With an average speed of 45 km/h, a train reaches its destination on time. If it goes with an average speed of 30 km/h, it is late by 15 minutes. The total journey is:
22.5 km
Understanding the Train Speed Problem
This problem involves a train traveling a certain distance. We are given two scenarios with different average speeds and the corresponding time taken, specifically the difference in time. The goal is to find the total distance of the journey.
The key relationship we will use is the formula connecting distance, speed, and time:
- Distance = Speed × Time
- Time = Distance / Speed
Let's define the unknowns:
- Let the total distance of the journey be \( d \) km.
- Let the scheduled time to reach the destination on time be \( t \) hours.
Setting up Equations from the Given Information
We have two different situations described:
- Scenario 1: Average speed is 45 km/h. The train reaches its destination on time.
- Scenario 2: Average speed is 30 km/h. The train is late by 15 minutes.
Let's convert the time in minutes to hours. 15 minutes is equal to \( \frac{15}{60} \) hours, which simplifies to \( \frac{1}{4} \) hours.
Now we can write equations based on the formula Time = Distance / Speed for each scenario:
- For Scenario 1: The speed is 45 km/h, and the time taken is the scheduled time \( t \). \[ t = \frac{d}{45} \quad \text{(Equation 1)} \]
- For Scenario 2: The speed is 30 km/h, and the time taken is 15 minutes (or \( \frac{1}{4} \) hour) more than the scheduled time \( t \). So, the time taken is \( t + \frac{1}{4} \). \[ t + \frac{1}{4} = \frac{d}{30} \quad \text{(Equation 2)} \]
Solving the Equations to Find the Distance
We now have a system of two equations with two variables (\( d \) and \( t \)). We can solve this system to find the value of \( d \).
Substitute the expression for \( t \) from Equation 1 into Equation 2:
\[ \frac{d}{45} + \frac{1}{4} = \frac{d}{30} \]To solve for \( d \), we need to eliminate the denominators. The least common multiple (LCM) of 45, 4, and 30 is 180. Multiply every term in the equation by 180:
\[ 180 \times \frac{d}{45} + 180 \times \frac{1}{4} = 180 \times \frac{d}{30} \]Simplify the terms:
\[ \frac{180d}{45} + \frac{180}{4} = \frac{180d}{30} \] \[ 4d + 45 = 6d \]Now, rearrange the equation to isolate the terms with \( d \):
Subtract \( 4d \) from both sides:
\[ 45 = 6d - 4d \] \[ 45 = 2d \]Finally, divide by 2 to find \( d \):
\[ d = \frac{45}{2} \] \[ d = 22.5 \]The total distance of the journey is 22.5 km.
Verification (Optional)
We can check if this distance works with the original conditions.
- If distance \( d = 22.5 \) km and speed is 45 km/h, time \( t = \frac{22.5}{45} = 0.5 \) hours (30 minutes).
- If distance \( d = 22.5 \) km and speed is 30 km/h, time taken is \( \frac{22.5}{30} = 0.75 \) hours (45 minutes).
The difference in time is 45 minutes - 30 minutes = 15 minutes, which matches the problem statement. So, the distance 22.5 km is correct.
Summary of the Calculation
| Scenario | Speed (km/h) | Time (hours) | Equation (Time = Distance / Speed) |
|---|---|---|---|
| On Time | 45 | \( t \) | \( t = \frac{d}{45} \) |
| 15 min Late | 30 | \( t + \frac{1}{4} \) | \( t + \frac{1}{4} = \frac{d}{30} \) |
Substitute \( t \) from the first equation into the second:
\[ \frac{d}{45} + \frac{1}{4} = \frac{d}{30} \]Multiply by LCM (180):
\[ 4d + 45 = 6d \]Solve for \( d \):
\[ 2d = 45 \] \[ d = 22.5 \]The total journey distance is 22.5 km.
Revision Table: Train Speed and Distance Concepts
| Concept | Description | Formula |
|---|---|---|
| Distance | The total length of the path traveled. | \( D = S \times T \) |
| Speed (Average) | The rate at which distance is covered over time. | \( S = \frac{D}{T} \) |
| Time | The duration taken to cover a certain distance. | \( T = \frac{D}{S} \) |
| Converting Minutes to Hours | Divide the number of minutes by 60. | \( \text{Hours} = \frac{\text{Minutes}}{60} \) |
Additional Information: Solving Speed, Time, Distance Problems
Speed, Time, and Distance problems are common in quantitative aptitude. They often involve setting up algebraic equations based on the given information and the fundamental relationship \( D = S \times T \).
- Identify Variables: Clearly define what each variable represents (distance, speed, time).
- Units: Ensure all quantities are in consistent units (e.g., km and hours, or meters and seconds). Convert units if necessary, as we did with minutes to hours in this problem.
- Formulate Equations: Write down equations that represent the different scenarios or conditions given in the problem.
- Solve the System: Use substitution or elimination methods to solve the system of equations for the unknown variable(s).
- Check the Answer: Plug the calculated value back into the original problem statement to see if it satisfies all conditions.
These problems often test your ability to translate a word problem into mathematical equations and solve them accurately.
Important Questions from Average Speed
A car runs first 275 km at an average speed of 50 km/h and the next 315 km at an average speed of 70 km/h. What is the average speed ( in km/h) for the entire journey?
Akhil rides first 12 km at a speed of 16 km/h and further 6 km at a speed of 20 km/h. Find his average speed (in km/h).
Shyam drives his car 30 km at a speed of 45 km/h and, for the next 1 h 20 m, he drives it at a speed of 51 km/h. Find his average speed (in km/h) for the entire journey.
X and Y travel a distance of 90 km each such that the speed of Y is greater than that of X. The sum of their speeds is 100 km/h and the total time taken by both is 3 hours 45 minutes. The ratio of the speed of X to that of Y is:
If a man travels at \(\frac{1}{x}\) km/h on a journey and returns at \(\rm \frac{1}{x^2}\) km/h, then his average speed for the journey is:
