Solving Train Speed and Time Journey Problems

This problem involves a train traveling a fixed distance at different speeds, resulting in different arrival times. We need to find the train's correct, scheduled journey time. The key principle here is that the distance covered is the same in both scenarios.

Understanding the Train Journey Scenarios

Let's break down the information given:

• Scenario 1: The train travels at a speed of 72 km/h and arrives 15 minutes late.
• Scenario 2: The train travels at a speed of 90 km/h and arrives 5 minutes late.

We are looking for the correct time or the scheduled time for the journey.

Setting up the Equations

Let:

• $D$ be the distance of the journey in kilometers.
• $T$ be the correct time for the journey in hours.

The formula relating distance, speed, and time is: $\text{Distance} = \text{Speed} \times \text{Time}$.

In scenario 1, the train is 15 minutes late. Since the correct time $T$ is in hours, we must convert 15 minutes to hours: $15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hours}$.

The actual time taken in scenario 1 is $T + \frac{1}{4}$ hours. Using the distance formula:

$\qquad D = 72 \times \left(T + \frac{1}{4}\right) \quad (1)$

In scenario 2, the train is 5 minutes late. Converting 5 minutes to hours: $5 \text{ minutes} = \frac{5}{60} \text{ hours} = \frac{1}{12} \text{ hours}$.

The actual time taken in scenario 2 is $T + \frac{1}{12}$ hours. Using the distance formula:

$\qquad D = 90 \times \left(T + \frac{1}{12}\right) \quad (2)$

Solving for the Correct Time

Since the distance $D$ is the same in both scenarios, we can equate the right-hand sides of equations (1) and (2):

$\qquad 72 \times \left(T + \frac{1}{4}\right) = 90 \times \left(T + \frac{1}{12}\right)$

Now, we solve this equation for $T$:

Expand both sides:

$\qquad 72T + 72 \times \frac{1}{4} = 90T + 90 \times \frac{1}{12}$

Simplify the multiplication:

$\qquad 72T + 18 = 90T + \frac{90}{12}$

Simplify the fraction $\frac{90}{12}$ by dividing both numerator and denominator by 6:

$\qquad 72T + 18 = 90T + \frac{15}{2}$

Rearrange the terms to group $T$ on one side and constants on the other:

$\qquad 18 - \frac{15}{2} = 90T - 72T$

Combine terms:

$\qquad \frac{36 - 15}{2} = 18T$

$\qquad \frac{21}{2} = 18T$

Solve for $T$:

$\qquad T = \frac{21}{2 \times 18} = \frac{21}{36}$

Simplify the fraction $\frac{21}{36}$ by dividing both numerator and denominator by 3:

$\qquad T = \frac{7}{12} \text{ hours}$

Converting Time to Minutes

The question asks for the correct time in minutes. To convert hours to minutes, we multiply by 60:

$\qquad \text{Correct Time} = T \times 60 \text{ minutes}$

$\qquad \text{Correct Time} = \frac{7}{12} \times 60 \text{ minutes}$

$\qquad \text{Correct Time} = 7 \times \frac{60}{12} \text{ minutes}$

$\qquad \text{Correct Time} = 7 \times 5 \text{ minutes}$

$\qquad \text{Correct Time} = 35 \text{ minutes}$

Thus, the correct time to cover the journey is 35 minutes.

Revision Table: Key Concepts

Additional Information: Relative Speed

While this problem didn't directly use relative speed, it's a crucial concept in many train-related problems. Relative speed is the speed of one object with respect to another.

• When two trains move in the same direction, their relative speed is the difference between their speeds.
• When two trains move in the opposite direction, their relative speed is the sum of their speeds.
• This concept is often used when calculating the time it takes for trains to pass each other or pass a pole/platform.

Understanding how time is affected by changes in speed for a fixed distance is fundamental to solving problems like the one above. If speed increases, time taken decreases, and vice versa, assuming the distance remains constant. This relationship is inverse proportionality ($T \propto 1/S$).