10
Q:

# Two trains of equal length , running in opposite directions , pass a pole in 18 and 12 seconds. The trains will cross each other in

 A) 14.4 sec B) 15.5 sec C) 18.8 sec D) 20.2 sec

Explanation:

Let the length of the train be L metres

Speed of first train = $\inline&space;\frac{3600}{18}\times&space;L$ m/hour

Speed of secxond train = $\inline&space;\frac{3600}{12}\times&space;L$ m/hour

When running in opposite directions, relative speed = 200 L + 300 L m/hour

Distance to be covered = L + L = 2L metre

Time taken = $\inline&space;\frac{2L}{500L}\times&space;3600$ sec

=14.4 sec

Q:

Two trains started at the same time, one from A to B and the other from B to A . If they arrived at B and A respectively 4 hours and 9 hours after they passed each other the ratio of the speeds of the two trains was

 A) 2:1 B) 3:2 C) 4:3 D) 5:4

Explanation:

Note : If two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take a and b sec in reaching B and A respectively, then: (A's speed) : (B's speed) = (b : a)

Therefore, Ratio of the speeds of two trains = $\inline \sqrt{9}:\sqrt{4}$ = 3 : 2

8 1540
Q:

A man sitting in a train which is traveling at 50 kmph observes that a goods train, traveling in opposite direction, takes 9 seconds to pass him. If the goods train is 280 m long, find its speed.?

 A) 60 B) 62 C) 64 D) 65

Explanation:

Relative speed =$\inline \fn_jvn \frac{280}{9}$  m / sec = $\inline \fn_jvn (\frac{280}{9}\times \frac{18}{5})$ kmph = 112 kmph.

Speed of goods train = (112 - 50) kmph = 62 kmph.

13 3965
Q:

Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction?

 A) 10 B) 25 C) 12 D) 20

Explanation:

Speed of train 1 = $\inline \fn_jvn \left ( \frac{120}{10} \right )m/sec$ = 12 m/sec

Speed of train 2 =$\inline \fn_jvn \left ( \frac{120}{15} \right )m/sec$  = 8 m/sec

if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec

distance covered = 120 + 120 = 240 m

time = distance/speed = 240/20 = 12 sec

6 2225
Q:

A jogger is running at 9 kmph alongside a railway track is 240 meters ahead of the engine of a 120 meters long train running at 45 kmph in the same direction. In how much time will the train  pass the jogger ?

 A) 48 sec B) 36 sec C) 18 sec D) 72 sec

Explanation:

Speed of the train relative to jogger = (45-9) km/hr = 36 km/hr

=$\inline \fn_jvn \left ( 36\times \frac{5}{18} \right )m/sec$ = 10 m/sec

Distance to be covered =(240 + 120)m = 360 m

$\inline \fn_jvn \therefore$ Time taken = $\inline \fn_jvn \left (\frac{360}{10}\right )$ sec = 36 sec

5 1671
Q:

A train is traveling at 48 kmph . It crosses another train having half of its length , traveling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. What is the length of the platform ?

 A) 500 B) 400 C) 360 D) 480

Explanation:

Speed of train 1 = 48 kmph
Let the length of train 1 = 2x meter

Speed of train 2 = 42 kmph
Length of train 2 = x meter (because it is half of train 1's length)

Distance = 2x + x = 3x
Relative speed= 48+42 = 90 kmph = $\inline \fn_jvn 90\times \frac{5}{18}$ m/s = 25 m/s

Time = 12 s

Distance/time = speed

$\inline \fn_jvn \Rightarrow \frac{3x}{12}=25 \Rightarrow x=\frac{25\times 12}{3}$$\inline \fn_jvn \Rightarrow$ $\inline \fn_jvn x=100 m$

Length of the first train = 2x = 200 meter
Time taken to cross the platform= 45 s

Speed of train 1 = 48 kmph = 480/36 = 40/3 m/s

Distance = 200 + y     [where y is the length of the platform]

$\inline \fn_jvn \Rightarrow 200+y = 45\times \frac{40}{3}$
$\inline \fn_jvn \Rightarrow$$\inline \fn_jvn y=400 m$