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.

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

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.