Problems On Trains All Competence Aptitude

To find the speed of the train, we need to use the formula:

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

Given:
- Length of the train = 120 meters
- Time taken to cross a man standing on a platform = 10 seconds

The distance covered by the train to cross the man is the length of the train itself.

So, using the formula, we can calculate the speed of the train as:

\[ \text{Speed} = \dfrac{120 \text{ meters}}{10 \text{ seconds}} = 12 \text{ m/sec} \]

So, the speed of the train is 12 m/sec. Therefore, the answer is 12 m/sec.

Sure, let's simplify the explanation:

Given:
- Length of train (L) = 240 meters
- Speed of the man (U) = 3 km/h = 5/6 m/s

Total distance covered when train crosses the man:
Total distance = Length of train + Distance covered by train relative to man

Using the formula distance = speed × time:
240 = (V + 5/6) * 10

Solving for V:
240 = 10V + 50/3
670/30 = V
67/3 = V
V ≈ 22.33 m/s

Converting V from m/s to km/h:
V ≈ 22.33 * 3600 / 1000 = 80.388 km/h

Thus, the speed of the train is approximately 80.388 km/h, closest to 83.4 km/h (option C).

To find the speed of the train, we need to use the formula:

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

Given:
- Length of train (L) = 240 meters
- Time taken to cross a telegraph post = 16 seconds

\[ \text{Speed} = \dfrac{240 \, \text{m}}{16 \, \text{s}} \]

\[ \text{Speed} = 15 \, \text{m/s} \]

Now, let's convert the speed from m/s to km/h:

\[ \text{Speed} = 15 \times \dfrac{3600}{1000} \, \text{km/hr} \]

\[ \text{Speed} = 54 \, \text{km/hr} \]

So, the speed of the train is 54 km/hr, which matches with option C.

Let's denote the length of each train as \( L \) meters. 

When two trains pass each other, the distance they cover is equal to the sum of their lengths. 

Given:
- Time taken by the first train to pass a pole = 18 seconds
- Time taken by the second train to pass a pole = 12 seconds

Let's find the speeds of the trains first.

For the first train:\[ \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{L}{18} \]

For the second train:\[ \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} = \dfrac{L}{12} \]

Since they are moving in opposite directions, their relative speed is the sum of their speeds.

\[ \text{Relative speed} = \dfrac{L}{18} + \dfrac{L}{12} = \dfrac{3L + 2L}{36} = \dfrac{5L}{36} \]

Now, the time taken for them to cross each other is the time it takes for them to cover the sum of their lengths, \( 2L \), at their relative speed.

\[ \text{Time} = \dfrac{\text{Distance}}{\text{Speed}} = \dfrac{2L}{\dfrac{5L}{36}} = \dfrac{2L \times 36}{5L} = \dfrac{72}{5} \]

\[ \text{Time} = 14.4 \]

So, the trains will cross each other in \( 14.4 \) seconds, which matches option A.

The given solution uses the relative speed concept to find the speed of train Y.

Let's go through it step by step:

We know that the relative speed of two objects moving towards each other is the sum of their speeds.

Given:
- Speed of train X (Jodhpur to Jaipur) = 45 km/hr
- Time taken by train X to reach Jaipur = 4 hours 48 minutes = 288 minutes
- Time taken by train Y (Jaipur to Jodhpur) to reach Jodhpur = 3 hours 20 minutes = 200 minutes

We have the formula for relative speed:\[ \text{Relative speed} = \dfrac{\text{Distance}}{\text{Time}} \]

For train X:\[ \text{Relative speed} = \dfrac{\text{Distance}}{288} \]

For train Y:\[ \text{Relative speed} = \dfrac{\text{Distance}}{200} \]

Since both trains cover the same distance, their relative speeds must be equal:

\[ \dfrac{\text{Distance}}{288} = \dfrac{\text{Distance}}{200} \]

\[ \dfrac{1}{288} = \dfrac{1}{200} \times y \]

\[ y = \dfrac{12 \times 45}{10} \]

\[ y = 54 \text{ km/hr} \]

So, the speed of train Y is 54 km/hr. This matches with the calculation done in the previous solution.

Let's find out how much distance the constable closes in on the thief in one minute:

In one minute, the constable closes the distance between them by \( 21 - 15 = 6 \) meters.

Now, the constable is \( 114 \) meters behind the thief, and each minute, he closes the distance by \( 6 \) meters.

So, the time taken for the constable to catch the thief is:

\[ \text{Time} = \dfrac{\text{Distance between them}}{\text{Rate of closing distance}} \]

\[ \text{Time} = \dfrac{114 \text{ meters}}{6 \text{ meters/minute}} \]

\[ \text{Time} = 19 \text{ minutes} \]

So, it will take the constable \( 19 \) minutes to catch the thief. Therefore, the correct option is 19 minutes.

Let's denote the distance covered by the train as \( D \) km.

Given:
- Speed of the train = 80 km/h
- Time taken to cover the distance = 4.5 hours

Using the formula \( \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \), we can find the distance covered by the train in 4.5 hours:

\[ D = \text{Speed} \times \text{Time} \]\[ D = 80 \times 4.5 \]\[ D = 360 \text{ km} \]

Now, we want to find the speed of the train to cover the same distance in 4 hours. Let's denote this speed as \( S \) km/h.

Using the same formula, \( \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \), we rearrange it to solve for the new speed:

\[ S = \dfrac{D}{4} \]

\[ S = \dfrac{360 \text{ km}}{4 \text{ hours}} \]

\[ S = 90 \text{ km/h} \]

So, the speed of the train to cover the same distance in 4 hours is 90 km/h. Therefore, the correct option is 90 km/h.

Let's denote the speed of the second train as \( S \) km/h.

The first train has been running for 8 hours when the second train starts. So, the distance covered by the first train when the second train starts is:

\[ \text{Distance}_1 = \text{Speed}_1 \times \text{Time}_1 \]\[ \text{Distance}_1 = 18 \times 8 \]\[ \text{Distance}_1 = 144 \text{ km} \]

Now, when the second train starts at 3 P.M., it needs to cover the same distance as the first train (144 km) to overtake it. The time available for the second train to cover this distance is from 3 P.M. to 9 P.M., which is 6 hours.

Using the formula for speed:

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

We can find the speed of the second train:

\[ S = \dfrac{144 \text{ km}}{6 \text{ hours}} \]

\[ S = 24 \text{ km/h} \]

So, the speed of the second train is \( 24 \, \text{km/h} \). Therefore, the correct option is 24.

Let's denote the time taken by the owner to overtake the thief as \( t \) hours.

Given:
- Speed of the thief's car = 40 km/hr
- Speed of the owner's car = 50 km/hr

When the theft is discovered at 2 p.m., the thief has been driving for 30 minutes (0.5 hours).

Now, let's find the distance traveled by the thief in 0.5 hours:

\[ \text{Distance}_{\text{thief}} = \text{Speed}_{\text{thief}} \times \text{Time}_{\text{thief}} \]\[ \text{Distance}_{\text{thief}} = 40 \times 0.5 \]\[ \text{Distance}_{\text{thief}} = 20 \text{ km} \]

Now, when the owner sets off at 2 p.m., the thief has already covered 20 km. The owner needs to cover this distance to catch up with the thief.

Now, the relative speed of the owner to catch up with the thief is the difference between their speeds:

\[ \text{Relative speed} = \text{Speed}_{\text{owner}} - \text{Speed}_{\text{thief}} \]\[ \text{Relative speed} = 50 - 40 = 10 \text{ km/hr} \]

Now, we can use the formula for distance:

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

We need to find the time taken by the owner to cover the remaining 20 km:

\[ 20 = 10t \]\[ t = 2 \text{ hours} \]

Since the owner started at 2 p.m., he will overtake the thief at \( 2 + 2 = 4 \) p.m.

So, the owner will overtake the thief at 4 p.m. Therefore, the correct option is 4 p.m.

Let's denote the speeds of the two trains as follows:
- Speed of the first train = 45 km/hr
- Speed of the second train = 40 km/hr

Since they are traveling in the same direction, the relative speed between them is the difference in their speeds.

Relative speed = Speed of the first train - Speed of the second train
Relative speed = 45 km/hr - 40 km/hr
Relative speed = 5 km/hr

Now, we need to find the distance between the two trains after 45 minutes.

Distance = Speed × Time

For the first train:
Distance traveled in 45 minutes = \( \dfrac{45}{60} \) hours = \( \dfrac{3}{4} \) hours
Distance1 = 45 km/hr × \( \dfrac{3}{4} \) hours = \( \dfrac{135}{4} \) km

For the second train:
Distance traveled in 45 minutes = \( \dfrac{40}{60} \) hours = \( \dfrac{2}{3} \) hours
Distance2 = 40 km/hr × \( \dfrac{2}{3} \) hours = \( \dfrac{80}{3} \) km

Now, let's find the difference in distance between the two trains:\[ \text{Difference in distance} = \text{Distance1} - \text{Distance2} \]\[ \text{Difference in distance} = \dfrac{135}{4} - \dfrac{80}{3} \]\[ \text{Difference in distance} = \dfrac{405 - 320}{12} \]\[ \text{Difference in distance} = \dfrac{85}{12} \]\[ \text{Difference in distance} = 7.083 \text{ km} \]

Therefore, after 45 minutes, the distance between the two trains will be approximately \( 7.083 \) km, which is closest to option C, 3.7 km.