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Mr-Crone-started-from-his-house-at-8-00-a-m-and-walked-t-to-his-office-at-an-average-speed-of-4-km-h-Mettle-started-from-Mr-Crones-at-8-30-a-m-and-travelled-by-cycle-in-the-same-direction-as-Mr-

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Question Number 35472 by Issifu1 last updated on 19/May/18
Mr. Crone started from his house at 8.00 a.m. and walked t to his office at an average speed of 4 km/h. Mettle started from Mr. Crones at 8.30 a.m. and travelled by cycle in the same direction as Mr. Crone at average speed of of 6 km/h. If the two arrive arrived at the office at the same time .(i) Find the time of arrival (ii) Find the distance between Mr. Crones house and the office
$${Mr}.\:{Crone}\:{started}\:{from}\:{his}\: \\ $$$${house}\:{at}\:\mathrm{8}.\mathrm{00}\:{a}.{m}.\:{and}\:{walked}\:{t} \\ $$$${to}\:{his}\:{office}\:{at}\:{an}\:{average}\:{speed} \\ $$$${of}\:\mathrm{4}\:{km}/{h}.\:{Mettle}\:{started}\:{from} \\ $$$${Mr}.\:{Crones}\:{at}\:\mathrm{8}.\mathrm{30}\:{a}.{m}.\:{and}\:{travelled} \\ $$$${by}\:{cycle}\:{in}\:{the}\:{same}\:{direction} \\ $$$${as}\:{Mr}.\:{Crone}\:{at}\:{average}\:{speed} \\ $$$${of}\:\:{of}\:\:\mathrm{6}\:{km}/{h}.\:{If}\:{the}\:{two}\:{arrive} \\ $$$${arrived}\:{at}\:{the}\:{office}\:{at}\:{the}\:{same}\:{time} \\ $$$$.\left({i}\right)\:{Find}\:{the}\:{time}\:{of}\:{arrival} \\ $$$$\left({ii}\right)\:{Find}\:{the}\:{distance}\:{between}\:{Mr}. \\ $$$${Crones}\:{house}\:{and}\:{the}\:{office} \\ $$
Answered by Rasheed.Sindhi last updated on 19/May/18
Let Mr Crone takes t hours and covers s km in arriving office. ∴ Mr Mettle takes t−(1/2) hours in covering s km(same distance. In case of Mr Crone: s=4 t In case of Mr Mettle: s=6 (t−(1/2))=6×((2t−1)/2)=6t−3 Comparing in both cases s=4t=6t−3⇒t=(3/2) hours=1(1/2) hours (i)So Mr Crone 8 am+1(1/2) hours=9:30 am (Mr Mettle starts at 8:30 am and he takes time 1(1/2)−(1/2)=1 hour and arrives at same time 9:30 am) (ii)Distence between Mr Crone and his office: s=4t=4×(3/2)=6 km
$$\mathrm{Let}\:\mathrm{Mr}\:\mathrm{Crone}\:\mathrm{takes}\:\mathrm{t}\:\mathrm{hours}\:\mathrm{and} \\ $$$$\mathrm{covers}\:\mathrm{s}\:\mathrm{km}\:\mathrm{in}\:\mathrm{arriving}\:\mathrm{office}. \\ $$$$\therefore\:\mathrm{Mr}\:\mathrm{Mettle}\:\mathrm{takes}\:\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{hours}\:\mathrm{in} \\ $$$$\mathrm{covering}\:\:\mathrm{s}\:\mathrm{km}\left(\mathrm{same}\:\mathrm{distance}.\right. \\ $$$$\mathrm{In}\:\mathrm{case}\:\mathrm{of}\:\mathrm{Mr}\:\mathrm{Crone}: \\ $$$$\:\:\mathrm{s}=\mathrm{4}\:\mathrm{t} \\ $$$$\mathrm{In}\:\mathrm{case}\:\mathrm{of}\:\mathrm{Mr}\:\mathrm{Mettle}: \\ $$$$\:\:\mathrm{s}=\mathrm{6}\:\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{6}×\frac{\mathrm{2t}−\mathrm{1}}{\mathrm{2}}=\mathrm{6t}−\mathrm{3} \\ $$$$\mathrm{Comparing}\:\mathrm{in}\:\mathrm{both}\:\mathrm{cases} \\ $$$$\mathrm{s}=\mathrm{4t}=\mathrm{6t}−\mathrm{3}\Rightarrow\mathrm{t}=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{hours}=\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{hours} \\ $$$$\left(\mathrm{i}\right)\mathrm{So}\:\mathrm{Mr}\:\mathrm{Crone}\:\mathrm{8}\:\mathrm{am}+\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{hours}=\mathrm{9}:\mathrm{30}\:\mathrm{am} \\ $$$$\left(\mathrm{Mr}\:\mathrm{Mettle}\:\mathrm{starts}\:\mathrm{at}\:\mathrm{8}:\mathrm{30}\:\mathrm{am}\:\mathrm{and}\:\mathrm{he}\right. \\ $$$$\mathrm{takes}\:\mathrm{time}\:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{1}\:\mathrm{hour}\:\mathrm{and}\:\mathrm{arrives} \\ $$$$\left.\mathrm{at}\:\mathrm{same}\:\mathrm{time}\:\mathrm{9}:\mathrm{30}\:\mathrm{am}\right) \\ $$$$\left(\mathrm{ii}\right)\mathrm{Distence}\:\mathrm{between}\:\mathrm{Mr}\:\mathrm{Crone}\:\mathrm{and}\:\mathrm{his}\:\mathrm{office}: \\ $$$$\:\:\:\:\:\:\mathrm{s}=\mathrm{4t}=\mathrm{4}×\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{6}\:\mathrm{km} \\ $$$$ \\ $$

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