Two trains running at 36 km/hr and 54 km/hr are running parallel in the opposite direction. If their lengths are 200m and 300m respectively. Then how much time will they take to cross each other?

Let the length of train be x meter and speed be y m/sec

Therefore, time taken to cross the pole = x/y sec

and time taken to cross the platform = ( Length of train + Length of platform) / speed of train

Therefore x/y = 10
And, (x+200) / y = 20
? 10y + 200 = 20 y
? speed of train, y = 20 m/sec
And length of rain = x = 10y = 200m

Speed in km/ hour =200

Speed in m/sec = 200 x 5/18 = 500/9 m/sec = 55.55 m/sec

Let the speed of first lady be (v + 2)km/h and speed of second lady will be v km/h.
Total distance = 24 km

For first lady, Speed = Distance/Time
v + 2 = 24/t ...(i)

Let time taken by first lady to cover distance of 24 km be t h.
Then, time taken by second lady to cover same distance will be(t + 1)h.
For second lady, Speed =Distance/time
v = 24/(t + 1) ...(ii)

Form Eqs.(i) and (ii), we get
24/t - 2 = 24/(t + 1)
? (24 - 2t)/t = 24/(t + 1)
? (24 - 2t)(t + 1) = 24t
? 24t - 2t² + 24 - 2t = 24t
? 2t² + 2t - 24 = 0
? t² + t - 12 = 0
? t² + 4t - 3t - 12 = 0
? t(t + 4) -3(t + 4) = 0
? (t - 3)(t + 4) = 0
? t = 3, -4
? t = 3 (t ?- 4)
? The distance travelled by the first lady in one hour
= 24/t = 24/3 = 8 km
And distance travelled by the second lady in one hour
= 24/t + 1
= 24/3 + 1 = 6 km

Let the time taken by car B to reach destination is T h.
So, the time taken by car A to reach destination is (T + 2) h.
Now, S₁T₁ = S₂T₂
? 30(T + 2) = 45 x T
? 30T + 60 = 45T
? 15T = 60
? T = 4 h

Now, distance between starting point and destination = S₂T₂
= 45 x 4 = 180 km

Let total distance = D
Now, according to the question,
40 x 3 + 60 x 4.5 = 3D/5
? 120 + 270 = 3D/5
? D = 390 x 5/3 = 650 km

? Remaining distance = [650 - (3/5) x 650]
= 650 - 390 = 260 km

Required average speed = 260/4 = 65 km/h

Speed of the train = (60 x 5/18) m/sec = 50/3 m/sec
Distance covered in passing the standing person = 200m
Required time taken= =200/(50/3) = 200 x (3/50) = 12 sec

Speed of the train = (264 x 5/18) m/sec = 220/3 m/sec
Distance covered in passing the platform = (330+330) m= 660m
Time taken= ( 660 / (220/3) ) sec = (660 x 3/ 220 ) = 9 sec

Length of train = 264 m

Time taken to cross the pole = 12 sec

Speed of the train, in m/sec = Length of the train/ Time taken to cross the pole

Therefore, speed in m/sec= (264 / 12) m/sec = 22 m /sec

Calculating speed in km/ hour = (22 m/sec ) x 18/5 = 396/5 km/hr = 79.2 km/ hour

Speed in m/sec = 36 km/hr x 5/18 = 10m/sec

Distance covered by train in crossing the bridge = Speed of train in m/sec x time in sec = 10 m/sec x 30 sec = 300 m

Now, we know that
Distance covered by train in crossing the bridge= Length of train + Length of bridge

Therefore, Length of bridge = 300m - 200m = 100m

Speed of train in m/sec = 360m / 36 sec = 10 m/sec

Distance covered by train in crossing the platform = Length of train + Length of Platform = 360m + 240m = 600m

Therefore, Time taken to cover this distance = 600m /(10 m/sec)
= 60 sec