Model 3 Train Vs Bridge/Platform Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Using Rule 10,

Let the length of the train be x metres.

When a train corsses a platform it covers a distance equal to the sum of lengths of train and platform.

Also, the speed of train is same.

${x + 162}/18 = {x + 120}/15$

6x + 720 = 5x + 810

6x - 5x = 810 - 720

x = 90

The length of the train = 90m.

Answer: (b)

Let, length of train = length of platform = x metre

Speed of train = 90 kmph

= $({90 × 5}/18)$ m/sec.

= 25 m/sec.

Speed of train

= $\text"Length of train and platform"/ \text"Time taken in crossing"$

25 = ${2x}/60$

2x = 25 × 60

$x = {25 × 60}/2$ = 750 metre

Answer: (a)

Using Rule 10,

36 kmph = $(36 × 5/18)$ m/sec.

= 10 m/sec.

Required time = ${270 + 180}/10$

= 45 seconds

Answer: (b)

Let the length of train be x metre Speed = 90 km/hr

= ${90 × 5}/18$ metre /sec. = 25 metre/sec.

Distance covered in 60 sec.

= 25 × 60 = 1500 metres

Now, according to question,

2x = 1500 ⇒ x = 750 metre

Answer: (a)

When a train crosses a railway platform, it travels a distance equal to sum of length of platform and its own length.

Speed = 132 kmph

= $132 × 5/18 = 110/3$ m/sec.

Required time

= ${110 + 165}/{110/3}$ seconds

= ${275 × 3}/110 = 7.5$ seconds

Using Rule 10,

Here, x = 110m, u = 132 km/hr.

= $132 × 5/18 = 110/3$ m/sec

y = 165m, t = ?

using t = ${x + y}/u$

t = ${110 + 165}/{110/3}$

= ${275 × 3}/110 = {25 × 3}/10$

= $15/2$ = 7.5 sec

Answer: (b)

Speed of train = 54 kmph

= $({54 × 5}/18)$ m/sec.

= 15 m/sec.

Required time

= $\text"Length of train and bridge"/ \text" Speed of train"$

= $({200 + 175}/15)$ seconds

= $(375/15)$ seconds = 25 seconds