Mr. Thomas invested an amount of Rs 13,900 divided in two different schemes A and B at the simple interest rate of 14% per annum and 11% per annum respectively. If the total amount of simple interest earned in 2 years be Rs 3508. What was the amount invested in scheme B?

Options

A. 7500
B. 6400
C. 6500
D. 7000

Answer

To solve the problem, we need to find out how much Mr. Thomas invested in scheme B. We know the total investment, the interest rates for both schemes, and the total interest earned over 2 years.

Let's denote:
- The amount invested in scheme A as \( x \)
- The amount invested in scheme B as \( y \)

From the problem, we have the following equations:

1. The total investment equation:
\[ x + y = 13900 \quad \text{(1)} \]

2. The total interest earned from both schemes after 2 years:
- The interest from scheme A at 14% for 2 years:
\[ \text{Interest from A} = \frac{x \times 14 \times 2}{100} = \frac{28x}{100} = 0.28x \]
- The interest from scheme B at 11% for 2 years:
\[ \text{Interest from B} = \frac{y \times 11 \times 2}{100} = \frac{22y}{100} = 0.22y \]
- The total interest from both schemes is given as Rs 3508:
\[ 0.28x + 0.22y = 3508 \quad \text{(2)} \]

Now, we can solve these equations step by step.

### Step 1: Solve Equation (1) for \( y \)
From equation (1):
\[ y = 13900 - x \quad \text{(3)} \]

### Step 2: Substitute Equation (3) into Equation (2)
Substituting \( y \) from equation (3) into equation (2):
\[ 0.28x + 0.22(13900 - x) = 3508 \]

### Step 3: Simplify the Equation
Expanding the equation:
\[ 0.28x + 0.22 \times 13900 - 0.22x = 3508 \]
Calculating \( 0.22 \times 13900 \):
\[ 0.22 \times 13900 = 3058 \]
So the equation becomes:
\[ 0.28x - 0.22x + 3058 = 3508 \]
Combining like terms:
\[ 0.06x + 3058 = 3508 \]

### Step 4: Isolate \( x \)
Subtract 3058 from both sides:
\[ 0.06x = 3508 - 3058 \]
\[ 0.06x = 450 \]

### Step 5: Solve for \( x \)
Dividing both sides by 0.06:
\[ x = \frac{450}{0.06} = 7500 \]

### Step 6: Find \( y \)
Now, substitute \( x \) back into equation (3) to find \( y \):
\[ y = 13900 - 7500 = 6400 \]

### Conclusion
The amount invested in scheme B is:
\[ \boxed{6400} \]