A, B and C can walk at the rates of 3, 4 and | vedclass

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(B) Let $t$ be the time taken by $B$ to catch $A$ after $B$ starts at $2 	ext{ o'clock}$.$A$ starts at $1 	ext{ o'clock}$,so at $2 	ext{ o'clock}$,

Vedclass · Accepted Answer

$5:15 	ext{ o'clock}$

Vedclass · Answer

(B) Let $t$ be the time taken by $B$ to catch $A$ after $B$ starts at $2 	ext{ o'clock}$.$A$ starts at $1 	ext{ o'clock}$,so at $2 	ext{ o'clock}$,$A$ has already covered $3 	ext{ km}$.Relative speed of $B$ with respect to $A = 4 - 3 = 1 	ext{ km/h}$.Time taken by $B$ to catch $A = \frac{	ext{Distance}}{	ext{Relative Speed}} = \frac{3 	ext{ km}}{1 	ext{ km/h}} = 3 	ext{ hours}$.So,$B$ catches $A$ at $2 + 3 = 5 	ext{ o'clock}$.At $5 	ext{ o'clock}$,the distance from Pune is $4 	imes 3 = 12 	ext{ km}$.Now,$B$ sends a message back to $C$ from this point ($12 	ext{ km}$ from Pune).At $5 	ext{ o'clock}$,$C$ has been walking for $5 - 3 = 2 	ext{ hours}$.Distance covered by $C$ at $5 	ext{ o'clock} = 5 	imes 2 = 10 	ext{ km}$ from Pune.Distance between $B$ (at $12 	ext{ km}$) and $C$ (at $10 	ext{ km}$) = $12 - 10 = 2 	ext{ km}$.Since they are moving towards each other,their relative speed = $4 + 5 = 9 	ext{ km/h}$.Time taken for them to meet = $\frac{	ext{Distance}}{	ext{Relative Speed}} = \frac{2}{9} 	ext{ hours} = \frac{2}{9} 	imes 60 	ext{ minutes} = 13.33 	ext{ minutes}$.This calculation suggests a discrepancy in the provided options. Re-evaluating: If $B$ sends $A$ back,$A$ travels at $3 	ext{ km/h}$ towards $C$. $C$ travels at $5 	ext{ km/h}$ towards $A$. Distance between them at $5 	ext{ o'clock}$ is $2 	ext{ km}$. Relative speed = $3 + 5 = 8 	ext{ km/h}$.Time = $\frac{2}{8} 	ext{ hours} = \frac{1}{4} 	ext{ hour} = 15 	ext{ minutes}$.Therefore,$C$ receives the message at $5:15 	ext{ o'clock}$.

$A, B$ and $C$ can walk at the rates of $3, 4$ and $5 \text{ km/h}$ respectively. They start from Pune at $1, 2$ and $3 \text{ o'clock}$ respectively. When $B$ catches $A$,$B$ sends him back with a message to $C$. When will $C$ get the message?

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