P and Q are 27 km away. Two trains will having speeds of 24 km/h and 18 km/h respectively starts simultaneously from P and Q and travel in the same direction. They meet at a point R beyond Q. Distance QR is

Given that, t₁ = 6 s and t₂ = 9 s
Then, time taken by the trains to cross each other = 2t₁ t₂ / (t₂ - t₁)
= (2 x 6 x 9)/(9 - 6) = 36 s

The length of the fast train = Relative speed x Time
= (40 - 20) x (5/18) x 5 = 27⁷/₉ m

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

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

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

Let the speed of both trains be V km/h and (V + 6) km/h, respectively
Then, according to the question.
160 = V x 4 + (V + 6) x 4
? 160 = 4V + 4V + 24
? 40 = V + V + 6
? 2V + 6 = 40
? 2V = 34
? V = 17
Hence, speeds of both the trains are 17 km/h and (17 + 6 ) km/h i,e 17 km/h and 23 km/h .

Let the two trains meet at D km from Mumbai. Then,
3:45 pm + D/50 = 2 : 35 pm + 510 - D/60
? (3:45 pm - 2:35 pm) + (D/50 + D/ 60) = 510/60
? (1¹/₆ h) + D{( 50 + 60 / 50 x 60)} = 17/2
? D(110/3000) = 17/2 - 7/6
? D(110/3000) = (51 - 7)/6 = 44/6
? D = (44/6) x (3000/110) = 200 km

Let the lengths of the trains A and B be a and 2a, respectively.
when a train crosses a pole, it covers the distance equal to its length.
? Required ratio of speeds = a / 25 : 2a /75 = 3 : 2

The distance covered by the train to cross the platform and bridge
= Length of the platform + Length of bridge + Distance between platform and bridge + Length of the train
= 165 + 135 + 30 + 110 = 440 m
Speed of train = 110 /3 m/s
? Required time taken = 440/(110/3) = 12 s

Let time taken to cover the distance = t
? Speed = (2t + 1)
? t(2t + 1) = 1830 ? 2t² + t = 1830
? 2t² + t - 1830 = 0
? t = -1 ± ?(1)² - 4 x 2 x(-1830) / 2 x 2
= -1 ± ?1 + 14640 / 4
= -1 ± 121 / 4
Taking '+' sign t = -1 + 121 / 4 = 120/4 = 30
? Required ratio = (2 x 30 + 1) : 30 = 61 : 30