A man starts running from point P at 11:00 a. | vedclass

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Question

(C) The first man starts at $11:00 \, a.m.$ and runs at $10 \, km/hr$.He runs for $2 \, hours$ ($11:00 \, a.m.$ to $1:00 \, p.m.$) covering $20 \, km$

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$6:00$

Vedclass · Answer

(C) The first man starts at $11:00 \, a.m.$ and runs at $10 \, km/hr$.He runs for $2 \, hours$ ($11:00 \, a.m.$ to $1:00 \, p.m.$) covering $20 \, km$.He then rests for $1 \, hour$ ($1:00 \, p.m.$ to $2:00 \, p.m.$).At $2:00 \, p.m.$,the first man is at a distance of $20 \, km$ from point $P$.At $2:00 \, p.m.$,the second man starts from point $P$ at $15 \, km/hr$.The relative speed between the two men is $15 - 10 = 5 \, km/hr$.The time taken by the second man to catch the first man is $\frac{	ext{Distance}}{	ext{Relative Speed}} = \frac{20 \, km}{5 \, km/hr} = 4 \, hours$.Therefore,the first man is caught at $2:00 \, p.m. + 4 \, hours = 6:00 \, p.m.$

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