Let the length of the stationary train Y be LY
Given that length of train X, LX = 300 m
Let the speed of Train X be V.
Since the train X crosses train Y and a pole in 60 seconds and 25 seconds respectively.
=> 300/V = 25 ---> ( 1 )
(300 + LY) / V = 60 ---> ( 2 )
From (1) V = 300/25 = 12 m/sec.
From (2) (300 + LY)/12 = 60
=> 300 + LY = 60 (12) = 720
=> LY = 720 - 300 = 420 m
Length of the stationary train = 420 m
Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.
Let us name the trains as A and B.
Then, (A's speed) : (B's speed)
= ?b : ?a = ?16 : ?9 = 4:3
Total time taken = k/40 + 2k/20 hours
= 5k/40 = k/8 hours
Average speed = 3k/(k/8) = 24 kmph.
Speed = 78 x 5/18 = 65/3 m/sec.
Time = 1 min = 60 sec.
Let the length of the train be x meters.
Then, (900 + x)/60 = 65/3
x = 400 m.
The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.
Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.
Time required = d/s = 500/10 = 50 sec.
We know that, Speed = Distance/time
Let the speed of the train be = x kmph
NORMAL SPEED SPEED 5 KMPH MORE
Distance = 360 Km Distance =360
speed = X Kmph Speed = (x+5)
Time =360/X Time =360/(x+15)
For the same journey if the speed increased 10 kmph more it will take 1 hour less
Time with original speed - time with increased speed = 1
=> (360/x) - (360/x+10) = 1
=> (360(x+5)-360x)/x(x+5) =1
=> 360X + 1800 -360X = X(X+5)
=> X^2 + 5x - 1800 = 0
X = 40 or -45
X cannot be negetive value
X = 40 kmph.
Let the original speed and time is S and T
then distance = S x T
Now the speed changes to 2/3S and T is T+20
As the distance is same
S x T = 2/3Sx(T+20)
solving this we get t = 40 minutes
=40/60 = 2/3 hour
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