A person invested in all 2600 at 4%, 6% and 8% per annum simple interest. At the end of the year, he got the same interest in all the three cases. The money invested at 4% is:

Options

A. 200
B. 600
C. 800
D. 1200

Answer

To solve the problem, we need to determine how much money was invested at 4% interest when a total of $2600 was invested at three different rates (4%, 6%, and 8%) and the interest earned from each investment was the same at the end of the year.

### Step-by-Step Solution:

1. **Define Variables**:
Let:
- \( A \) = amount invested at 4%
- \( B \) = amount invested at 6%
- \( C \) = amount invested at 8%

We know that:
\[ A + B + C = 2600 \quad \text{(1)} \]

2. **Set Up Interest Equations**:
The interest earned from each investment can be expressed as:
- Interest from \( A \) at 4% = \( \frac{4}{100} \times A = 0.04A \)
- Interest from \( B \) at 6% = \( \frac{6}{100} \times B = 0.06B \)
- Interest from \( C \) at 8% = \( \frac{8}{100} \times C = 0.08C \)

Since the interest earned from all three investments is the same, we can set up the following equations:
\[ 0.04A = 0.06B \quad \text{(2)} \]
\[ 0.04A = 0.08C \quad \text{(3)} \]

3. **Express B and C in terms of A**:
From equation (2):
\[ B = \frac{0.04A}{0.06} = \frac{2}{3}A \quad \text{(4)} \]

From equation (3):
\[ C = \frac{0.04A}{0.08} = \frac{1}{2}A \quad \text{(5)} \]

4. **Substitute B and C into equation (1)**:
Substitute equations (4) and (5) into equation (1):
\[ A + \frac{2}{3}A + \frac{1}{2}A = 2600 \]

To combine the terms, we need a common denominator. The least common multiple of 3 and 2 is 6:
\[ A + \frac{4}{6}A + \frac{3}{6}A = 2600 \]
\[ A + \frac{7}{6}A = 2600 \]
\[ \frac{13}{6}A = 2600 \]

5. **Solve for A**:
Multiply both sides by \( \frac{6}{13} \):
\[ A = 2600 \times \frac{6}{13} \]
\[ A = 1200 \]

6. **Conclusion**:
The amount invested at 4% is \( A = 1200 \).

### Final Answer:
The money invested at 4% is **$1200**.