Explanation: To find the average speed of a car traveling four equal distances at different speeds, we need to use the concept of the harmonic mean. The harmonic mean is particularly useful when dealing with rates or speeds where the time taken for each segment is different.
The formula for the harmonic mean \( H \) of \( n \) speeds \( s_1, s_2, \ldots, s_n \) is given by:
\[ H = \frac{n}{\frac{1}{s_1} + \frac{1}{s_2} + \cdots + \frac{1}{s_n}} \]
In this problem, the car travels four equal distances at speeds of 2 kmph, 3 kmph, 4 kmph, and 6 kmph. Let's denote these speeds as \( s_1 = 2 \) kmph, \( s_2 = 3 \) kmph, \( s_3 = 4 \) kmph, and \( s_4 = 6 \) kmph. The number of speeds \( n \) is 4.
Substituting the values into the harmonic mean formula, we get:
\[ H = \frac{4}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6}} \]
First, we need to find the sum of the reciprocals of the speeds:
\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \]
To add these fractions, we need a common denominator. The least common multiple of 2, 3, 4, and 6 is 12. Converting each fraction to have a denominator of 12, we get:
\[ \frac{1}{2} = \frac{6}{12}, \quad \frac{1}{3} = \frac{4}{12}, \quad \frac{1}{4} = \frac{3}{12}, \quad \frac{1}{6} = \frac{2}{12} \]
Adding these fractions together:
\[ \frac{6}{12} + \frac{4}{12} + \frac{3}{12} + \frac{2}{12} = \frac{15}{12} \]
Now, substituting this sum back into the harmonic mean formula:
\[ H = \frac{4}{\frac{15}{12}} = \frac{4 \times 12}{15} = \frac{48}{15} = 3.2 \]
Therefore, the average speed of the car is 3.2 kmph, which corresponds to option A.
This method ensures that the average speed is correctly calculated by taking into account the varying times taken for each segment of the journey, rather than simply averaging the speeds. The harmonic mean is the appropriate measure for such scenarios, as it gives more weight to the slower speeds, which take more time to cover the same distance.