A person invested some amount at the rate of 12% simple interest and the remaining at 10%. He received yearly an interest ofX 130. Had he interchanged the amount invested, he would have received an interest ofX 134. How much money did he invest at different rates?

Options

A. ? 500 @ 10%, ? 800 @ 12%
B. ? 700 @ 10%, f 600 @ 12%
C. ? 800 @ 10%, ? 400 @ 12%
D. ? 700 @ 10%, ? 500 @ 12%

Answer

To solve the problem step by step, we will define the variables, set up the equations based on the information given, and then solve for the amounts invested at different rates.

### Step 1: Define the Variables
Let:
- \( X \) = amount invested at 12%
- \( Y \) = amount invested at 10%

### Step 2: Set Up the First Equation
From the information given, the total interest received from both investments is \( 130 \). The interest from the amount invested at 12% for one year is \( \frac{12}{100} \times X \), and the interest from the amount invested at 10% is \( \frac{10}{100} \times Y \). Therefore, we can write the first equation as:
\[ \frac{12}{100}X + \frac{10}{100}Y = 130 \]
Multiplying through by 100 to eliminate the fractions gives:
\[ 12X + 10Y = 13000 \quad \text{(Equation 1)} \]

### Step 3: Set Up the Second Equation
If the amounts invested were interchanged, the interest received would be \( 134 \). In this case, the interest from \( Y \) at 12% is \( \frac{12}{100}Y \) and the interest from \( X \) at 10% is \( \frac{10}{100}X \). Thus, we can write the second equation as:
\[ \frac{12}{100}Y + \frac{10}{100}X = 134 \]
Again, multiplying through by 100 gives:
\[ 12Y + 10X = 13400 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have a system of linear equations:
1. \( 12X + 10Y = 13000 \)
2. \( 10X + 12Y = 13400 \)

To eliminate one variable, we can multiply Equation 1 by 6 and Equation 2 by 5:
\[ 72X + 60Y = 78000 \quad \text{(Equation 3)} \]
\[ 50X + 60Y = 67000 \quad \text{(Equation 4)} \]

### Step 5: Subtract the Equations
Now, subtract Equation 4 from Equation 3:
\[ (72X + 60Y) - (50X + 60Y) = 78000 - 67000 \]
This simplifies to:
\[ 22X = 11000 \]
Dividing both sides by 22 gives:
\[ X = 500 \]

### Step 6: Substitute Back to Find Y
Now that we have \( X \), we can substitute it back into Equation 1 to find \( Y \):
\[ 12(500) + 10Y = 13000 \]
This simplifies to:
\[ 6000 + 10Y = 13000 \]
Subtracting 6000 from both sides gives:
\[ 10Y = 7000 \]
Dividing by 10 gives:
\[ Y = 700 \]

### Conclusion
The amounts invested are:
- Amount invested at 12%: \( \text{X} = 500 \)
- Amount invested at 10%: \( \text{Y} = 700 \)