Speed of the train = 12000 / 10*60 = 20 m/s
Length of the train = Speed X time = 20 X 6 = 120m
Let length of platform be ?y?metres.
Then, at the platform,Distance travelled = y + (length of train) = (y + 480)m
Then, Speed of train = (y + 480)m / (3x60) sec = (y+480)/180 ...(1)
Also, at the pole,Distance travelled = length of train = 480m
Then, Speed of train = 480m / 30sec = 16 m/s ...(2)
Equating eq.(1) & eq.(2), we get,
(y+480)/180 = 16(y+480) = 16 x 180 = 2880
y = 2880 ?480 = 2400m or 2.4 km long platform.
Ans.
Given that length of first train + length of second train = 660
Speed of first train : Speed of second train = 5:8 =5x : 8x
Time taken to cross the poll by two trains = 4:3 =4y:3y
Now ,
(5x*4y) + (8x*3y) = 660
44xy = 660
xy = 15
L1 - L2 = 24xy -20xy = 4xy = 60 m
Let length of train A be ?L? m and speed be ?V? m/s
ATQ ?
V =
Ans.
Distance covered by 1st train in 4 hours = 80*4 = 320 km
Relative speed of 2nd train = 120-80 = 40 km
It will overtake another train in 320/40 = 8 hours
Distance between overtake point and station A = 8*120 = 960 km
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