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A person travelled the first half of the distance from her house to her office at 24 km/h, exactly half of the remaining distance at 20 km/h and the rest at 15 km/h. Her average speed (km//h) in entire journey was :

A

20

B

`18(1)/(3)`

C

`19(2)/(3)`

D

18

Text Solution

AI Generated Solution

The correct Answer is:
To find the average speed of the person during her entire journey, we can follow these steps: ### Step 1: Define the total distance Let the total distance from the house to the office be \( D \). For ease of calculation, we can express \( D \) as \( 4x \), where \( x \) is a unit distance. ### Step 2: Break down the journey into segments 1. The first half of the distance is \( \frac{D}{2} = 2x \). 2. The remaining distance after the first half is also \( 2x \). Half of this remaining distance is \( \frac{2x}{2} = x \). 3. The rest of the distance is also \( x \). ### Step 3: Calculate the time taken for each segment - **Time for the first segment (2x at 24 km/h)**: \[ t_1 = \frac{2x}{24} = \frac{x}{12} \text{ hours} \] - **Time for the second segment (x at 20 km/h)**: \[ t_2 = \frac{x}{20} \text{ hours} \] - **Time for the third segment (x at 15 km/h)**: \[ t_3 = \frac{x}{15} \text{ hours} \] ### Step 4: Calculate the total time taken Now, we sum the times for all segments: \[ \text{Total time} = t_1 + t_2 + t_3 = \frac{x}{12} + \frac{x}{20} + \frac{x}{15} \] ### Step 5: Find a common denominator and simplify The least common multiple (LCM) of 12, 20, and 15 is 60. We convert each term: - \( \frac{x}{12} = \frac{5x}{60} \) - \( \frac{x}{20} = \frac{3x}{60} \) - \( \frac{x}{15} = \frac{4x}{60} \) Now, adding these: \[ \text{Total time} = \frac{5x + 3x + 4x}{60} = \frac{12x}{60} = \frac{x}{5} \text{ hours} \] ### Step 6: Calculate the average speed The average speed is given by the formula: \[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{4x}{\frac{x}{5}} = 4x \times \frac{5}{x} = 20 \text{ km/h} \] ### Conclusion Thus, the average speed of the person during her entire journey is **20 km/h**. ---
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