Answer

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**Hint:**We have been given that the total sum after 18 months when a principal sum of money is compounded half yearly at the rate of $2\dfrac{1}{2}\%$ per annum is equal to Rs. 9826. For this we will assume the principal sum to be x and then we will use the formula:

$A=P{{\left( 1+\dfrac{r}{100n} \right)}^{nt}}$

A= final amount

P= initial principal balance

r= interest rate (in percent)

n= number of times interest applied per time period

t= number of time periods elapsed

According to the formula, we will calculate the required values by taking the time period as 1 year, i.e. 12 months. Then we will put all these values in this formula and obtain an equation in x. by solving that equation, we will get the required value of x.

**Complete step by step answer:**

Here, we have been given the total sum when a principal amount is compounded half yearly at the rate of $2\dfrac{1}{2}\%$ per annum for 18 months as Rs. 9826 and we have been asked to find the principal amount of money.

For this, we will first assume the principal amount to be as ‘x’.

Now, we know that the final amount after compound interest is calculated by the formula:

$A=P{{\left( 1+\dfrac{r}{100n} \right)}^{nt}}$

A= final amount

P= initial principal balance

r= interest rate (in percent)

n= number of times interest applied per time period

t= number of time periods elapsed

Here, from the question, we can see that:

A= Rs. 9826

P= x

r= $2\dfrac{1}{2}\%$

Simplifying the value of r, we get:

$\begin{align}

& r=2\dfrac{1}{2}=\dfrac{4+1}{2} \\

& \Rightarrow r=\dfrac{5}{2}\% \\

\end{align}$

Now, we have been given here that the rate of interest is compounded half-yearly and the total time is 18 months.

So, if we take the time to be 1 year, i.e. 12 months, we can calculate the number of time periods, i.e. t as:

$\begin{align}

& t=\dfrac{18}{12} \\

& \Rightarrow t=\dfrac{3}{2} \\

\end{align}$

It has also been given to us that the interest is compounded half yearly, i.e. the rate of interest ts compounded twice every year.

Since, we have taken the time period to 1 year, we can say that the rate of interest is compounded twice in one time period.

Thus, we get the value of n= 2.

Now, if we put all these values in the formula, we will obtain as equation in ‘x’.

Putting all the values in the formula for compound interest mentioned above we get:

$\begin{align}

& A=P{{\left( 1+\dfrac{r}{100n} \right)}^{nt}} \\

& \Rightarrow 9826=x{{\left( 1+\dfrac{\dfrac{5}{2}}{100\times 2} \right)}^{2\times \dfrac{3}{2}}} \\

\end{align}$

Now, solving this equation, we will get:

$\begin{align}

& 9826=x{{\left( 1+\dfrac{\dfrac{5}{2}}{100\times 2} \right)}^{2\times \dfrac{3}{2}}} \\

& \Rightarrow 9826=x{{\left( 1+\dfrac{1}{80} \right)}^{3}} \\

& \Rightarrow 9826=x{{\left( \dfrac{81}{80} \right)}^{3}} \\

& \Rightarrow 9826\times {{\left( \dfrac{80}{81} \right)}^{3}}=x \\

\end{align}$

Finally cubing and comparing them into decimals we get:

$\begin{align}

& \Rightarrow 9826\times {{\left( \dfrac{80}{81} \right)}^{3}}=x \\

& \Rightarrow 9826\times 0.96=x \\

& \therefore x=9466.54 \\

\end{align}$

Hence, the required principal sum is Rs. 9466.54

**So, the correct answer is “Option A”.**

**Note:**We have here taken the time period to be 1 year. But we can also do this question by taking it as 6 months.

The values according to will change as follows:

A and P will remain the same.

Here, the time is given to be 18 months. Therefore, the number of time periods, i.e. t will be:

$\begin{align}

& t=\dfrac{18}{6} \\

& \Rightarrow t=3 \\

\end{align}$

Now, in the question, the rate of interest is given as “per annum” but as we are taking the time period to be 6 months, we need to change it accordingly.

Rate of interest per annum= $2\dfrac{1}{2}\%=\dfrac{5}{2}\%$

Hence, the rate of interest per 6 months will be half of the rate of interest per annum.

Thus, r=$\dfrac{1}{2}\left( \dfrac{5}{2}\% \right)=\dfrac{5}{4}\%$

Now, in the question it is also given that the interest is compounded half yearly. Hence, it will be compounded once every 6 months. This will give us n=1.

Now, putting all these values in the formula for compound interest we get:

$\begin{align}

& A=P{{\left( 1+\dfrac{r}{100n} \right)}^{nt}} \\

& \Rightarrow 9826=x{{\left( 1+\dfrac{\dfrac{5}{4}}{100\times 1} \right)}^{1\times 3}} \\

& \Rightarrow 9826=x{{\left( 1+\dfrac{1}{80} \right)}^{3}} \\

& \Rightarrow 9826=x{{\left( \dfrac{81}{80} \right)}^{3}} \\

\end{align}$

Now, we can see that the same equation is formed as above. Hence, the answer will be the same in both the cases.

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