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(B) Let the speeds of the first and second trains be $x \; m/s$ and $y \; m/s$ respectively.Length of the first train $= x 	imes 27 = 27x \; 	ext{me

Vedclass · Accepted Answer

$3:2$

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(B) Let the speeds of the first and second trains be $x \; m/s$ and $y \; m/s$ respectively.Length of the first train $= x 	imes 27 = 27x \; 	ext{meters}$.Length of the second train $= y 	imes 17 = 17y \; 	ext{meters}$.When the trains cross each other in opposite directions, the relative speed is $(x + y) \; m/s$ and the total distance covered is the sum of their lengths $(27x + 17y) \; 	ext{meters}$.Time taken to cross each other $= \frac{	ext{Total Distance}}{	ext{Relative Speed}} = \frac{27x + 17y}{x + y} = 23$.Multiplying both sides by $(x + y)$, we get:$27x + 17y = 23(x + y)$$27x + 17y = 23x + 23y$Rearranging the terms to solve for the ratio $x/y$:$27x - 23x = 23y - 17y$$4x = 6y$$\frac{x}{y} = \frac{6}{4} = \frac{3}{2}$.Thus, the ratio of their speeds is $3:2$.

Two trains running in opposite directions cross a man standing on the platform in $27 \;seconds$ and $17 \;seconds$ respectively, and they cross each other in $23 \;seconds$. The ratio of their speeds is

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