Question

A train crosses a platform 90 long in 30 seconds and a man standing on the platform in 15 seconds. Find the speed of the train​

Answer 1

$\sf {let \: the \: speed \: of \: train \: be \: x}$

$\sf \dfrac{length \: of \: the \: speed}{speed \: of \: the \: train} = time \: taken \: by \: train \: to \: pa - ss \: the \: man$

$\implies \sf \dfrac{y }{x} = 15$

$\implies \sf {y} = 15x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

$\sf \dfrac{length \: of \: train \: + length \: of \: platform}{speed \: of \: train \: } = \: time \: taken \: by \: the \: train \: to \: cross \: the \: platform \:$

$\implies\sf \dfrac{y + 90}{x} = 30$

$\implies\sf{y + 90 = 30x} \: \: \: \: \: \: \: \: \: \: \: (2)$

$\sf{put \: the \: value \: of \: y \: from(1) \: in \: (2)}$

$\implies \sf15x + 90 = 30x$

$\implies \sf{15x = 90}$

$\implies \sf{ x \: = 6m \ / \sec}$

$\sf {speed \: of \: train = (6 \times \dfrac{18}{5}})km/hr = \: 21.6km \: / \: hr$