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

Answer: (c)

Using Rule 10,

Speed of train = 90 kmph

= $90 × 5/18$ = 25 m/sec

Distance covered

= 230 + 120 = 350 m

Time taken = $350/25$

= 14 seconds

Answer: (c)

Using Rule 10,

Let the length of train be x

According to the question,

${x + 600}/30$ = 30

x + 600 = 900

x = 900 - 600 = 300 m

Answer: (c)

Using Rule 10,

Let the length of the train be x

According to the question,

${x + 122}/17 = {x + 210}/25$

25x + 3050 = 17x + 3570

25x - 17x = 3570 - 3050

8x = 520 ⇒ x = $520/8$ = 65 metres

Speed of the train

= ${65 + 122}/17 = 187/17$ metre/second

= 11 metre/second

= ${11 × 18}/5$ kmph = 39.6 kmph

Answer: (a)

Using Rule 10,

Length of train = x metre (let)

Speed of train

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

${x + 500}/100 = {x + 250}/60$

${x + 500}/5 = {x + 250}/3$

5x + 1250 = 3x + 1500

5x - 3x = 1500 - 1250

2x = 250 ⇒ x = $250/2$ = 125 metre

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)

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