Let the length of the train be x metres and its speed by y m/sec.
Then,
x
= 15 y =
x
.
y
15
x + 100
=
x
25
15
15(x + 100) = 25x
15x + 1500 = 25x
1500 = 10x
x = 150 m.
Then,
x
= 15 y =
x
.
y
15
x + 100
=
x
25
15
15(x + 100) = 25x
15x + 1500 = 25x
1500 = 10x
x = 150 m.
45 x
5
m/sec
=
25
m/sec.
18
2
Time = 30 sec.
Let the length of bridge be x metres.
Then,
130 + x
=
25
30
2
2(130 + x) = 750
x = 245 m.
Video Explanation: https://youtu.be/M_d8WufJWKc
4.5 x
5
m/sec =
5
m/sec = 1.25 m/sec, and
18
4
5.4 km/hr =
5.4 x
5
m/sec =
3
m/sec = 1.5 m/sec.
18
2
Let the speed of the train be x m/sec.
Then, (x – 1.25) x 8.4 = (x – 1.5) x 8.5
8.4x – 10.5 = 8.5x – 12.75
0.1x = 2.25
x = 22.5
Speed of the train =
22.5 x
18
km/hr = 81 km/hr.
5
27x + 17y
= 23
x+ y
27x + 17y = 23x + 23y
4x = 6y
x
=
3
.
y
2
X x
5
m/s.
18
Therefore, Speed =
45 x
5
m/sec
=
25
m/sec.
18
2
Total distance to be covered = (360 + 140) m = 500 m.
Formula for finding Time =
Distance
Speed
Required time =
500 x 2
sec
= 40 sec.
25
=
66 x
5
m/sec
18
=
55
m/sec.
3
Time taken to pass the man =
110 x
3
sec = 6 sec.
55
78 x
5
m/sec
=
65
m/sec.
18
3
Time = 1 minute = 60 seconds.
Let the length of the tunnel be x metres.
Then,
800 + x
=
65
60
3
3(800 + x) = 3900
x = 500.
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.