A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to 4,950, find the sum invested.

Answer

To solve the problem step by step, we will follow the process of calculating the compound interest for each year and then find the initial sum invested.

### Step 1: Define the variables
Let the sum invested be \( P \) (the principal amount).

### Step 2: Calculate the interest for the first year
The interest rate for the first year is 10%. The simple interest for the first year can be calculated using the formula:
\[ \text{SI} = \frac{P \times \text{Rate} \times \text{Time}}{100} \]
For the first year:
\[ \text{SI}_1 = \frac{P \times 10 \times 1}{100} = \frac{P}{10} \]

### Step 3: Calculate the amount at the end of the first year
At the end of the first year, the amount \( A_1 \) will be:
\[ A_1 = P + \text{SI}_1 = P + \frac{P}{10} = \frac{10P + P}{10} = \frac{11P}{10} \]

### Step 4: Calculate the interest for the second year
The principal for the second year is now \( A_1 = \frac{11P}{10} \). The interest rate for the second year is 15%. The simple interest for the second year is:
\[ \text{SI}_2 = \frac{A_1 \times 15 \times 1}{100} = \frac{\frac{11P}{10} \times 15}{100} = \frac{11P \times 15}{1000} = \frac{165P}{1000} = \frac{33P}{200} \]

### Step 5: Set up the equation with the given information
According to the problem, the compound interest for the second year is given as 4950. Therefore, we can set up the equation:
\[ \frac{33P}{200} = 4950 \]

### Step 6: Solve for \( P \)
To find \( P \), we can rearrange the equation:
\[ P = 4950 \times \frac{200}{33} \]
Calculating this gives:
\[ P = 4950 \times \frac{200}{33} = 4950 \times 6.06060606 \approx 30,000 \]

### Final Answer
The sum invested is \( P = 30,000 \) rupees.

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