a) ₹x

b) ₹100x

c) ₹\left(\frac{100}x\right)

d) ₹\left(\large\frac{100}{x^2}\right)

correct answer is: d) ₹\left(\large\frac{100}{x^2}\right)

#### Explanation

Here ‘*a sum*’ means ‘*principal*’.

According to the question,

Principle*(p)* = ?

Rate*(r)* = x %

Time*(t)* = x years

S.I. = ₹ x

So, We know, S.I. =\large\frac{p\times t\times r}{100}

\rightarrow ₹x=\left(\large\frac{p\times x\times x}{100}\right)

\rightarrow px^2=₹100x _{[cross multiply]}

\rightarrow p=₹\large\frac{100x}{x^2}

\therefore Sum will be =₹\large\frac{100x}{x^2} **Ans:** Sum will be =₹\large\frac{100x}{x^2}

##### Another method to articulate this particular math query is available:

**How much principal sum will yield ₹x in simple interest at x% per annum for x years?****Calculate the principal amount needed to generate ₹x in simple interest at x% per annum for x years.****Find the sum required to earn ₹x in simple interest at an annual rate of x% for x years.****Determine the principal sum that results in ₹x as simple interest over x years at an interest rate of x% per annum.****What is the initial amount needed to accrue ₹x in simple interest with an interest rate of x% per annum for x years?**