Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, com pounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is:

The correct option is 20808.

To find the initial investment made by Sunil, we analyze the growth of both investments over the 6-year period and set their final amounts equal.

Step 1: Identify the interest rate per compounding period
The scheme has an annual interest rate of 4% compounded half-yearly.
Since interest compounding occurs twice a year, the rate of interest per half-year period is:
$$r=\frac{4\%}{2}=2\%=0.02$$

Step 2: Calculate the final amount received by Anil
Anil invests Rs 22,000 for 6 years in this scheme.
The number of half-yearly compounding periods over 6 years is:
$$n_A=6\times 2=12$$
Using the compound interest formula, the total amount received by Anil at the end of 6 years is:
$$A_A=22000\times {(1+0.02)}^{12}=22000\times {\left(1.02\right)}^{12}$$

Step 3: Calculate the final amount received by Sunil
Let Sunil's initial investment be P. He invests this amount in the same scheme for 5 years.
The number of compounding periods for Sunil's initial investment is:
$$n_S=5\times 2=10$$
The accumulated amount Sunil receives at the end of 5 years is:
$$A_{S,5}=P\times {\left(1.02\right)}^{10}$$
Sunil then reinvests this entire amount $A_{S,5}$ for 1 year at 10% simple interest.
The final amount received by Sunil at the end of 6 years is:
$$A_{S,6}=A_{S,5}\times (1+\frac{R\times T}{100})$$
Substituting $R=10$ and $T=1$ year:
$$A_{S,6}=A_{S,5}\times (1+0.10)=1.1\times A_{S,5}$$
Substituting the expression for $A_{S,5}$:
$$A_{S,6}=1.1\times P\times {\left(1.02\right)}^{10}$$

Step 4: Equate both amounts and solve for P
We are given that the amounts received by both Anil and Sunil at the end of 6 years are equal:
$$A_A=A_{S,6}$$
Substitute the values:
$$22000\times {\left(1.02\right)}^{12}=1.1\times P\times {\left(1.02\right)}^{10}$$
Divide both sides by ${\left(1.02\right)}^{10}$:
$$22000\times {\left(1.02\right)}^2=1.1\times P$$
Now, solve for P:
$$P=\frac{22000}{1.1}\times {\left(1.02\right)}^2$$
Simplify the division:
$$\frac{22000}{1.1}=20000$$
This gives:
$$P=20000\times {\left(1.02\right)}^2$$
Compute the square:
$${\left(1.02\right)}^2=1.0404$$
Finally, calculate the value of P:
$$P=20000\times 1.0404=20808$$
Therefore, the initial investment made by Sunil is Rs 20,808.