Understanding the Distance, Speed, and Time Problem

This problem involves calculating the ratio of two speeds given information about the distances covered and the relationship between the time taken for each part of the journey. Karan travels a certain distance, part by car and part on foot after a breakdown. We are given the fraction of the distance covered by car and the relationship between the time spent walking and the time spent driving.

Analyzing the Given Information

• Total distance is unknown. Let's denote it by \(D\).
• Distance covered by car = Two third of the total distance.
• Distance covered on foot = Remaining distance.
• Time taken on foot = 9 times the time taken by car.

Let's break down the distances:

• Distance covered by car (\(D_{car}\)) = \(\frac{2}{3} \times D\)
• Remaining distance = \(D - \frac{2}{3}D = \frac{1}{3}D\)
• Distance covered on foot (\(D_{foot}\)) = \(\frac{1}{3}D\)

Let's denote the speeds and times:

• Speed of the car = \(S_{car}\)
• Speed while walking = \(S_{walk}\)
• Time taken by car = \(T_{car}\)
• Time taken on foot = \(T_{foot}\)

We are given the relationship between the times:

• \(T_{foot} = 9 \times T_{car}\)

Using the Speed, Distance, and Time Formula

The fundamental relationship is:

Distance = Speed \(\times\) Time

From this, we can express Time as:

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

Applying this formula to the car journey:

\(T_{car} = \frac{D_{car}}{S_{car}}\)

Substituting the distance: \(T_{car} = \frac{\frac{2}{3}D}{S_{car}}\)

Applying this formula to the journey on foot:

\(T_{foot} = \frac{D_{foot}}{S_{walk}}\)

Substituting the distance: \(T_{foot} = \frac{\frac{1}{3}D}{S_{walk}}\)

Setting up the Equation from the Time Relation

We are given \(T_{foot} = 9 \times T_{car}\). Let's substitute the expressions for \(T_{foot}\) and \(T_{car}\) into this equation:

\(\frac{\frac{1}{3}D}{S_{walk}} = 9 \times \left(\frac{\frac{2}{3}D}{S_{car}}\right)\)

Solving for the Speed Ratio

Now, we need to simplify the equation and find the ratio \(S_{walk} : S_{car}\) or \(\frac{S_{walk}}{S_{car}}\).

The distance \(D\) appears on both sides of the equation. Assuming the distance is not zero (\(D \neq 0\)), we can cancel \(D\) from both sides:

\(\frac{\frac{1}{3}}{S_{walk}} = 9 \times \left(\frac{\frac{2}{3}}{S_{car}}\right)\)

\(\frac{1}{3 \times S_{walk}} = 9 \times \frac{2}{3 \times S_{car}}\)

\(\frac{1}{3 \times S_{walk}} = \frac{18}{3 \times S_{car}}\)

\(\frac{1}{3 \times S_{walk}} = \frac{6}{S_{car}}\)

To find the ratio \(\frac{S_{walk}}{S_{car}}\), we can cross-multiply or rearrange the terms:

\(1 \times S_{car} = 6 \times (3 \times S_{walk})\)

\(S_{car} = 18 \times S_{walk}\)

Now, to get the ratio \(\frac{S_{walk}}{S_{car}}\), divide both sides by \(S_{car}\) (assuming \(S_{car} \neq 0\)):

\(\frac{S_{car}}{S_{car}} = \frac{18 \times S_{walk}}{S_{car}}\)

\(1 = 18 \times \frac{S_{walk}}{S_{car}}\)

Finally, divide both sides by 18:

\(\frac{1}{18} = \frac{S_{walk}}{S_{car}}\)

This means the ratio of his walking speed to his car's speed is 1 : 18.

Conclusion

The ratio of Karan's walking speed to his car's speed is 1:18. This makes sense because he covered a much shorter distance (1/3 D) on foot but took much longer time (9 times the car time) compared to the car which covered a larger distance (2/3 D) in a shorter time. A lower speed for the foot journey compared to the car journey is expected.

Revision Table: Key Concepts

Additional Information: Solving Ratio Problems

When solving problems involving ratios of speeds or times, it is often helpful to express all quantities in terms of a few variables (like total distance \(D\) and one of the times or speeds). Using the fundamental formula \(D = S \times T\) allows you to set up equations based on the given information and then solve for the desired ratio. In this problem, setting up the equation based on the time relationship (\(T_{foot} = 9 \times T_{car}\)) was the key step after determining the distances.

Remember that if the distance is constant, speed is inversely proportional to time (\(S \propto \frac{1}{T}\)). If the time is constant, speed is directly proportional to distance (\(S \propto D\)). If the speed is constant, distance is directly proportional to time (\(D \propto T\)). This problem combines different distances and a relationship between times, requiring careful application of the formula for each segment.