Correct Option: C

Let the speed of train be x kmph and its length be y km.
When the train crosses a man, it covers its own length
According to he question,

=	y					= 10
	(x − 3) ×	5
		3

⇒  18 y = 10 × 5(x –3)
⇒  18y = 50x –150   ..... (i)

and,	y					= 11
	(x − 5) ×	5
		18

⇒  18y = 55(x–5)
⇒  18y = 55x –275   .... (ii)
From equations (i) and (ii),
55x –275 = 50x–150
⇒  55x –50x = 275 – 150
⇒  5x = 125

⇒  x =	125	= 25
	5

∴  Speed of the train = 25 kmph
Second Method :
Here, S₁ = 3, S₂ = 5

t1 =	10	,  t2 =	11
	3600		3600

Speed of train =	t1S1 − t2S2
	t1 − t2

=	3 × 10	−	5 × 11
	3600		3600
	10	−	11
	3600		3600

=	− 25	×	3600
	3600		− 1

= 25 m/sec