If a person travels x% faster than normal, he reaches y minutes earlier than normal. What is his normal time of travel?

Correct option is A

​Given:
A person travels x% faster than normal.
He reaches y minutes earlier.
We are to find his normal time of travel.

Formula Used:
If speed increases by x%, the time taken becomes:

​$$\text{New Time} = \frac{\text{Original Time}}{1 + \frac{x}{100}} \\[10pt]\text{Original Time} - \frac{\text{Original Time}}{1 + \frac{x}{100}} = y$$​​

Concept:

Distance = speed × time.

Solution:

$$\text{Let normal time be } T \text{ minutes} \\\text{Person travels } x\% \text{ faster and reaches } y \text{ minutes earlier} \\[5pt]T - \frac{T}{1 + \frac{x}{100}} = y \\[10pt]T \left(1 - \frac{1}{1 + \frac{x}{100}} \right) = y \\[10pt]T \left( \frac{ \frac{x}{100} }{1 + \frac{x}{100}} \right) = y \\[10pt]T = \frac{y \left(1 + \frac{x}{100} \right)}{ \frac{x}{100} } \\[10pt]T = y \cdot \left( \frac{100 + x}{x} \right)$$​

$$T=\Big(\frac{100}{x}+1\Big)y$$

Final Answer: (A)

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