A train leaves station A at 7 , am and reache | vedclass

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(D) Let the distance between station $A$ and $B$ be $28 \, km$ ($LCM$ of $4$ and $3.5$).Speed of train from $A = \frac{28}{4} = 7 \, km/h$.Speed of tr

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$9.24$

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(D) Let the distance between station $A$ and $B$ be $28 \, km$ ($LCM$ of $4$ and $3.5$).Speed of train from $A = \frac{28}{4} = 7 \, km/h$.Speed of train from $B = \frac{28}{3.5} = 8 \, km/h$.The train from $A$ starts at $7 \, am$. By $8 \, am$,it has traveled for $1$ hour.Distance covered by train from $A$ in $1$ hour $= 7 \, km/h 	imes 1 \, h = 7 \, km$.Remaining distance at $8 \, am = 28 - 7 = 21 \, km$.Since the trains are moving towards each other,their relative speed $= 7 + 8 = 15 \, km/h$.Time taken to cover the remaining distance $= \frac{21}{15} \, hours = \frac{7}{5} \, hours = 1 \, hour \, 24 \, minutes$.Therefore,the trains cross each other at $8 \, am + 1 \, hour \, 24 \, minutes = 9:24 \, am$.

$A$ train leaves station $A$ at $7 \, am$ and reaches station $B$ at $11 \, am$. Another train leaves station $B$ at $8 \, am$ and reaches station $A$ at $11.30 \, am$. At what time do the two trains cross each other?

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