Walking at $\frac{5}{8}$ of his normal speed from his home, Tarun is 21 minutes late. The usual time taken by him to cover the distance between his home and office is:
35 minutes
To find the usual time Tarun takes to reach his office, let's follow these steps:
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Let's denote Tarun's normal speed as V and the usual time to his office as T (in minutes).
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Tarun walks at \frac{5}{8} of his normal speed; hence, his new speed is \frac{5}{8}V.
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The distance between his home and office remains the same. Hence, using the formula for speed:
\text{Distance} = \text{Speed} \times \text{Time}
The time taken at this reduced speed is:
\frac{\text{Distance}}{\frac{5}{8}V} = \frac{8 \times \text{Distance}}{5V}
This can also be written as \frac{8}{5}T given that usual speed \text{Distance} = VT.
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According to the problem, Tarun is 21 minutes late, so the time taken at reduced speed is:
\frac{8}{5}T = T + 21
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Solving the equation for T:
\frac{8}{5}T = T + 21
Multiply through by 5 to clear the fraction:
8T = 5T + 105
Subtract 5T from both sides:
3T = 105
Dividing both sides by 3 gives:
T = 35
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Thus, the usual time taken by Tarun to cover the distance is 35 minutes.
Therefore, the correct answer is 35 minutes.
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