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

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

(Speed of the train)

${x + 162}/6 = {x + 120}/5$

6x + 720 = 5x + 810

x = 810 - 720 = 90

Speed of the train

=${90 + 162}/18$ m/sec.

= $252/18 × 18/5$ kmph = 50.4 kmph

Answer: (d)

Speed of train

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

=$({450 +150}/20)$ m/sec.

= $(600/20)$ m/sec.

= $(30 × 18/5)$ kmph = 108 kmph.

Answer: (a)

Using Rule 10,

Speed of train

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

= $({100 + 50}/10)$ metre/second

= 15 metre/second

= $(15 × 18/5)$ kmph = 54 kmph

Answer: (d)

Using Rule 10,

Speed of train = 36 kmph

= $({36 × 5}/18)$ m/sec.

= 10 m/sec

Required time

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

= ${120 + 360}/10$

= $480/10$ = 48 seconds

Answer: (a)

When a train crosses a tunnel, it covers a distance equal to the sum of its own length and tunnel.

Let the length of tunnel be x Speed = 78 kmph

= ${78 × 1000}/{60 × 60}$ m/sec. = $65/3$ m/sec.

Speed = $\text"Distance"/\text"Time"$

$65/3 = {800 + x}/60$

(800 + x ) × 3 = 65× 60

800 + x = 65 × 20 m

x = 1300 - 800 = 500

Length of tunnel = 500 metres.

Using Rule 10,

Here, x = 800 m, u = 78 km/hr

= 78$5/18 = 65/3$ m/sec

t = 1 min = 60 sec, y = ?

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

$60 = {800 + y}/{65/3}$

60 × $65/3$ = 800 + y

1300 - 800 = y ⇒ y = 500 metres

Answer: (c)

Using Rule 10,

Time = 5 minutes = $1/12$ hour

Length of bridge = Speed × Time

= 15 × $1/12 = 5/4$ km.

= $(5/4 × 1000)$ metre

= 1250 metre