Two trains, in which one train is 50 meters longerthan another, are ru

1.	Two trains, in which one train is 50 meters longerthan another, are running in opposite directions andcrosses each other in 10 seconds. If they run in thesame direction, the fast moving train crosses thesecond train in 1 minute & 30 seconds. If the speedof the fast moving train is 90 kmph, then what is thelength of each train?
Answer» Let the LENGTH of trains be X m and (x + 50)m and the speed of other TRAIN be y m per sec. The speed of the FIRST train = 90 km per hr. = 90 × 5 = 25 m per sec. 18 Case : I Opposite direction, Their relative speed = (y + 25)m per sec. Distance covered = x + x + 50 = 2x + 50 metres ∴ TIME taken = 2x + 50 = 10 y + 25 ⇒ 2x + 50 = 10y + 250 ...(i) Case II. Direction is Same Their relative speed = (25 – y) m per sec. Distance covered = x + x + 50 = 2x + 50m ∴ Time taken = 2x + 50 = 90 25 − y ⇒ 2x + 50 = 90 (25 – y) ...(ii) From equations (i) and (ii) 10y + 250 = 2250 – 90y ⇒ 10y + 90y = 2250 – 250 ⇒ y = 2000 = 20 100 Putting y = 20 in equation (i), we have 2x + 50= 10 × 20 + 250 = 450 ⇒ 2x = 450 – 50 = 400 ⇒ x = 400 = 200 2 ∴ x + 50 = 200 + 50 = 250 metres. Hence, The length of the 1st train = 200 metres. The length of the 2nd train = 250 metres. The speed of the 2nd train = 20 m per sec.

Two trains, in which one train is 50 meters longerthan another, are running in opposite directions andcrosses each other in 10 seconds. If they run in thesame direction, the fast moving train crosses thesecond train in 1 minute & 30 seconds. If the speedof the fast moving train is 90 kmph, then what is thelength of each train?

Answer»

Let the LENGTH of trains be X m and (x + 50)m and the speed of other TRAIN be y m per sec.

The speed of the FIRST train = 90 km per hr.

= 90 ×

= 25 m per sec.

Case : I Opposite direction,

Their relative speed = (y + 25)m per sec.

Distance covered = x + x + 50 = 2x + 50 metres

∴ TIME taken =

2x + 50

= 10

y + 25

⇒ 2x + 50 = 10y + 250 ...(i)

Case II. Direction is Same

Their relative speed = (25 – y) m per sec.

Distance covered = x + x + 50 = 2x + 50m

∴ Time taken =

2x + 50

= 90

25 − y

⇒ 2x + 50 = 90 (25 – y) ...(ii)

From equations (i) and (ii)

10y + 250 = 2250 – 90y

⇒ 10y + 90y = 2250 – 250

⇒ y =

2000

= 20

100

Putting y = 20 in equation (i), we have

2x + 50= 10 × 20 + 250 = 450

⇒ 2x = 450 – 50 = 400

⇒ x =

400

= 200

∴ x + 50 = 200 + 50 = 250 metres.

Hence,

The length of the 1st train = 200 metres.

The length of the 2nd train = 250 metres.

The speed of the 2nd train = 20 m per sec.