Find the amount when Rs 10,000 is invested for `2(1)/(2)` years at 10% intersect compounded yearly.

Answer

To find the amount when Rs 10,000 is invested for \(2 \frac{1}{2}\) years at 10% interest compounded yearly, we can follow these steps:

### Step 1: Identify the values
- Principal (P) = Rs 10,000
- Rate of interest (R) = 10%
- Time (T) = \(2 \frac{1}{2}\) years = 2.5 years

### Step 2: Calculate the amount for the first 2 years
Using the compound interest formula:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
Where:
- \(A\) = Amount
- \(P\) = Principal
- \(R\) = Rate of interest
- \(n\) = Number of years

For the first 2 years:
\[ A = 10,000 \left(1 + \frac{10}{100}\right)^2 \]
\[ A = 10,000 \left(1 + 0.1\right)^2 \]
\[ A = 10,000 \left(1.1\right)^2 \]
\[ A = 10,000 \times 1.21 = 12,100 \]

### Step 3: Calculate the amount for the remaining \( \frac{1}{2} \) year
Now, we will use the amount from the first 2 years as the principal for the remaining \( \frac{1}{2} \) year.

New Principal (P) = Rs 12,100
Time (T) = \( \frac{1}{2} \) year
Using the formula again:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]
\[ A = 12,100 \left(1 + \frac{10}{100}\right)^{\frac{1}{2}} \]
\[ A = 12,100 \left(1 + 0.1\right)^{\frac{1}{2}} \]
\[ A = 12,100 \left(1.1\right)^{\frac{1}{2}} \]
Calculating \( (1.1)^{\frac{1}{2}} \):
\[ (1.1)^{\frac{1}{2}} \approx 1.0488 \]
Now substituting back:
\[ A = 12,100 \times 1.0488 \approx 12,705.68 \]

### Step 4: Final Amount
Thus, the total amount after \(2 \frac{1}{2}\) years is approximately Rs 12,705.68.

### Summary
The amount when Rs 10,000 is invested for \(2 \frac{1}{2}\) years at 10% interest compounded yearly is approximately Rs 12,705.68.

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