Nitesh borrows a sum a Rs. 1200 at the beginning of a year. After 4 months, Rs. 1800 more is borrowed at a rate of interest double the previous one. At the end of a year, the sum of interest on both the loans is Rs. 216. What is the first rate of interest per annum?

Options

A. `9%`
B. `6%`
C. `8%`
D. `12%`

Answer

To solve the problem, we need to calculate the interest on both loans and then find the rate of interest for the first loan.

### Step 1: Identify the loans and their durations
- The first loan is Rs. 1200, borrowed for the entire year (12 months).
- The second loan is Rs. 1800, borrowed after 4 months, which means it will be borrowed for 8 months (12 - 4 = 8).

### Step 2: Let the rate of interest for the first loan be \( r \% \) per annum
- The interest for the first loan can be calculated using the formula:
\[ \text{Interest} = \frac{P \times R \times T}{100} \]
where \( P \) is the principal amount, \( R \) is the rate of interest, and \( T \) is the time in years.

- For the first loan:
\[ \text{Interest}_1 = \frac{1200 \times r \times 1}{100} = 12r \]

### Step 3: Calculate the interest for the second loan
- The second loan has a rate of interest that is double the first loan, which is \( 2r \% \).
- The interest for the second loan is calculated for 8 months (which is \( \frac{8}{12} = \frac{2}{3} \) years):
\[ \text{Interest}_2 = \frac{1800 \times 2r \times \frac{2}{3}}{100} = \frac{1800 \times 2r \times 2}{300} = \frac{7200r}{300} = 24r \]

### Step 4: Set up the equation for total interest
- According to the problem, the total interest from both loans is Rs. 216:
\[ \text{Interest}_1 + \text{Interest}_2 = 216 \]
Substituting the values we calculated:
\[ 12r + 24r = 216 \]
\[ 36r = 216 \]

### Step 5: Solve for \( r \)
- Dividing both sides by 36:
\[ r = \frac{216}{36} = 6 \]

### Conclusion
- The first rate of interest per annum is **6%**.