Let us name the trains as A and B. Then,
(A’s speed) : (B’s speed) = b : a = 16 : 9 = 4 : 3.
=
75 x
5
m/sec
18
=
125
m/sec.
6
We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
Required time =
500 x
6
= 24 sec.
125
Relative speed
= (x + 50) km/hr
=
(x + 50) x
5
m/sec
18
=
250 + 5x
m/sec.
18
Distance covered = (108 + 112) = 220 m.
220
= 6
250 + 5x
18
250 + 5x = 660
x = 82 km/hr.
Then,
x
= 8 x = 8y
y
Now,
x + 264
= y
20
8y + 264 = 20y
y = 22.
Speed = 22 m/sec =
22 x
18
km/hr = 79.2 km/hr.
5
125
m/sec
10
=
25
m/sec.
2
=
25
x
18
km/hr
2
5
= 45 km/hr.
Let the speed of the train be x km/hr. Then, relative speed = (x – 5) km/hr.
x – 5 = 45 x = 50 km/hr.
=
36 x
5
m/sec
18
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Time taken =
360
sec
= 36 sec.
10
=
7
+
1
miles
2
4
=
15
miles.
4
Time taken
=
15
hrs
4 x 75
=
1
hrs
20
=
1
x 60
min.
20
= 3 min.
100 x
5
m/sec
=
250
m/sec.
18
9
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time =
300 x
9
sec
=
54
sec = 10.8 sec.
250
5
54 x
5
m/sec = 15 m/sec.
18
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
Then,
x + 300
= 15
36
x + 300 = 540
x = 240 m.