# Problems on Trains Questions

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.

15 4293
Q:

Two cogged wheels of which one has 32 cogs and other 54 cogs, work into each other. If the latter turns 80 times in three quarters of a minute, how often does the other turn in 8 seconds?

 A) 48 B) 24 C) 38 D) 36

Explanation:

Less Cogs $\inline \fn_jvn \Rightarrow$ more turns and less time $\inline \fn_jvn \Rightarrow$ less turns

$\inline \fn_jvn \begin{matrix} & Cogs& Time&Turns \\ A& 54& 45 &80 \\ B& 32& 8& ? \end{matrix}$

Number of turns required=80 ×$\inline \fn_jvn \frac{54}{32}$ ×$\inline \fn_jvn \frac{8}{45}$ = 24 times

12 3622
Q:

A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?

 A) 69.5 km/hr B) 70 km/hr C) 79 km/hr D) 79.2 km/hr

Explanation:

Let the length of the train be x metres and its speed by y m/sec.
Then,$\inline \fn_jvn \frac{x}{y}=8\Rightarrow x=8y$
Now, $\inline \fn_jvn \frac{x+264}{20}=y$
$\inline \fn_jvn \Rightarrow$8y + 264 = 20y
$\inline \fn_jvn \Rightarrow$y = 22.
$\inline \fn_jvn \therefore$Speed = 22 m/sec =$\inline \fn_jvn \left ( 22\times \frac{18}{5} \right )$km/hr = 79.2 km/hr.

24 3392
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 2490
Q:

A train of length 110 meter is running at a speed of 60 kmph. In what time, it will pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

 A) 10 B) 8 C) 6 D) 4

Explanation:

Distance = 110 m

Relative speed = 60 + 6 = 66 kmph (Since both the train and the man are in moving in opposite direction)

$\inline \fn_jvn 66\times \frac{5}{18}$ m/sec = $\inline \fn_jvn \frac{55}{3}$ m/sec

$\inline \fn_jvn \therefore$Time taken to pass the man = $\inline \fn_jvn 100\times \frac{3}{55}$ = 6 s