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**Hint:**In the given question, we have to find the compound interest where the principal amount is borrowed and the rate of annual interest is given. We can use the formula \[A = P{(1 + \dfrac{r}{n})^{n\,t}}\] to calculate the amount payable after three years and then deduct the principal amount to arrive at the compound interest i.e. \[CI = A - P\].

**Complete step by step solution:**

Compound interest (also known as compounding interest) is the interest on a loan or deposit that is measured using both the original principal and the interest accrued over time.

The total compounded amount payable after a specific period can be calculated with the following formula:

\[A = P{(1 + \dfrac{r}{n})^{n\,t}}\]

Where,

\[A = \]Final Amount

\[P = \]Principal Amount

\[r = \]Rate of interest

\[n = \]Number of times interest is to be compounded per year/annum

\[t = \]Time (in years)

For example, Rs.\[100\] compounded at \[10\% \] per annum for \[2\] year will result in:

\[ = 100{(1 + \dfrac{{10\% }}{1})^{(1)(2)}}\]

\[ = 100{(1 + 0.1)^2}\]

On simplifying in further we get

\[ = 100{(1.1)^2}\]

\[ = 100(1.21)\]

\[ = 121\]

Here, if the interest was to be compounded half-yearly, then \[n = 4\]as interest will be compounded four times over the two years half-yearly. Similarly, if it was quarterly, then \[n = 8\].

Now we can solve the sum as follows:

The loan taken will become our principal amount and rate of interest is given to be compounded annually. Hence-

\[P = 16,000\]

\[n = 1\] (annually)

\[t = 3\] (years)

\[r = 12\dfrac{1}{2}\% = 0.125\]

Applying the formula, we get,

\[A = P{(1 + \dfrac{r}{n})^{n\,t}}\]

\[A = 16000{(1 + \dfrac{{0.125}}{1})^{(1)(3)}}\]

\[A = 16000{(1 + 0.125)^3}\]

On simplifying in further we get

\[A = 16000{(1.125)^3}\]

\[A = 16000(1.4238)\]

\[A = 22781.25\]

Now Compound Interest (CI) can be found out as follows:

\[CI = A - P\]

\[CI = 22781.25 - 16000\]

\[CI = 6781.25\]

**Hence compounded Interest will be \[Rs.6781.25\] over the period of three years. Therefore, Option (A) is correct.**

**Note:**

We can also apply the following formula to solve the sum directly:

\[CI = P[{(1 + \dfrac{r}{{100}})^n} - 1]\] where \[n = \]number of years

\[CI = 16000[{(1 + \dfrac{{12.5}}{{100}})^3} - 1]\]

\[CI = 16000[{(1 + 0.125)^3} - 1]\]

On simplifying in further we get

\[CI = 16000[{(1.125)^3} - 1]\]

\[CI = 16000[1.4238 - 1]\]

\[CI = 16000[0.4238]\]

\[CI = 6780.8\]

Difference of \[6781.25 - 6780.8 = 0.45\] in the answer is due to decimal points.

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