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

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: (b)

Speed of train = 36kmph

= $36 × 5/18$ = 10 m/sec.

If the length of bridge be x metre, then

10 = ${200 + x}/55$

200 + x = 550

x = 550 - 200 = 350 metre.

Using Rule 10,

Here, x = 200 m

u = 36 km/hr, ${36 × 5}/18$ m/second

= 10 m/sec

y = ?, t = 55 sec

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

55 = ${200 + y}/10$

y = 550 - 200 = 350 m

Answer: (b)

In crossing the bridge, the train travels its own length plus the length of the bridge.

Total distance (length)

= 300 + 200 = 500 m.

Speed = 25m/sec.

The required time

= 500 ÷ 25 = 20 seconds

Here, x = 300m, y= 200m, t = ? u= 25 m/sec

t = ${x + y}/u$

= ${300 + 200}/25$

t = $500/25$ = 20 seconds

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)

Using Rule 10,

Let the length of the train be x

Speed of train

${x + 300}/21 ={x + 240}/18$

${x + 300}/7 ={x + 240}/6$

7x + 1680 = 6x + 1800

x = 120

Speed of train

= ${x + 300}/21 = 420/21$= 20 m/sec

= $({20 × 18}/5)$ kmph = 72 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