# Problems on Trains Questions

FACTS  AND  FORMULAE  FOR  PROBLEMS  ON  TRAINS

1. a km/hr = [a x (5/18)] m/s.

2. a m/s = [a x (18/5)] km/hr.

3. Time taken by a train of length l metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover l metres.

4. Time taken by a train of length 1 metres to pass a stationary object of length b metres is the time taken by the train to cover (1 + b) metres.

5. Suppose two trains or two bodies are moving in the same direction at u m/s and v m/s, where u > v, then their relatives speed = (u - v) m/s.

6. Suppose two trains or two bodies are moving in opposite directions at u m/s and v m/s, then their relative speed = (u + v) m/s.

7. If two trains of length a metres and b metres are moving in opposite directions at u m/s and v m/s, then time taken by the trains to cross each other = $\inline \frac{(a+b)}{(u+v)}$ sec.

8. If two trains of length a metres and b metres are moving in the same direction at u m/s and v m/s, then the time taken by the faster train to cross the slower train = $\inline \frac{(a+b)}{(u-v)}$sec.

9. 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) = $\inline \left ( \sqrt{b}:\sqrt{a} \right )$

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

Answer & Explanation Answer: B) 62

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.

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19 6385
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

Answer & Explanation Answer: B) 24

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

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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

Answer & Explanation Answer: C) 6

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

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7 4853
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

Answer & Explanation Answer: 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.

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29 4778
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

Answer & Explanation Answer: C) 12

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

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