Understanding the Distance, Speed, and Time Problem

This problem involves calculating the distance based on how changes in speed affect the travel time. We are given two scenarios with different speeds and the corresponding arrival times relative to a scheduled time (late or early).

Analyzing the Given Information

• Speed 1: \(3 \, \text{km/h}\)
• Arrival Time 1: 15 minutes late
• Speed 2: \(4 \, \text{km/h}\)
• Arrival Time 2: 15 minutes early

The key to solving this type of problem is recognizing the relationship between speed, distance, and time, and how the difference in arrival times in the two scenarios relates to the time taken in each case.

Relating Speed, Distance, and Time

The fundamental formula is:

\(\text{Distance} = \text{Speed} \times \text{Time}\)

This can be rearranged to find time:

\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)

Let the distance Ram has to cover be \(d\) km.

Let the scheduled time to cover the distance be \(t\) hours.

Calculating Time in Each Scenario

Scenario 1: Speed = 3 km/h

Time taken (\(t_1\)) = \(\frac{d}{3}\) hours.

Since Ram reaches 15 minutes late, this time is the scheduled time plus 15 minutes.

15 minutes needs to be converted to hours: \(15 \, \text{minutes} = \frac{15}{60} \, \text{hours} = \frac{1}{4} \, \text{hours} = 0.25 \, \text{hours}\).

So, \(t_1 = t + 0.25\).

Thus, \(\frac{d}{3} = t + 0.25\) (Equation 1)

Scenario 2: Speed = 4 km/h

Time taken (\(t_2\)) = \(\frac{d}{4}\) hours.

Since Ram reaches 15 minutes early, this time is the scheduled time minus 15 minutes.

So, \(t_2 = t - 0.25\).

Thus, \(\frac{d}{4} = t - 0.25\) (Equation 2)

Finding the Time Difference

The difference between the time taken in the first scenario and the second scenario is the difference between being 15 minutes late and 15 minutes early.

Time difference = \(t_1 - t_2\)

Using the relations with scheduled time \(t\):

\(t_1 - t_2 = (t + 0.25) - (t - 0.25)\)

\(t_1 - t_2 = t + 0.25 - t + 0.25\)

\(t_1 - t_2 = 0.50\) hours.

0.50 hours is equivalent to 30 minutes.

Alternatively, using the distance formula:

\(\frac{d}{3} - \frac{d}{4} = 0.5\)

Solving for the Distance

We have the equation: \(\frac{d}{3} - \frac{d}{4} = 0.5\)

To solve for \(d\), find a common denominator for 3 and 4, which is 12.

Multiply both sides of the equation by 12:

\(12 \left(\frac{d}{3} - \frac{d}{4}\right) = 12 \times 0.5\)

\(\frac{12d}{3} - \frac{12d}{4} = 6\)

\(4d - 3d = 6\)

\(d = 6\)

So, the distance Ram has to cover is 6 km.

Verification

Let's check if the distance 6 km satisfies the conditions.

• At 3 km/h, time taken = \(\frac{6 \, \text{km}}{3 \, \text{km/h}} = 2\) hours.
• At 4 km/h, time taken = \(\frac{6 \, \text{km}}{4 \, \text{km/h}} = 1.5\) hours (or 1 hour 30 minutes).

The difference in time is \(2 \, \text{hours} - 1.5 \, \text{hours} = 0.5\) hours (or 30 minutes).

If 2 hours is 15 minutes late, the scheduled time is \(2 \, \text{hours} - 15 \, \text{minutes} = 2 \, \text{hours} - 0.25 \, \text{hours} = 1.75\) hours (or 1 hour 45 minutes).

If 1.5 hours is 15 minutes early, the scheduled time is \(1.5 \, \text{hours} + 15 \, \text{minutes} = 1.5 \, \text{hours} + 0.25 \, \text{hours} = 1.75\) hours (or 1 hour 45 minutes).

Both scenarios give the same scheduled time (1.75 hours), and the time difference is 30 minutes, which matches the difference between being 15 minutes late and 15 minutes early. The calculation is correct.

The distance Ram has to cover is 6 km.

Revision Table: Key Concepts

Additional Information: Distance, Speed, and Time Problems

Distance, speed, and time problems are common in quantitative aptitude. They often involve:

• Constant speed travel.
• Changes in speed affecting time.
• Relative speed (for objects moving towards or away from each other).
• Problems involving trains, boats (upstream/downstream), or planes (with wind).

A key principle used here is that for a fixed distance, speed is inversely proportional to time. If speed increases, time decreases proportionally, and vice versa.

In this problem, we used the difference in times to set up an equation based on the unknown distance. This method is effective when you know the speeds and the resulting time difference for a constant distance.