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Question

A train moving at speed of 36 km/hr crosses a platform in 30 secs. If the length of platform is equal to 20% of length of train, then find the length of platform.

This question was previously asked in
ESIC UDC Mains MBT (30 Apr 2022)
The correct answer is

50 m

Understanding the Train Crossing Problem

This problem involves calculating the length of a platform based on a train's speed, the time it takes to cross the platform, and a relationship between the train's length and the platform's length.

Step 1: Convert Speed to Consistent Units

The train's speed is given as 36 km/hr. Since the time is in seconds, we need to convert the speed to meters per second (m/sec) for consistency.

The conversion factor from km/hr to m/sec is $\frac{5}{18}$.

Speed = $36 \, \text{km/hr} \times \frac{5}{18} \, \frac{\text{m/sec}}{\text{km/hr}}$

Speed = $2 \times 5 \, \text{m/sec}$

Speed = 10 m/sec

Step 2: Define Variables and Formulate Equations

Let the length of the train be denoted by $L_T$ (in meters).

Let the length of the platform be denoted by $L_P$ (in meters).

When a train crosses a platform, the total distance the train covers is the sum of its own length and the length of the platform. The formula relating distance, speed, and time is: Distance = Speed $\times$ Time.

In this case:

$L_T + L_P = \text{Speed} \times \text{Time}$

Substituting the known values:

$L_T + L_P = 10 \, \text{m/sec} \times 30 \, \text{secs}$

$L_T + L_P = 300 \, \text{meters}$ (Equation 1)

We are also given that the length of the platform is 20% of the length of the train.

$L_P = 20\% \, \text{of} \, L_T$

$L_P = \frac{20}{100} \times L_T$

$L_P = \frac{1}{5} L_T$ (Equation 2)

Step 3: Solve for the Length of the Platform

Now we can substitute Equation 2 into Equation 1 to solve for the lengths.

Substitute $L_P = \frac{1}{5} L_T$ into $L_T + L_P = 300$:

$L_T + \frac{1}{5} L_T = 300$

To combine the terms, find a common denominator:

$\frac{5 L_T}{5} + \frac{1 L_T}{5} = 300$

$\frac{6 L_T}{5} = 300$

Now, solve for $L_T$:

$6 L_T = 300 \times 5$

$6 L_T = 1500$

$L_T = \frac{1500}{6}$

$L_T = 250 \, \text{meters}$

Now that we have the length of the train, we can find the length of the platform using Equation 2:

$L_P = \frac{1}{5} L_T$

$L_P = \frac{1}{5} \times 250 \, \text{meters}$

$L_P = 50 \, \text{meters}$

Conclusion

The length of the platform is 50 meters. This matches option 3.

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Similar Questions

  1. The speed of boat in still water is 6 km/hr and speed of stream is 3 km/hr. Calculate the time taken to cover 27 km distance downstream and returning on it upstream.

  2. A train moving with the speed of 25 m/s crosses a platform which is one third of its length in 20 seconds. What would be the length of the train?

  3. Sohan can reach destination in 15 hours. If he reduces his speed of walking by 1/4th then he travels 5 km less then what should be actually covered. Find total distance covered by sohan

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  5. A can rows a certain distance in downstream in 10 hours and returns from the place in 20 hours. If the speed of current is 5 km/h then find the speed of A in still water.


Important Questions from Speed Time and Distance

  1. A journey of 900 km is completed in 11 h. If two-fifth of the journey is completed at the speed of 60 km/h, at what speed (in km/h) is the remaining journey completed?

  2. A car starts from point A towards point B, travelling at the speed of 20 km/h. 1 \(\frac{1}{2}\) hours later, another car starts from point A and travelling at the speed of 30 km/h and reaches 2 \(\frac{1}{2}\) hours before the first car. Find the distance between A and B.

  3. A bus covered a distance of 162 km. If speed of this bus is 15 m/s, then what will be the time taken ?

  4. An athlete runs an 800 m race in 96 seconds. His speed (in km / h) is:

  5. A person has to cover a distance of 150 km in 15 hours. If he traveled with the speed of 11.8 km/hr for 10 hours. At what speed he has to travel to cover the remaining distance in the remaining time?

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