A person invests Rs 10,000 for two years at a certain rate of intersect compounded annually. At the end of one years this sum amounts to Rs 11,200. Calcaulte.
the amount at the end fo the second year.

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Identify the given values
- Principal (P) = Rs 10,000
- Amount after 1 year (A1) = Rs 11,200
- Time (t) = 2 years

### Step 2: Calculate the rate of interest (R) for the first year
We know that the amount after the first year can be calculated using the formula:
\[ A = P(1 + R) \]
Where:
- A = Amount after 1 year
- P = Principal
- R = Rate of interest (in decimal)

Substituting the known values:
\[ 11,200 = 10,000(1 + R) \]

### Step 3: Solve for R
To isolate R, divide both sides by 10,000:
\[ 1 + R = \frac{11,200}{10,000} \]
\[ 1 + R = 1.12 \]

Now, subtract 1 from both sides:
\[ R = 1.12 - 1 \]
\[ R = 0.12 \]

### Step 4: Convert R to percentage
To express R as a percentage:
\[ R = 0.12 \times 100 = 12\% \]

### Step 5: Calculate the amount at the end of the second year
Now, we will use the compound interest formula to find the amount after 2 years:
\[ A = P(1 + R)^t \]

Substituting the known values:
\[ A = 10,000(1 + 0.12)^2 \]

### Step 6: Calculate (1 + R)^2
Calculate:
\[ (1 + 0.12)^2 = (1.12)^2 = 1.2544 \]

### Step 7: Calculate the final amount
Now substitute back into the formula:
\[ A = 10,000 \times 1.2544 \]
\[ A = 12,544 \]

### Final Answer
The amount at the end of the second year is Rs 12,544.

---