A train meets with an accident after travelling 30 kms,

A train meets with an accident after travelling 30 kms, after which it moves
with 4 th of its original speed and arrives
5
at the destination 45 minutes late. Had the accident happened 18 kms further on, it would have been 9 minutes before. Find the distance of journey and original speed of the train.

Let the original speed be x and distance be y
Case I.
Time taken by train to travel

30 km =	30
	x

Time taken by train after accident =	y − 30
	4/5x

Total time taken =	30	+	y − 30
	x		4/5x

Case II :
Time taken by train to travel

48 km =	48
	x

Time taken by train after accident =	y − 48
	4/5x

Total time taken =	48	+	y − 48
	x		4/5x

	30	+	y − 30		−		48	+	y − 48
	x		4/5x				x		4/5x

=	9	[∵ Difference between time is 9 minutes]
	60

	y − 30	−	y − 48		+		30	−	48
	4/5x		4/5x				x		x

=	9
	60

y − y − 30 + 48	+	(− 18)	=	9
4/5x	+	x	=	60

5(18)	−	18	=	9
4x	−	4x	=	60

⇒	90 − 72	=	9
	4x		60

x =	18 × 60	= 30
	4 × 9

Hence, original speed = 30 kmph
Also,

30	+	y − 30	=	y	+	45
x	+	4/5x	=	x	+	60

[Original time + 45 minute = New time]
⇒ 3x – y = –30
⇒ 3(30) – y = –30
⇒ i.e. Distance = 120 km