Solving the Overtaking Travel Problem

This problem involves three individuals, X, Y, and Z, traveling from the same starting point at different uniform speeds. We need to determine the time difference in starting between Y and Z such that Y and Z overtake X at the exact same moment.

Understanding the Problem Setup

We are given the uniform speeds of X, Y, and Z:

• Speed of X = 4 km/hr
• Speed of Y = 5 km/hr
• Speed of Z = 6 km/hr

We know that Y starts 2 hours after X. Let's denote the time X travels until being overtaken as \(t_X\) hours. Since Y starts 2 hours later, the time Y travels until the same instant is \(t_Y = t_X - 2\) hours.

Let the time Z travels until the same instant be \(t_Z\) hours. Z starts some time after Y. We need to find this time difference.

When Y and Z overtake X at the same instant, they must all be at the same distance from the starting point. Let this distance be \(D\).

Setting up Equations based on Speed, Time, and Distance

The relationship between distance, speed, and time is: \(\text{Distance} = \text{Speed} \times \text{Time}\).

Applying this to each person at the moment Y and Z overtake X at distance \(D\):

• For X: \(D = \text{Speed}_X \times t_X \implies D = 4 \times t_X\) (Equation 1)
• For Y: \(D = \text{Speed}_Y \times t_Y \implies D = 5 \times t_Y\) (Equation 2)
• For Z: \(D = \text{Speed}_Z \times t_Z \implies D = 6 \times t_Z\) (Equation 3)

We also have the relationship between X's and Y's travel time:

• \(t_Y = t_X - 2\) (Equation 4)

Solving for Travel Times and Distance

We can use Equations 1, 2, and 4 to find the travel times for X and Y, and the distance \(D\).

From Equation 1 and Equation 2, since both equal \(D\):

\(4 t_X = 5 t_Y\)

Substitute \(t_Y\) from Equation 4 into this equation:

\(4 t_X = 5 (t_X - 2)\)

Now, let's solve for \(t_X\):

\(4 t_X = 5 t_X - 10\)

\(10 = 5 t_X - 4 t_X\)

\(t_X = 10\) hours

So, X travels for 10 hours until being overtaken.

Now we can find \(t_Y\) using Equation 4:

\(t_Y = t_X - 2 = 10 - 2 = 8\) hours

Y travels for 8 hours.

We can also find the distance \(D\) using either Equation 1 or 2:

\(D = 4 \times t_X = 4 \times 10 = 40\) km

or

\(D = 5 \times t_Y = 5 \times 8 = 40\) km

The distance where they all meet is 40 km from the start.

Finding Z's Travel Time and Start Time

Now consider Z. Z travels the same distance \(D = 40\) km at a speed of 6 km/hr. Using Equation 3:

\(D = 6 \times t_Z\)

\(40 = 6 \times t_Z\)

\(t_Z = \frac{40}{6} = \frac{20}{3}\) hours

So, Z travels for \(\frac{20}{3}\) hours until reaching the overtaking point.

We need to find out how long after Y must Z start. We know Y travels for 8 hours and Z travels for \(\frac{20}{3}\) hours. Since they reach the same point at the same instant, Z must have started later than Y because Z's travel time (\(\frac{20}{3}\) hours) is less than Y's travel time (8 hours).

The time Z starts after Y is the difference between Y's travel time and Z's travel time:

Time Z starts after Y \( = t_Y - t_Z\)

Time Z starts after Y \( = 8 - \frac{20}{3}\)

To subtract, we find a common denominator (3):

\(8 = \frac{8 \times 3}{3} = \frac{24}{3}\)

Time Z starts after Y \( = \frac{24}{3} - \frac{20}{3} = \frac{24 - 20}{3} = \frac{4}{3}\) hours

Therefore, Z must start \(\frac{4}{3}\) hours after Y in order for them to overtake X at the same instant.

Conclusion

By calculating the travel times for X and Y based on their relative start times and speeds, we found the distance at which the overtaking occurs. Then, using this distance and Z's speed, we determined Z's travel time. The difference between Y's and Z's travel times gives us how much later Z must start than Y.

The final answer is \(\frac{4}{3}\) hours.

This table summarizes the key values calculated for each person at the moment of overtaking.

Revision Table: Speed, Time, and Distance Problems

Speed, time, and distance problems are common in quantitative aptitude. Here's a quick review of key formulas and concepts:

• Formula: \(\text{Distance} = \text{Speed} \times \text{Time}\)
• From this, you can derive: \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\) and \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
• When dealing with multiple objects or different segments of a journey, ensure units are consistent (e.g., km/hr and hours, or m/s and seconds).
• Problems often involve relative speed when objects are moving towards or away from each other, or in the same direction (like in overtaking scenarios).

Additional Information: Relative Speed Concept

While we solved this problem using absolute speeds and travel times, the concept of relative speed is also useful in overtaking problems. When two objects move in the same direction, their relative speed is the difference between their speeds. The time it takes for the faster object to overtake the slower one is the distance between them divided by their relative speed.

In this problem, at the moment Y starts, X has a head start. The distance X has covered in 2 hours is \(4 \text{ km/hr} \times 2 \text{ hours} = 8 \text{ km}\). The relative speed of Y with respect to X is \(5 - 4 = 1\) km/hr. The time it takes for Y to cover the 8 km head start is \(8 \text{ km} / 1 \text{ km/hr} = 8\) hours. This is the time Y travels until overtaking X, which matches our \(t_Y = 8\) hours. This confirms our method.

Similarly, Z needs to overtake X. When Z starts, X has a certain head start. The calculation ensures Z reaches the same point as Y and X at the same time instant.