X, Y and Z travel from the same place with uniform speeds 4 km/hr, 5 km/hr and 6 km/hr respectively. Y starts 2 hours after X. How long after Y must Z start in order that they overtake X at the same instant?
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.
| Person | Speed (km/hr) | Travel Time (hours) | Distance (km) |
|---|---|---|---|
| X | 4 | \(t_X = 10\) | \(D = 4 \times 10 = 40\) |
| Y | 5 | \(t_Y = 8\) (\(t_X - 2\)) | \(D = 5 \times 8 = 40\) |
| Z | 6 | \(t_Z = \frac{20}{3}\) (\(\frac{40}{6}\)) | \(D = 6 \times \frac{20}{3} = 40\) |
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.
Similar Questions
A passenger train departs from Delhi at 6 p.m. for Mumbai. At 9 p.m., an express train, whose average speed exceeds that of the passenger train by 15 km/hour leaves Mumbai for Delhi. Two trains met each other mid-route. At what time do they meet, given that the distance between the cities is 1080 km?
Two men, A and B run a 4 km race on a course 0.25 km round. If their speeds are in the ratio 5 : 4, how often does the winner pass the other?
A car did a journey in t hours. Had the average speed been x kmph greater, the journey would have taken y hours less. How long was the journey?
In a race of 1000 m, A beats B by 150 m, while in another race of 3000 m, C beats D by 400 m. Speed of B is equal to that of D. (Assume that A, B, C and D run with uniform speed in all the events). If A and C participate in a race of 6000 m, then which one of the following is correct?
A thief is spotted by a policeman from a distance of 100 m. When the policeman starts the chase, the thief also starts chasing. If the speed of the thief is 8 km/hr and that of the policeman is 10 km/hr, then how far will the thief have to run before he is overtaken?
When the speed of a train is increased by 20%, it takes 20 minutes less to cover the same distance. What is the time taken to cover the same distance with the original speed?
The speeds of three cars are in the ratio 2 : 3 : 4. What is the ratio between the time taken by these cars to travel the same distance?
A bike consumes 20 mL of petrol per kilometre, if it is driven at a speed in the range of 25 – 50 km/hour and consumes 40 mL of petrol per kilometre at any other speed. How much petrol is consumed by the bike in travelling a distance of 50 km, if the bike is driven at a speed of 40 km/hour for the first 10 km, at a speed of 60 km/hour for the next 30 km and at a speed of 30 km/hour for the last 10 km?
In a race of 1000 m, A beats B by 100 m or 10 seconds. If they start a race of 1000 m simultaneously from the same point and if B gets injured after running 50 m less than half the race length and due to which his speed gets halved, then by how much time will A beat B?
Radha and Hema are neighbors and study in the same school. Both of them use bicycles to go to the school. Radha’s speed is 8 km/hr. whereas Hema’s speed is 10 km/hr. Hema takes 9 minutes less than Radha to reach the school. How far is the school from the locality of Radha and Hema?
Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
A train, running at a speed of 36 kmph, takes 60 seconds to cross an electric pole. How much time it will take to cross a platform of length 250 meter?
A truck covered a distance of 780 km in 12 hours. What distance would a car cover in 11 hours travelling at a speed which is seven kmph more than that of the truck?
