A man deposited Rs. 300 for 2 years, Rs. 600 for 4 years and Rs. 1200 for 6 years. He received Rs. 1020 as the simple interest. The rate of interest per annum is

Options

A. (a) 10 %
B. (b) 15 %
C. (c) 20 %
D. (d) 30 %

Answer

To find the rate of interest per annum, we will use the formula for simple interest and the information given in the problem.

### Step-by-Step Solution:

1. **Identify the Principal Amounts and Time Periods:**
- Let \( P_1 = 300 \) (for 2 years)
- Let \( P_2 = 600 \) (for 4 years)
- Let \( P_3 = 1200 \) (for 6 years)
- Let \( T_1 = 2 \) years
- Let \( T_2 = 4 \) years
- Let \( T_3 = 6 \) years

2. **Set Up the Simple Interest Formula:**
The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]
where \( SI \) is the simple interest, \( P \) is the principal amount, \( R \) is the rate of interest, and \( T \) is the time in years.

3. **Calculate the Simple Interest for Each Deposit:**
- For the first deposit:
\[ SI_1 = \frac{P_1 \times R \times T_1}{100} = \frac{300 \times R \times 2}{100} = \frac{600R}{100} = 6R \]
- For the second deposit:
\[ SI_2 = \frac{P_2 \times R \times T_2}{100} = \frac{600 \times R \times 4}{100} = \frac{2400R}{100} = 24R \]
- For the third deposit:
\[ SI_3 = \frac{P_3 \times R \times T_3}{100} = \frac{1200 \times R \times 6}{100} = \frac{7200R}{100} = 72R \]

4. **Sum the Simple Interests:**
The total simple interest received is given as Rs. 1020. Therefore, we can write:
\[ SI_1 + SI_2 + SI_3 = 1020 \]
Substituting the values we calculated:
\[ 6R + 24R + 72R = 1020 \]
Combining like terms:
\[ 102R = 1020 \]

5. **Solve for R:**
To find the rate \( R \), divide both sides by 102:
\[ R = \frac{1020}{102} = 10 \]

6. **Conclusion:**
The rate of interest per annum is \( R = 10\% \).

### Final Answer:
The required rate of interest is **10% per annum**.