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

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: (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: (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: (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

Answer: (c)

Speed of train

= $\text"Lengthof(train + platform)"/ \text"Timetaken tocross"$

= $({500 + 700}/10)$ feet/second

= 120 feet/second

Using Rule 10,

Here, x = 500 feet, y = 700 feet

t = 10 seconds, u = ?

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

u = ${500 + 700}/10$

= 120ft/second

Answer: (b)

Speed of train

= $\text"Length of (train + platform)"/ \text"Time taken in crossing"$

= ${(50 + 100)}/10 = 150/10$ = 15 m/sec