Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received 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 the same, then the initial investment made by Sunil, in rupees, is:

Question

Accepted Answer

#### Step 1: Calculate Anil's total amount after 6 years
Anil invests Rs. 22000 at **4% per annum compounded half-yearly**. The formula for compound interest is:  
$$ A = P \left(1 + \frac{r}{n}ight)^{n \cdot t} $$  
Here:  
- $$ P = 22000 $$ (principal),  
- $$ r = 4\% = \frac{4}{100} $$ (annual rate),  
- $$ n = 2 $$ (compounding frequency per year),  
- $$ t = 6 $$ years.

Substitute values:  
$$ A = 22000 \left(1 + \frac{4}{2 \cdot 100}ight)^{2 \cdot 6} = 22000 \left(1 + \frac{2}{100}ight)^{12} = 22000 	imes (1.02)^{12}. $$

Using calculations:  
$$ A = 22000 	imes 1.26824 = Rs. 27891.28. $$

#### Step 2: Calculate Sunil's total amount after 6 years
Let Sunil's initial investment be $$ P $$.

**Step A:** Sunil invests for **5 years** in the same scheme with compound interest:  
$$ A_5 = P 	imes (1.02)^{10}. $$

**Step B:** At the end of 5 years, Sunil reinvests the entire amount for **1 year** at **10% simple interest**:  
$$ A_6 = A_5 	imes (1 + r_{simple}) = P 	imes (1.02)^{10} 	imes (1 + 0.10). $$  
$$ A_6 = P 	imes (1.02)^{10} 	imes 1.1. $$

#### Step 3: Equate Anil's and Sunil's amounts
Since both amounts are equal after 6 years:  
$$ Anil's Amount = Sunil's Amount $$.  
$$ 22000 	imes (1.02)^{12} = P 	imes (1.02)^{10} 	imes 1.1. $$

Simplify to solve for $$ P $$:  
$$ P = \frac{22000 	imes (1.02)^{12}}{(1.02)^{10} 	imes 1.1}. $$  
$$ P = \frac{22000 	imes (1.02)^2}{1.1}. $$

Calculate:  
$$ P = \frac{22000 	imes 1.0404}{1.1} = Rs. 20808. $$

#### Final Answer:
The initial investment made by Sunil is **Rs. 20808**.