Quantitative Aptitude - SPLessons

Train Problems

Chapter 19

SPLessons 5 Steps, 3 Clicks
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Train Problems

shape Introduction

Train problems are totally based on four topics including conversion, distance formula, relativity, and train theory.


shape Methods

Conversion: It includes conversion of kilometer per hour (kmph) into meter per second (mps) or vice -versa.


Distance formula: D = S x T
where,

    D ⇒ Distance
    S ⇒ Speed
    T ⇒ Time


Relativity: It is a broad term. It describes about objects moving in either direction or in the same direction, speed of the objects, and time taken by the objects, etc.


Example 1:
A train 100 m long is running at the speed of 30 km/hr. find the time taken by in to pass a man standing near the railway line.


Solution:

    Speed of the train = (30 x \(\frac{5}{8}\)) m/sec = (\(\frac{25}{3}\)) m/sec

    Distance moved in passing the standing man = 100 m.

    Required time taken = (\(\frac{100}{\frac{25}{3}}\)) = (100 x \(\frac{3}{25}\)) sec = 12 sec.

Example 2:
A train is moving at a speed of 132 km/hr. If the length of the train is 110 metres, how long will it take to cross a railway platform 165 metre long?


Solution:

    Speed of the train = (132 x \(\frac{5}{18}\)) m/sec = (\(\frac{110}{3}\)) m/sec

    Distance covered in passing the platform = (110+165) m = 275 m.

    Time taken = (275 x \(\frac{3}{110}\)) sec = (\(\frac{15}{2}\)) sec = 7\(\frac{1}{2}\) sec


Example 1:
A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed.


Solution:

    Let the length of the train be x metres.

    Then, the train covers x metres in 8 seconds and (\(x\) + 180) metres in 20 seconds.

    ∴ \(\frac{x}{8}\) = \(\frac{(x + 180)}{20}\) ⇔ 20\(x\) = 8(\(x\) + 180) ⇔ \(x\) = 120.

    ∴ Length of the train = 120 m.

    Speed of the train = (\(\frac{120}{8}\)) m/sec = m/sec = (15 x \(\frac{18}{5}\)) kmph = 54 kmph.


Example 2:
A man sitting in a train which is travelling at 50 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. if the goods train is 280 m long, find its speed.


Solution:

    Relative speed = (\(\frac{280}{9}\)) m/sec = (\(\frac{280}{9}\) x \(\frac{18}{5}\)) kmph = 112 kmph.

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

shape Formulae

1. \(x\) km/hr = \(x\) x \(\frac{5}{18}\) m/s

2. \(x\) m/s = \(x\) x \(\frac{18}{5}\) km/hr

3. Time taken by a train of length \(y\) meters to pass a pole or a standing man or a signal post or an object of negligible width would be equal to the time taken by the train to cover \(y\) meters which is primarily the length of the train.

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

5. Suppose two trains or two bodies are moving in the same direction at \(x\) m/s and \(y\) m/s, where \(x\) > \(y\), then their
Relative speed = (\(x\) – \(y\)) m/s

6. Suppose two trains or two bodies are moving in the opposite direction at \(x\) m/s and \(y\) m/s, then their
Relative speed = (\(x\) + \(y\)) m/s

7. If two trains of length \(a\) meters and \(b\) meters are moving in the opposite direction at \(x\) m/s \(y\) m/s, then the time taken by the faster train to cross the slower train = \(\frac{(a + b)}{(x + y)}\) sec.

8. If two trains of length \(a\) meters and \(b\) meters are moving in the same direction at \(x\) m/s \(y\) m/s, then the time taken by the faster train to cross the slower train = \(\frac{(a + b)}{(x – y)}\) sec.

9. If the two trains (or bodies) start at the same time from points A and B towards each other and after crossing they take \(x\) and \(y\) sec in reaching B and A respectively, then
(A’s speed) : (B’s speed) = \(\sqrt{y}\) : \(\sqrt{x}\)

shape Samples

1. A train of length 200 m is running at the speed of 20 km/hr. Find the time taken by the train to pass a man standing near the railway line?

Solution:

    Given,

    Distance moved in passing the standing man = 200 m

    Speed = 20 kmph

    Convert kmph to mps

    Speed = 20 x \(\frac{5}{18}\) = \(\frac{50}{9}\)

    So, Required time = \(\frac{distance}{speed}\)

    ⇒ Time = \(\frac{200}{\frac{50}{9}}\)

    ⇒ Time = \(\frac{200 * 9}{50}\)

    ⇒ Time = 36 sec

    Therefore, the time taken by it to pass a man standing near the railway line = 36 sec


2. A man is standing on a railway bridge which is 200m long. He finds that a train crosses the bridge in 10 seconds but himself in 6 seconds. Find the length of the train and its speed?

Solution:

    Let,

    The length of the train = \(x\) m

    Then, the train covers \(x\) m in 6 seconds and \(x + 200\) m in 10 seconds.

    So, Length of the train is

    \(\frac{x}{6}\) = \(\frac{x + 200}{10}\)

    ⇒ 10 x \(x\) = 6(\(x\) + 200)

    ⇒ 10 x \(x\) – 6 x \(x\) = 1200

    ⇒ 4 x \(x\) = 1200

    ⇒ \(x\) = 300

    ⇒ Length of train = 300 m

    Then, Speed of the train = \(\frac{300}{6}\) m/s = 50 m/s

    ⇒ 50 x \(\frac{18}{5}\) km/hr

    ⇒ 180 km/hr

    Therefore, length of train = 300m and

    Speed of train = 180 km/hr.


3. A train 320 m long is running with speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the opposite direction in which the train is going?

Solution:

    Given,

    Relative Speed of the train and man = (60 + 6) = 66 kmph

    Converting it into m/s, i.e.

    ⇒ 66 x \(\frac{5}{18}\)

    ⇒ \(\frac{55}{3}\) m/s

    Time taken by the train to cross the man is

    = time taken by it to cover 320 m at \(\frac{55}{3}\) m/s

    = 320 x \(\frac{3}{55}\)

    = 17.45 sec

    Therefore, time taken by the train to cross the man is 17.45 sec


4. Two trains 200 m and 320 m long are running in the same direction with speeds of 72 km/hr and 54 km/hr. In how much time will the first train cross the second?

Solution:

    Relative Speed of the trains = (72 – 54) km/hr = 18 km/hr

    Convert it into m/sec, i.e.

    ⇒ 18 x \(\frac{5}{18}\)

    ⇒ 5 m/s

    Time taken by the trains to cross each other is

    = time taken to cover (200 + 320)m at 5 m/s

    = \(\frac{520}{5}\)

    = 104

    Therefore, time taken by the trains to cross each other is 104 sec.


5. A man sitting in a train which is travelling at 60 kmph observes that a goods train, travelling in opposite direction, takes 9 seconds to pass him. If the goods train is 300 m long, find its speed?

Solution:

    Given,

    The goods train travels 60 kmph

    Relative speed = \(\frac{300}{9}\) m/s

    Convert it into kmph, i.e.

    \(\frac{300}{9}\) x \(\frac{18}{5}\) = 120kmph

    Now, Speed of goods train is

    = (120 – 60) kmph

    = 60 kmph

    Therefore, Speed of goods train = 60 kmph