Let speed of 1st train be 8S and speed of 2nd train be 9S
According to the question.
Speed of second train = 360/4 = 90 km/h
9S = 90
? S = 10
So, speed of 1st train = 8 x 10 = 80 km/h
? Required distance = 80 x 3 = 240 km
Given that, T1 = 9 h and T 2 = 4 h
According to the formula.
(1st train's speed) : (2nd train's speed) = ?4 : ?9
= 2 : 3
Let length of each train be L m.
Then. (65 + 85 ) x 5/18 = (L + L)/6
? (150 x 5 x 6)/18 = 2L
? 2L = 250
? L = 125 m
Let the length of each train be L m.
Relative speed = 46 - 36 = 10 km/h = 10 x 5/18 m/s = 25/9 m/s
Sum of length of train / Relative speed of train = Time taken
? 2L/(25/9) = 36
? 2L = (36 x 25)/9 = 100
? L = 50 m.
Speed = 60 km/h = 60 x (5/18) m/s = 50/3 m/s
Required time = (110 + 170) / (50/3) = (280 x 3)/50 = 16.8 s
Total length of the train = 105 + 90 = 195 m
Relative speed = 72 + 45 km/h = 117 km/h = 117 x (5/18) = 585/18 m/s
? Required time = (195 x (18/585) s = 6 s
Let the trains meet after time t at a distance x from station N.
Then another train coming from station M covers a distance of (x + 50)
For station M,
(x + 50) = 125t
? x = 125t - 50 ..(i)
For station N,
x = 75t .....(ii)
From Eqs. (i) and (ii), we get
75t = 125t - 50
? t = 1h
Distance between station M and N = 125 + 75t = 200 x 1 = 200 km
Because of stoppages, train covers 18 km less per hour .
? Time taken to cover 18 km = 18/108 = 1/6 h = (1/6) x 60 = 10 min
Because of stoppages, train covers 50 km less per hour.
? Time taken taken to cover 50 km = 50/150 h = (50/150) x 60 = 20 min
Let the speeds of two trains be x and y, respectively .
? Length of 1st train = 54x
Length of the 2nd train = 34y
According to the question.
(54x + 34y) / (x + y) = 46
? 54x + 34y = 46x + 46y
? 27x + 17y = 23x + 23y
? 4x = 6y
? x/y = 3/2
? x : y = 3 : 2
The length of the fast train = Relative speed x Time
= (40 - 20) x (5/18) x 5 = 277/9 m
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