Three cars A, B and C started from a point at 5 p.m., 6 p.m. and 7 p.m. respectively and travelled at uniform speeds of 60 km/hr., 80 km/hr. and x km/hr. respectively in the same direction. If all the three met at another point at the same instant during their journey, then what is the value of x?
120
Solving Three Cars Meeting Point Problem
This problem involves three cars starting at different times from the same point and travelling in the same direction at uniform speeds. The key information is that all three cars meet at another point at the same instant during their journey. We need to find the speed of the third car, Car C.
Understanding the Problem
Let's denote the starting point as Point S and the meeting point as Point M. The distance between S and M is the same for all three cars. Let's also define a reference time, say 5 p.m., as time 0.
- Car A starts at 5 p.m. (Time 0) with a speed of 60 km/hr.
- Car B starts at 6 p.m. (Time 1 hour after 5 p.m.) with a speed of 80 km/hr.
- Car C starts at 7 p.m. (Time 2 hours after 5 p.m.) with a speed of x km/hr.
Let T be the time (in hours) from 5 p.m. when all three cars meet at Point M. Since they meet at the same instant, they all reach Point M at time T.
Calculating Distances Travelled
The formula for distance is: Distance = Speed × Time.
- Car A travels for T hours. Distance covered by A = \(60 \times T\) km.
- Car B starts 1 hour later (at 6 p.m.). So, Car B travels for \((T - 1)\) hours. Distance covered by B = \(80 \times (T - 1)\) km.
- Car C starts 2 hours later (at 7 p.m.). So, Car C travels for \((T - 2)\) hours. Distance covered by C = \(x \times (T - 2)\) km.
Since all three cars meet at the same point M, the distances covered by them must be equal.
Distance A = Distance B = Distance C
Finding the Meeting Time (T)
We can use the equality of distances covered by Car A and Car B to find the meeting time T.
\(60T = 80(T - 1)\)
Let's solve this equation for T:
\(60T = 80T - 80\)
Subtract \(60T\) from both sides:
\(0 = 80T - 60T - 80\)
\(0 = 20T - 80\)
Add 80 to both sides:
\(80 = 20T\)
Divide by 20:
\(T = \frac{80}{20}\)
\(T = 4\) hours.
So, the three cars meet 4 hours after 5 p.m., which is at 9 p.m.
Calculating the Speed of Car C (x)
Now that we know the meeting time T = 4 hours, we can use the distance equality involving Car C to find its speed, x.
Let's use Distance A = Distance C:
\(60T = x(T - 2)\)
Substitute T = 4 into the equation:
\(60 \times 4 = x(4 - 2)\)
\(240 = x(2)\)
\(240 = 2x\)
Divide by 2:
\(x = \frac{240}{2}\)
\(x = 120\)
The speed of Car C is 120 km/hr.
Alternatively, we could use Distance B = Distance C:
\(80(T - 1) = x(T - 2)\)
Substitute T = 4 into the equation:
\(80(4 - 1) = x(4 - 2)\)
\(80(3) = x(2)\)
\(240 = 2x\)
\(x = 120\)
Both comparisons give the same speed for Car C.
Summary of Calculations
Let's summarize the key steps and results.
| Car | Start Time | Speed (km/hr) | Time Travelled (hours relative to 5 p.m. = T) | Actual Time Travelled (hours) | Distance Covered (km) |
|---|---|---|---|---|---|
| A | 5 p.m. (T=0) | 60 | T | T | \(60T\) |
| B | 6 p.m. (T=1) | 80 | T | \(T-1\) | \(80(T-1)\) |
| C | 7 p.m. (T=2) | x | T | \(T-2\) | \(x(T-2)\) |
Equating distances for A and B:
\(60T = 80(T - 1) \implies T = 4\) hours.
Equating distances for A and C (using T=4):
\(60(4) = x(4 - 2) \implies 240 = 2x \implies x = 120\) km/hr.
Equating distances for B and C (using T=4):
\(80(4 - 1) = x(4 - 2) \implies 80(3) = x(2) \implies 240 = 2x \implies x = 120\) km/hr.
The value of x is 120.
Revision Table: Key Concepts
| Concept | Explanation | Formula |
|---|---|---|
| Distance, Speed, Time | Relationship between distance travelled, the speed of travel, and the time taken. | Distance = Speed × Time |
| Uniform Speed | The speed remains constant throughout the journey. | N/A |
| Meeting Point | A location where two or more moving objects are at the same position at the same time. | Distances covered are equal. |
| Relative Time | Accounting for differences in start times when calculating time travelled. | Time Travelled = Meeting Time - Start Time Offset |
Additional Information: Time and Distance Problems
Time and distance problems are a common topic in quantitative aptitude. They often involve objects moving at uniform speeds and require calculating distance, speed, or time based on given conditions. Key considerations include:
- Ensuring all units (distance, speed, time) are consistent (e.g., km, km/hr, hours).
- Understanding relative speed when objects move towards or away from each other (though not directly used in this problem as they move in the same direction and we are focused on their individual times and distances to a common point).
- Carefully accounting for differences in start times or breaks during the journey.
This specific problem is a good example of using the equality of distance travelled to solve for an unknown speed or time, especially when objects start at different times but meet at the same point.
Similar Questions
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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?
The distance between Delhi and Patna is $480$ km. A train starts from Delhi and travels towards Patna at a speed of $60$ kmph. Another train starts from Patna towards Delhi at a speed of $80$ kmph, but starts $1$ hour after the first train. In how much time, after the first train started, will they meet each other?
Ram traveling at the speed of 3 km/h reaches 15 minutes late. Had he walked at 4 km/h, he would have reached 15 minutes earlier. How much distance does Ram have to cover?
