Problems on Trains Questions and Answers

Answer: C

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

\[ \text{relative speed} = 30 + 36 = 66 \text{ km/hr} = 66 \times \frac{5}{18} = 18.33 \text{ m/s} \]

\[ t = \frac{156.62 + 100}{18.33} = \frac{256.62}{18.33} \approx 14 \text{ seconds} \]

Both lengths are covered, since the trains must clear each other completely.

Answer: D

Passing the driver - a single point - means covering only the faster train's own length.

\[ \text{relative speed} = 50 - 20 = 30 \text{ kmph} = 30 \times \frac{5}{18} = \frac{25}{3} \text{ m/s} \]

\[ L = \frac{25}{3} \times 20 = \frac{500}{3} = 166.67 \approx 166 \text{ m} \]

Had it passed the whole slower train, both lengths would have counted.

Answer: C

Speed of train = 48 km/h = (48 * 1000)/60 m/minute
= 800 m/minute
Thus we have to find the number of poles situated along a distance of 800 with each pole at a distance of 40 m from each other.

|--------40 m-------|--------40 m ---------|
Consider a distance of 80 m. There are 3 poles situated along this distance as shown.
Thus number of poles = 80/40 + 1 = 2 + 1 = 3
Similarly along a distance of 900 m, number of poles situated
= 800/40 + 1 = 21

Answer: B

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

\[ 71 \text{ km/hr} = 71 \times \frac{5}{18} = 19.72 \text{ m/s} \]

\[ t = \frac{90 + 132}{19.72} = \frac{222}{19.72} = 11.26 \text{ seconds} \]

Crossing a bridge covers the train plus the bridge; the train's own 90 m must not be dropped.

Answer: B

Three stops of 20 minutes cost exactly one hour, so the journey with stops takes an hour longer.

\[ \frac{D}{40} - \frac{D}{60} = 1 \]

\[ D\left(\frac{3 - 2}{120}\right) = 1 \quad\Longrightarrow\quad D = 120 \text{ km} \]

Checking: 3 hours with stops against 2 hours of pure running.

Answer: D

Each crossing covers the train's length plus the station's.

\[ L + 160 = 20v, \qquad L + 110 = 15v \]

Subtracting:

\[ 50 = 5v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 20 \times 10 - 160 = 40 \text{ m} \]

Subtracting the two equations eliminates the unknown length in one step.

Answer: C

Total distance = addition of length of the two trains = 140 + 120 = 260 metres
As the two trains are travelling in the same direction, their relative speed is:
v = | v1 - v2 | = | 40 - 60 | = 20 km/hr = 20*1000/60 = 1000/3 metres/min
t = 260/ 1000*3
t = 0.78 minutes

Answer: C

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

\[ \text{relative speed} = \frac{100 + 100}{8} = 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/hr} \]

With speeds \( v \) and \( 2v \):

\[ 3v = 90 \quad\Longrightarrow\quad v = 30 \text{ km/hr} \]

\[ \text{faster train} = 60 \text{ km/hr} \]

Answer: D

Let us assume for this question that the length of train is x metres and it is assumed to be running at the speed of y m/sec.

A pole is assumed as a point object.

=>Time taken by the train to pass the pole= x/y

=>15=x/y

=>y=x/15

Now, the train passes the platform which is 100 m in length in 25 seconds.

=>x+100/y=25

=>x+100/25=y

Equating speed generated from both cases:

x+100/25=x/15

Therefore x = 150 metres.

Answer: B

The first train has a half-hour head start.

\[ \text{lead at 9:00} = 60 \times 0.5 = 30 \text{ km} \]

The second closes at \( 75 - 60 = 15 \) km/hr:

\[ t = \frac{30}{15} = 2 \text{ h} \]

\[ \text{distance from Jaipur} = 75 \times 2 = 150 \text{ km} \]

Checking with the first train: \( 60 \times 2.5 = 150 \) km.