A train leaves station A at 5, AM and reaches | vedclass

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(D) Let the distance between station $A$ and $B$ be $D$.Time taken by the $1^{st}$ train from $A$ to $B = 4$ hours.Speed of the $1^{st}$ train $(v_1)$

Vedclass · Accepted Answer

$7.56$

Vedclass · Answer

(D) Let the distance between station $A$ and $B$ be $D$.Time taken by the $1^{st}$ train from $A$ to $B = 4$ hours.Speed of the $1^{st}$ train $(v_1)$ = $D/4$.Time taken by the $2^{nd}$ train from $B$ to $A = 3.5$ hours = $7/2$ hours.Speed of the $2^{nd}$ train $(v_2)$ = $D / (7/2) = 2D/7$.The $1^{st}$ train starts at $5\, AM$. By $7\, AM$,it has traveled for $2$ hours.Distance covered by the $1^{st}$ train in $2$ hours = $(D/4) 	imes 2 = D/2$.Remaining distance at $7\, AM = D - D/2 = D/2$.Since the trains are moving towards each other,their relative speed = $v_1 + v_2 = D/4 + 2D/7 = (7D + 8D) / 28 = 15D/28$.Time taken to meet after $7\, AM$ $(t)$ = $	ext{Remaining Distance} / 	ext{Relative Speed} = (D/2) / (15D/28) = (D/2) 	imes (28/15D) = 14/15$ hours.Convert $14/15$ hours to minutes: $(14/15) 	imes 60 = 56$ minutes.Therefore,the trains meet at $7\, AM + 56$ minutes = $7:56\, AM$.

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