Speed =
300
m/sec =
50
m/sec.
18
3
Let the length of the platform be x metres.
Then,
x + 300
=
50
39
3
3(x + 300) = 1950
x = 350 m.
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300
m/sec =
50
m/sec.
18
3
Let the length of the platform be x metres.
Then,
x + 300
=
50
39
3
3(x + 300) = 1950
x = 350 m.
  =
10 x
5
m/sec
18
  =
25
m/sec
9
2x
=
25
36
9
2x = 100
x = 50.
(100 + 100)
= 3x
8
24x = 200
x =
25
.
3
So, speed of the faster train =
50
m/sec
3
  =
50
x
18
km/hr
3
5
  = 60 km/hr.
= 60 km/hr
=
60 x
5
m/sec
18
=
50
m/sec.
3
Time taken to pass the man
=
500 x
3
sec
50
= 30 sec.
  =
150 x
5
m/sec
18
  =
125
m/sec.
3
Distance covered = (1.10 + 0.9) km = 2 km = 2000 m.
Required time =
2000 x
3
sec = 48 sec.
125
120
m/sec = 12 m/sec.
10
Speed of the second train =
120
m/sec = 8 m/sec.
15
Relative speed = (12 + 8) = 20 m/sec.
Required time =
(120 + 120)
sec = 12 sec.
20
Then, the length of the second train is
x
metres.
2
Relative speed = (48 + 42) kmph =
90 x
5
m/sec = 25 m/sec.
18
[x + (x/2)]
= 12 or
3x
= 300   or   x = 200.
25
2
Length of first train = 200 m.
Let the length of platform be y metres.
Speed of the first train =
48 x
5
m/sec =
40
m/sec.
18
3
(200 + y) x
3
= 45
40
600 + 3y = 1800
y = 400 m.
240
m/sec = 10 m/sec.
24
Required time =
240 + 650
sec = 89 sec.
10
20 x
5
m/sec =
50
m/sec.
18
9
Length of faster train =
50
x 5
m =
250
m = 27
7
m.
9
9
9
So, 2x =
(120 + 120)
12
2x = 20
x = 10.
Speed of each train = 10 m/sec =
10 x
18
km/hr = 36 km/hr.
5