Problems on Trains Questions and Answers

Answer: D

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 90 \text{ km/hr} = 25 \text{ m/s} \]

In one minute the train covers

\[ 25 \times 60 = 1500 \text{ m} \]

That is the train plus the platform, and the two are equal:

\[ 2L = 1500 \quad\Longrightarrow\quad L = 750 \text{ metres} \]

Answer: B

The man runs the same way, so the train closes on him at the difference of the speeds and covers only its own length.

\[ \text{relative speed} = \frac{125}{10} = 12.5 \text{ m/s} = 12.5 \times \frac{18}{5} = 45 \text{ km/hr} \]

\[ v_{\text{train}} = 45 + 5 = 50 \text{ km/hr} \]

The man's speed is added back because it had been subtracted to give the relative speed.

Answer: B

Using the formula, speed=distance/time

Speed of truck = 550/60 m/s

Speed of train= 33000/(45*60) m/s

Ratio of speed of truck to train= (550/60 ) / [33000/(45*60)]

=3:4

Answer: B

Let the speeds be \( a \) and \( b \) m/s. Crossing a man gives each length:

\[ L_{1} = 27a, \qquad L_{2} = 17b \]

Running opposite ways they cross in 23 seconds, covering both lengths at \( a + b \):

\[ \frac{27a + 17b}{a + b} = 23 \]

\[ 27a + 17b = 23a + 23b \quad\Longrightarrow\quad 4a = 6b \]

\[ \frac{a}{b} = \frac{3}{2} \quad\Longrightarrow\quad 3 : 2 \]

Neither actual speed is needed - the lengths cancel out into the ratio.

Answer: D

Passing a man sitting in the slower train means covering only the fast train's own length, at the difference of the speeds.

\[ \text{relative speed} = 40 - 20 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]

\[ L = \frac{50}{9} \times 5 = \frac{250}{9} = 27\tfrac{7}{9} \text{ m} \]

The slower train's length never enters, because the reference point is one seated person.

Answer: D

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 72 \text{ kmph} = 20 \text{ m/s} \]

\[ \text{total distance} = 20 \times 26 = 520 \text{ m} \]

\[ \text{train} = 520 - 250 = 270 \text{ m} \]

Answer: B

A signal pole is a point, so that crossing gives the speed.

\[ v = \frac{300}{18} = \frac{50}{3} \text{ m/s} \]

\[ \text{total distance over the platform} = \frac{50}{3} \times 39 = 650 \text{ m} \]

\[ \text{platform} = 650 - 300 = 350 \text{ m} \]

The data are complete, so "data inadequate" is a trap; 650 m is the combined distance, not the platform.

Answer: A

Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.

Let the first train be \( L \) long; the second is \( \tfrac{L}{2} \).

\[ \text{relative speed} = 48 + 42 = 90 \text{ kmph} = 25 \text{ m/s} \]

\[ L + \frac{L}{2} = 25 \times 12 = 300 \quad\Longrightarrow\quad 1.5L = 300 \]

\[ L = 200 \text{ m} \]

Crossing the platform, the train travels alone at 48 kmph \( = \tfrac{40}{3} \) m/s:

\[ 200 + P = \frac{40}{3} \times 45 = 600 \quad\Longrightarrow\quad P = 400 \text{ m} \]

Answer: C

Let the train's speed be \( v \) km/hr. Passing each boy covers the train's own length at the relative speed.

\[ L = (v - 8) \times \frac{5}{18} \times 36 = (v - 16) \times \frac{5}{18} \times 40 \]

The factor \( \tfrac{5}{18} \) is common:

\[ 36(v - 8) = 40(v - 16) \quad\Longrightarrow\quad 36v - 288 = 40v - 640 \]

\[ 4v = 352 \quad\Longrightarrow\quad v = 88 \text{ km/hr} \]

\[ L = (88 - 8) \times \frac{5}{18} \times 36 = 80 \times 10 = 800 \text{ m} \]

Answer: D

The pole crossing covers the train alone, the platform crossing the train plus 100 m.

\[ L = 15v, \qquad L + 100 = 25v \]

Subtracting:

\[ 100 = 10v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 15 \times 10 = 150 \text{ m} \]