{travelled}\mathrm{by}\mathrm{bus}\mathrm{from}$$$$\mathrm{Mombasa}:$$$$3\frac{3}{4}\times 60=\frac{15}{4}\times 60=225\mathrm{km}$$$$\mathrm{a}\left(\mathrm{i}\right)\mathrm{The}\mathrm{distance}\mathrm{of}\mathrm{bus}\mathrm{from}\mathrm{Nairobi}:$$$$=500-225=275\mathrm{km}$$$$\mathrm{a}\left(\mathrm{ii}\right)\mathrm{The}\mathrm{car}\mathrm{travelled}225\mathrm{km}\mathrm{to}$$$$\mathrm{catch}\mathrm{up}\mathrm{the}\mathrm{bus}.$$$$\left(\mathrm{c}\right)\mathrm{If}\mathrm{the}\mathrm{car}\mathrm{would}\mathrm{have}\mathrm{continued}$$$$\mathrm{his}\mathrm{first}\mathrm{average}\mathrm{speed}\mathrm{it}\mathrm{would}\mathrm{have}$$$$\mathrm{completed}\mathrm{its}\mathrm{journey}\mathrm{in}5\mathrm{hours}.$$$$\bullet \mathrm{In}\mathrm{case}\mathrm{car}\mathrm{stays}\mathrm{for}25\mathrm{minutes}$$$$(=\frac{5}{12}\mathrm{hours})\mathrm{let}\mathrm{the}\mathrm{remaining}\mathrm{journey}$$$$\mathrm{is}\mathrm{completed}\mathrm{in}\mathrm{y}\mathrm{hours}$$$$\mathrm{x}+\frac{5}{12}+\mathrm{y}=5$$$$3\frac{3}{4}+\frac{5}{12}+\mathrm{y}=5$$$$\mathrm{y}=5-\frac{25}{6}=\frac{5}{6}\mathrm{hours}$$$$\mathrm{In}\mathrm{this}\mathrm{duration}\left(\frac{5}{6}\mathrm{hours}\right)\mathrm{the}\mathrm{car}$$$$\mathrm{has}\mathrm{to}\mathrm{travel}275\mathrm{km}$$$$\mathrm{v}=\frac{225}{5/6}=225\times \frac{6}{5}=270\mathrm{km}/\mathrm{hour}$$$$\left(\mathrm{c}\right)270\mathrm{km}/\mathrm{hour}$$

Commented by mr W last updated on 05/Jan/23

$$HappyNewYeartoyoutoo!$$$$Youranswerissurelycorrect.$$$$Ijustlikethefastcarwhichcan$$$$catchupthebus.onlyfewcarsin$$$$theworldcandrivesofast.$$

Commented by mr W last updated on 04/Jan/23

Commented by Rasheed.Sindhi last updated on 04/Jan/23

$$\mathbf{Sir}\mathbf{Mr}\mathbf{W}!$$$$\overline{)\begin{array}{c}{\overline{)\begin{array}{c}{\mathbf{HAPPY}}_{{\mathbf{NEW}}_{\mathbf{YEAR}!}}^{}\end{array}}}_{}^{}\end{array}}$$

Commented by Rasheed.Sindhi last updated on 04/Jan/23

$$\mathrm{Are}\mathrm{the}\mathrm{above}\mathrm{answers}\mathrm{correct}?$$

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