Problems On Trains All Competence Aptitude

Let's denote the speed of each train as \( v \) m/s.

Given:
- Length of each train = 120 meters
- Time taken for the trains to cross each other = 12 seconds

The distance covered by each train in 12 seconds is the sum of its length and the length of the other train:\[ 120 \, \text{m} + 120 \, \text{m} = 240 \, \text{m} \]

Using the formula for speed:\[ \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{240 \, \text{m}}{12 \, \text{s}} = 20 \, \text{m/s} \]

Since the trains are moving in opposite directions, the relative speed between them is the sum of their individual speeds. Therefore:\[ 2v = 20 \]\[ v = \dfrac{20}{2} = 10 \, \text{m/s} \]

Now, let's convert this speed from meters per second to kilometers per hour:\[ v = 10 \times \dfrac{3600}{1000} = 36 \, \text{km/hr} \]

So, the speed of each train is \( \textbf{36 km/hr} \).

Therefore, the correct answer is \( \textbf{36 km/hr} \).

Let's first find the time it takes for the first train, starting from city A, to meet the second train, starting from city B.

Let \( t \) be the time in hours it takes for the two trains to meet.

Since the first train started at 8 a.m., it has been traveling for \( t \) hours.
Since the second train started at 9 a.m., it has been traveling for \( t - 1 \) hours.

The distance covered by the first train in \( t \) hours is \( 60t \) km.
The distance covered by the second train in \( t - 1 \) hours is \( 75(t - 1) \) km.

According to the problem, the sum of these distances is the total distance between the two cities, which is 330 km.

\[ 60t + 75(t - 1) = 330 \]

Now, let's solve for \( t \):

\[ 60t + 75t - 75 = 330 \]\[ 135t - 75 = 330 \]\[ 135t = 405 \]\[ t = \dfrac{405}{135} = 3 \]

So, it takes 3 hours for the two trains to meet.

Now, let's find out at what time they meet.

Since the first train started at 8 a.m., they meet at \( 8 \text{ a.m.} + 3 \text{ hours} = 11 \text{ a.m.} \).

Therefore, they meet at \(\textbf{11 a.m.}\).

Let's denote the distance between stations A and B as \( D \) kilometers.

Given:
- Speed of the first train = 50 km/hr
- Speed of the second train = 60 km/hr
- At the time of their meeting, the second train has traveled 120 km more than the first.

Let \( t \) be the time it takes for the two trains to meet.

Since they are moving towards each other, their relative speed is the sum of their individual speeds:\[ \text{Relative speed} = 50 \, \text{km/hr} + 60 \, \text{km/hr} = 110 \, \text{km/hr} \]

The distance covered by the first train in \( t \) hours is \( 50t \) km.
The distance covered by the second train in \( t \) hours is \( 60t \) km.

According to the problem, at the time of their meeting, the second train has traveled 120 km more than the first:\[ 60t = 50t + 120 \]\[ 10t = 120 \]\[ t = \dfrac{120}{10} = 12 \, \text{hours} \]

Now, we can find the distance between stations A and B using the time it took for the trains to meet:\[ D = \text{Speed} \times \text{Time} = 110 \times 12 = 1320 \, \text{km} \]

So, the distance between stations A and B is \( \textbf{1320 km} \).

Therefore, the correct answer is \( \textbf{1320 km} \).

Let's break down the problem step by step.

1. When the two trains cross each other:
   - Relative speed = Speed of first train + Speed of second train = (48 + 42) km/hr = 90 km/hr = 90 * (5/18) m/s (converting to m/s)
   - Length of the first train = L
   - Length of the second train = L/2 (as given)
   - Time taken to cross = 12 seconds

Using the formula: Distance = Speed * Time,\[ \text{Distance} = \text{Relative speed} \times \text{Time} \]\[ L + \dfrac{L}{2} = (90 \times \dfrac{5}{18}) \times 12 \]

2. When the train passes a platform:
   - Length of the platform = P
   - Total distance covered = L (length of the train) + P (length of the platform)
   - Speed = 48 km/hr = 48 * (5/18) m/s (converting to m/s)
   - Time taken = 45 seconds

\[ \text{Distance} = \text{Speed} \times \text{Time} \]\[ L + P = (48 \times \dfrac{5}{18}) \times 45 \]

Now, you have two equations:

1. \( L + \dfrac{L}{2} = (90 \times \dfrac{5}{18}) \times 12 \)
2. \( L + P = (48 \times \dfrac{5}{18}) \times 45 \)

Solve these equations to find the values of \(L\) and \(P\).

\[ L + \dfrac{L}{2} = 300 \]\[ \dfrac{3L}{2} = 300 \]\[ L = \dfrac{600}{3} = 200 \text{ meters} \]

Now that you know the length of the train is 200 meters, substitute it into the second equation:

\[ 200 + P = 600 \]\[ P = 600 - 200 = 400 \text{ meters} \]

So, the length of the railway platform is 400 meters. Therefore, the answer is 400 m.

Let's denote the speed of each train as \(v\), and we'll find it using the formula:

\[ \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \]

Given that the length of each train is 120 meters:

1. For the first train, crossing the telegraph post in 10 seconds:\[ v_1 = \dfrac{120 \text{ meters}}{10 \text{ seconds}} = 12 \text{ m/s} \]

2. For the second train, crossing the telegraph post in 15 seconds:\[ v_2 = \dfrac{120 \text{ meters}}{15 \text{ seconds}} = 8 \text{ m/s} \]

When two trains move towards each other, their relative speed is the sum of their individual speeds. Hence, the relative speed of the two trains is \(v_1 + v_2 = 12 \text{ m/s} + 8 \text{ m/s} = 20 \text{ m/s}\).

Now, when two trains cross each other, they effectively cover the sum of their lengths. Thus, the time taken for them to cross each other is:

\[ \text{Time} = \dfrac{\text{Total distance}}{\text{Relative speed}} \]

\[ = \dfrac{2 \times 120 \text{ meters}}{20 \text{ m/s}} = \dfrac{240 \text{ meters}}{20 \text{ m/s}} = 12 \text{ seconds} \]

So, the time taken for the trains to cross each other traveling in opposite directions is 12 seconds. Therefore, the answer is 12.

Let's find out the time when the two trains meet.

Let \( t \) be the time after 8 AM when the trains meet. 

The distance covered by the first train in \( t \) hours is \( 70t \) km. Since it started at 8 AM, it travels for \( t \) hours.

The distance covered by the second train in \( t - 2 \) hours is \( 110(t - 2) \) km. It started at 10 AM, so it has been traveling for \( t - 2 \) hours.

Since they meet when the sum of their distances is equal to the total distance between A and B, we can set up the equation:

\[ 70t + 110(t - 2) = 500 \]

Let's solve for \( t \):

\[ 70t + 110t - 220 = 500 \]\[ 180t - 220 = 500 \]\[ 180t = 720 \]\[ t = \dfrac{720}{180} \]\[ t = 4 \]

So, the two trains meet \( 4 \) hours after the first train starts, which is at \( 8 \text{ AM} + 4 \text{ hours} = 12 \text{ Noon} \).

Therefore, the two trains meet at 12 Noon.

Let's denote the distance between the two places as \( D \) kilometers.

Since P and Q are moving towards each other, the combined speed of P and Q is \( 20 + 30 = 50 \) km/h.

Let \( t \) be the time it takes for them to meet.

Distance covered by P in \( t \) hours = \( 20t \) km
Distance covered by Q in \( t \) hours = \( 30t \) km

Given that by the time they meet, Q has covered 36 km more than P:\[ 30t = 20t + 36 \]\[ 10t = 36 \]\[ t = \dfrac{36}{10} = 3.6 \text{ hours} \]

Now, we can find the distance between the two places using the time \( t \):

\[ \text{Distance} = \text{Speed} \times \text{Time} \]\[ \text{Distance} = 50 \times 3.6 \]\[ \text{Distance} = 180 \text{ km} \]

So, the distance between the two places is 180 km. Therefore, the answer is 180.

To find the time it takes for the two trains to cross each other, we need to calculate the time it takes for them to cover the sum of their lengths while moving towards each other.

Let's denote:
- \( L_1 = 150 \) meters (length of the first train)
- \( L_2 = 180 \) meters (length of the second train)
- \( S_1 = 50 \) km/hr (speed of the first train)
- \( S_2 = 58 \) km/hr (speed of the second train)

When two trains are moving towards each other, their relative speed is the sum of their speeds. Therefore, relative speed \( S_r = S_1 + S_2 \).

Converting the speeds from km/hr to m/s:
- \( S_1 = 50 \times \dfrac{5}{18} \) m/s
- \( S_2 = 58 \times \dfrac{5}{18} \) m/s

So, the relative speed \( S_r = (50 \times \dfrac{5}{18}) + (58 \times \dfrac{5}{18}) \) m/s.

The total distance the two trains need to cover to cross each other is the sum of their lengths, \( L_1 + L_2 \).

The time taken to cover this distance is given by the formula:\[ \text{Time} = \dfrac{\text{Total distance}}{\text{Relative speed}} \]

\[ \text{Time} = \dfrac{L_1 + L_2}{S_r} \]

Substituting the given values:\[ \text{Time} = \dfrac{150 + 180}{(50 \times \dfrac{5}{18}) + (58 \times \dfrac{5}{18})} \]

Now, calculate the time to find the answer. 

\[ \text{Time} = \dfrac{330}{(\dfrac{250}{18}) + (\dfrac{290}{18})} \]

\[ \text{Time} = \dfrac{330}{\dfrac{540}{18}} \]

\[ \text{Time} = \dfrac{330 \times 18}{540} \]

\[ \text{Time} = \dfrac{5940}{540} \]

\[ \text{Time} = 11 \text{ seconds} \]

So, it will take 11 seconds for the two trains to cross each other. Therefore, the answer is 11 seconds.

Let's denote the distance between stations A and B as \( D \) miles.

The combined speed of both trains is the sum of their speeds: \( 16 + 21 = 37 \) miles/hour.

Let \( t \) be the time it takes for the two trains to meet each other.

Using the formula: Distance = Speed × Time, we can write equations for the distances traveled by each train:

1. Distance traveled by the first train \( = 16t \) miles
2. Distance traveled by the second train \( = 21t \) miles

Given that the second train has traveled 60 miles more than the first train when they meet, we can write the equation:

\[ 21t = 16t + 60 \]

Solving for \( t \):

\[ 21t - 16t = 60 \]\[ 5t = 60 \]\[ t = \dfrac{60}{5} \]\[ t = 12 \] hours

Now that we have found \( t \), we can find the distance between stations A and B:

\[ D = \text{Speed} \times \text{Time} \]\[ D = 37 \times 12 \]\[ D = 444 \] miles

So, the distance between stations A and B is 444 miles. Therefore, the answer is 444.

To find the time it takes for the trains to pass each other, we need to calculate the time it takes for them to cover the sum of their lengths while moving towards each other.

Let's denote:
- \( L_1 = 70 \) meters (length of the first train)
- \( L_2 = 80 \) meters (length of the second train)
- \( S_1 = 68 \) km/hr (speed of the first train)
- \( S_2 = 40 \) km/hr (speed of the second train)

When two trains are moving towards each other, their relative speed is the sum of their speeds. Therefore, relative speed \( S_r = S_1 + S_2 \).

Converting the speeds from km/hr to m/s:
- \( S_1 = 68 \times \dfrac{5}{18} \) m/s
- \( S_2 = 40 \times \dfrac{5}{18} \) m/s

So, the relative speed \( S_r = (68 \times \dfrac{5}{18}) + (40 \times \dfrac{5}{18}) \) m/s.

The total distance the two trains need to cover to pass each other is the sum of their lengths, \( L_1 + L_2 \).

The time taken to cover this distance is given by the formula:\[ \text{Time} = \dfrac{\text{Total distance}}{\text{Relative speed}} \]

\[ \text{Time} = \dfrac{L_1 + L_2}{S_r} \]

Substituting the given values:\[ \text{Time} = \dfrac{70 + 80}{(68 \times \dfrac{5}{18}) + (40 \times \dfrac{5}{18})} \]

Now, calculate the time to find the answer. 

\[ \text{Time} = \dfrac{150}{(\dfrac{340}{18}) + (\dfrac{200}{18})} \]

\[ \text{Time} = \dfrac{150}{(\dfrac{540}{18})} \]

\[ \text{Time} = \dfrac{150 \times 18}{540} \]

\[ \text{Time} = \dfrac{2700}{540} \]

\[ \text{Time} = 5 \text{ seconds} \]

So, it will take 5 seconds for the two trains to pass each other. Therefore, the answer is 5 seconds.