If compound interest on some amount is 4050 for 1st year and 4723.92 for 3rd year then find the rate of interest?

Options

A. `11%`
B. `9%`
C. `10%`
D. `8%`

Answer

To solve the problem, we need to find the rate of interest (r) given the compound interest for the first year and the third year. Let's break it down step by step.

### Step 1: Define the Variables
Let the principal amount be \( P \) and the rate of interest be \( r \% \).

### Step 2: Write the Equations for Compound Interest
1. For the first year, the compound interest is given as \( 4050 \):
\[ \text{CI}_1 = P \times \left(1 + \frac{r}{100}\right) - P = 4050 \]
This simplifies to:
\[ P \times \frac{r}{100} = 4050 \quad \text{(Equation 1)} \]

2. For the third year, the compound interest is given as \( 4723.92 \):
\[ \text{CI}_3 = P \times \left(1 + \frac{r}{100}\right)^3 - P \times \left(1 + \frac{r}{100}\right)^2 \]
This simplifies to:
\[ P \times \left(1 + \frac{r}{100}\right)^3 - P \times \left(1 + \frac{r}{100}\right)^2 = 4723.92 \]
Rearranging gives:
\[ P \times \left(1 + \frac{r}{100}\right)^2 \times \left(1 + \frac{r}{100} - 1\right) = 4723.92 \]
Which further simplifies to:
\[ P \times \left(1 + \frac{r}{100}\right)^2 \times \frac{r}{100} = 4723.92 \quad \text{(Equation 2)} \]

### Step 3: Divide Equation 2 by Equation 1
Now, we can divide Equation 2 by Equation 1 to eliminate \( P \):
\[ \frac{P \times \left(1 + \frac{r}{100}\right)^2 \times \frac{r}{100}}{P \times \frac{r}{100}} = \frac{4723.92}{4050} \]
This simplifies to:
\[ \frac{\left(1 + \frac{r}{100}\right)^2}{1} = \frac{4723.92}{4050} \]

### Step 4: Calculate the Right Side
Calculating the right side:
\[ \frac{4723.92}{4050} \approx 1.165 \]

### Step 5: Solve for \( r \)
Now we have:
\[ \left(1 + \frac{r}{100}\right)^2 = 1.165 \]
Taking the square root of both sides:
\[ 1 + \frac{r}{100} = \sqrt{1.165} \approx 1.080 \]
Subtracting 1 from both sides:
\[ \frac{r}{100} \approx 0.080 \]
Multiplying by 100 gives:
\[ r \approx 8 \]

### Final Answer
The rate of interest \( r \) is \( 8\% \).

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