Question

The train travel at the speed of X km/h. Crossed a 200 metre long platform in 30 seconds hand overtook a man walking in the same direction at the speed of 6 km/h in 20 seconds. what is the value of x ?

Answer 1

=> FIND

=> value of x

=> GIVEN

=> train travel with a speed of x km/hr

=> crossed 200 metre long platform in 30 seconds.

let the length of the train be = L m

so, distance = speed x time

=> ( l + 200 ) = 30x

=> l = 30x - 200 .....(1)

=> speed of man = 6 km/hr

=> change in m/s = 6 x 5/18

=> 5/3 m/s

=> relative speed = x - 5/3

=> l / 20 = 3x - 5/3

=> 30x - 200 /20 = 3x - 5/3

=> 90x - 600 = 60x - 100

=> 30x = 500

=> x = 50/3 m/s

=> x = 50 / 3 x 18 / 5 = 60 km/ hr

Answer 2

$\fbox{Given}$

→$\tt{Train\:travel\:with\:a\:speed\:of\:x\:km/hr}$

→$\tt{Crossed\:200m\:long\:platform\:in\:30\:sec}$

$\fbox{To\:Find}$

$\tt{The\:value\:of\:x}$

$\fbox{Solution}$

$\tt{Let\:the\:length\:of\:the\:train\:be\:L\:m}$

$\tt{we\:know\:that}$

$\fbox{Distance\:=\:Speed×Time}$

$→(l + 200) = 30x \\ →l = 30x - 200.............(1) \\ →speed \: of \: man \: = \: 6kmhr^{ - 1} \\ →change \: in \: ms^{ - 1} = 6 \times \frac{5}{18} \\ →change \: in \: ms ^{ - 1} = \frac{5}{3} {ms}^{ - 1} \\ →relative \: speed \: = x - \frac{5}{3} \\ → \frac{l}{20} = 3x - \frac{5}{3} \\ →30x - \frac{200}{20} = 3x - \frac{5}{3} \\ →90x - 600 = 60x - 100 \\ →30x = 500 \\ →x = \frac{50}{3} {ms}^{ - 1} \\ →x = \frac{50}{3} \times \frac{18}{5 } \\ →x = 60 {kmhr}^{ - 1}$

$\fbox{x=60km/hr}$