A money-lender, lends a part of his money at 10% per annum and the rest at 15% per annum. His annual income is ` 1900. However, if he had interchanged the rate of interest on the two sums, he would have earned 200 more. The amount lent will fetch what 15%?

Options

A. 6000
B. 4000
C. 10000
D. 4400

Answer

To solve the problem step by step, we will define the amounts lent at different interest rates and set up equations based on the information provided.

### Step 1: Define the Variables
Let:
- \( x \) = amount lent at 10%
- \( y \) = amount lent at 15%

### Step 2: Set Up the First Equation
According to the problem, the total income from these amounts is Rs. 1900. The income from each part can be calculated as follows:
- Income from \( x \) at 10% = \( 0.10x \)
- Income from \( y \) at 15% = \( 0.15y \)

Thus, we can write the first equation:
\[ 0.10x + 0.15y = 1900 \]

### Step 3: Set Up the Second Equation
If the interest rates are interchanged, the income would be Rs. 2100 (which is Rs. 200 more than Rs. 1900). The new income calculations would be:
- Income from \( x \) at 15% = \( 0.15x \)
- Income from \( y \) at 10% = \( 0.10y \)

This gives us the second equation:
\[ 0.15x + 0.10y = 2100 \]

### Step 4: Simplify the Equations
To eliminate decimals, we can multiply both equations by 100:
1. From the first equation:
\[ 10x + 15y = 190000 \quad \text{(Equation 1)} \]
2. From the second equation:
\[ 15x + 10y = 210000 \quad \text{(Equation 2)} \]

### Step 5: Solve the System of Equations
Now, we will solve these two equations simultaneously. We can start by multiplying Equation 1 by 3 and Equation 2 by 2 to align the coefficients of \( y \):
1. \( 30x + 45y = 570000 \) (Equation 3)
2. \( 30x + 20y = 420000 \) (Equation 4)

Next, we subtract Equation 4 from Equation 3:
\[ (30x + 45y) - (30x + 20y) = 570000 - 420000 \]
This simplifies to:
\[ 25y = 150000 \]

### Step 6: Solve for \( y \)
Now, divide both sides by 25:
\[ y = 6000 \]

### Step 7: Find \( x \)
Now that we have \( y \), we can substitute it back into Equation 1 to find \( x \):
\[ 10x + 15(6000) = 190000 \]
\[ 10x + 90000 = 190000 \]
\[ 10x = 190000 - 90000 \]
\[ 10x = 100000 \]
\[ x = 10000 \]

### Conclusion
The amount lent at 15% is \( y = 6000 \).