A person borrowed a certain sum of money at `16(2)/(3)`% per annum compound interest. He cleared the debt by paying Rs 20,825 at the end of 2 years. Find the sum borrowed.

Options

A. Rs 15,300
B. Rs 15,800
C. Rs 14,300
D. Rs 14,800

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the Given Information
- The rate of interest is \( 16 \frac{2}{3} \% \) per annum.
- The total amount paid after 2 years is Rs 20,825.
- We need to find the principal amount (the sum borrowed).

### Step 2: Convert the Rate of Interest
Convert \( 16 \frac{2}{3} \% \) into an improper fraction:
\[ 16 \frac{2}{3} = \frac{50}{3} \% \]

### Step 3: Use the Compound Interest Formula
The formula for compound interest is:
\[ A = P \left(1 + \frac{r}{100}\right)^t \]
Where:
- \( A \) = Amount after time \( t \)
- \( P \) = Principal amount (sum borrowed)
- \( r \) = Rate of interest
- \( t \) = Time in years

Substituting the known values:
\[ 20,825 = P \left(1 + \frac{50/3}{100}\right)^2 \]

### Step 4: Simplify the Equation
First, simplify \( \frac{50/3}{100} \):
\[ \frac{50/3}{100} = \frac{50}{300} = \frac{1}{6} \]
So the equation becomes:
\[ 20,825 = P \left(1 + \frac{1}{6}\right)^2 \]

### Step 5: Calculate \( \left(1 + \frac{1}{6}\right)^2 \)
Calculate \( 1 + \frac{1}{6} \):
\[ 1 + \frac{1}{6} = \frac{6}{6} + \frac{1}{6} = \frac{7}{6} \]
Now square it:
\[ \left(\frac{7}{6}\right)^2 = \frac{49}{36} \]

### Step 6: Substitute Back into the Equation
Now substitute back into the equation:
\[ 20,825 = P \cdot \frac{49}{36} \]

### Step 7: Solve for \( P \)
To find \( P \), multiply both sides by \( \frac{36}{49} \):
\[ P = 20,825 \cdot \frac{36}{49} \]

### Step 8: Calculate \( P \)
Now calculate \( P \):
\[ P = 20,825 \cdot \frac{36}{49} = \frac{20,825 \cdot 36}{49} \]

Calculating \( 20,825 \cdot 36 \):
\[ 20,825 \cdot 36 = 749,700 \]
Now divide by 49:
\[ P = \frac{749,700}{49} = 15,300 \]

### Conclusion
The sum borrowed (the principal amount) is Rs 15,300.

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