A train leaves a station A at 7:00 text{ am} | vedclass

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(D) Time taken by the $1^{st}$ train $= 4 	ext{ hours}.$Time taken by the $2^{nd}$ train $= 3.5 	ext{ hours} = \frac{7}{2} 	ext{ hours}.$Let the di

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$9:24 	ext{ am}$

Vedclass · Answer

(D) Time taken by the $1^{st}$ train $= 4 	ext{ hours}.$Time taken by the $2^{nd}$ train $= 3.5 	ext{ hours} = \frac{7}{2} 	ext{ hours}.$Let the distance between $A$ and $B$ be $D.$ Speed of $1^{st}$ train $(v_1) = \frac{D}{4}.$Speed of $2^{nd}$ train $(v_2) = \frac{D}{3.5} = \frac{2D}{7}.$The $1^{st}$ train starts at $7:00 	ext{ am}$ and the $2^{nd}$ train starts at $8:00 	ext{ am}.$ In $1 	ext{ hour}$ (from $7:00 	ext{ am}$ to $8:00 	ext{ am}$),the $1^{st}$ train covers distance $= v_1 	imes 1 = \frac{D}{4}.$Remaining distance at $8:00 	ext{ am} = D - \frac{D}{4} = \frac{3D}{4}.$Since the trains are moving towards each other,their relative speed $= v_1 + v_2 = \frac{D}{4} + \frac{2D}{7} = \frac{7D + 8D}{28} = \frac{15D}{28}.$Time taken to meet after $8:00 	ext{ am} = \frac{	ext{Remaining Distance}}{	ext{Relative Speed}} = \frac{3D/4}{15D/28} = \frac{3}{4} 	imes \frac{28}{15} = \frac{7}{5} 	ext{ hours}.$$\frac{7}{5} 	ext{ hours} = 1 	ext{ hour and } 24 	ext{ minutes}.$Meeting time $= 8:00 	ext{ am} + 1 	ext{ hour } 24 	ext{ minutes} = 9:24 	ext{ am}.$

$A$ train leaves a station $A$ at $7:00 \text{ am}$ and reaches another station $B$ at $11:00 \text{ am}.$ Another train leaves station $B$ at $8:00 \text{ am}$ and reaches $A$ at $11:30 \text{ am}.$ At what time do the two trains cross each other?

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