Find the compound interest on Rs 8100 for 1 year and 6 months at the rate `11 1/(9) %?`

Options

A. ₹ 2050
B. ₹3050
C. ₹ 1400
D. ₹2250

Answer

To find the compound interest on Rs 8100 for 1 year and 6 months at the rate of \(11 \frac{1}{9}\% \), we will follow these steps:

### Step 1: Convert the mixed fraction to an improper fraction
The rate \(11 \frac{1}{9}\%\) can be converted to an improper fraction:
\[ 11 \frac{1}{9} = \frac{11 \times 9 + 1}{9} = \frac{99 + 1}{9} = \frac{100}{9} \% \]

### Step 2: Calculate the rate for 1 year and 6 months
The total time is 1 year and 6 months, which is equivalent to 1.5 years. We will calculate the compound interest for 1 year and then for the additional 6 months.

### Step 3: Calculate the total amount after 1 year
Using the formula for compound interest:
\[ A = P \left(1 + \frac{r}{100}\right)^n \]
Where:
- \(A\) = total amount
- \(P\) = principal amount (Rs 8100)
- \(r\) = rate of interest (\(\frac{100}{9}\%\))
- \(n\) = number of years (1 year)

Substituting the values:
\[ A = 8100 \left(1 + \frac{100/9}{100}\right)^1 \]
\[ A = 8100 \left(1 + \frac{1}{9}\right) \]
\[ A = 8100 \left(\frac{10}{9}\right) \]
\[ A = 9000 \]

### Step 4: Calculate the total amount after the additional 6 months
Now we will calculate the amount for the additional 6 months using the amount we just calculated as the principal for the next 6 months:
\[ A = P \left(1 + \frac{r}{200}\right)^n \]
Where \(P = 9000\), \(r = \frac{100}{9}\%\), and \(n = 0.5\) years.

Substituting the values:
\[ A = 9000 \left(1 + \frac{100/9}{200}\right)^{0.5} \]
\[ = 9000 \left(1 + \frac{1}{18}\right) \]
\[ = 9000 \left(\frac{19}{18}\right) \]
\[ = 9500 \]

### Step 5: Calculate the compound interest
Now, we can find the compound interest by subtracting the principal from the total amount after 1 year and 6 months:
\[ \text{Compound Interest} = A - P \]
\[ = 9500 - 8100 \]
\[ = 1400 \]

### Final Answer
The compound interest on Rs 8100 for 1 year and 6 months at the rate of \(11 \frac{1}{9}\%\) is Rs 1400.

---