Answer :
To solve this problem, we need to determine how to split an [tex]$11,000 investment between two different interest rates (6% and 10%) so that the total interest earned in one year is $[/tex]820. Follow these steps:
1. Define Variables:
- Let's denote the amount invested at 6% as \( x \) dollars.
- The remaining amount, which is invested at 10%, will then be \( 11000 - x \) dollars.
2. Set Up the Interest Equations:
- The interest earned from the amount invested at 6% will be \( 0.06x \).
- The interest earned from the amount invested at 10% will be \( 0.10 (11000 - x) \).
3. Total Interest Equation:
- The total interest earned from both investments should equal $820. Therefore, we can set up the following equation:
[tex]\[ 0.06x + 0.10(11000 - x) = 820 \][/tex]
4. Solve the Equation:
- First, distribute the 0.10 in the second term:
[tex]\[ 0.06x + 0.10 \cdot 11000 - 0.10x = 820 \][/tex]
[tex]\[ 0.06x + 1100 - 0.10x = 820 \][/tex]
- Combine like terms:
[tex]\[ 0.06x - 0.10x + 1100 = 820 \][/tex]
[tex]\[ -0.04x + 1100 = 820 \][/tex]
- Isolate the variable term:
[tex]\[ -0.04x = 820 - 1100 \][/tex]
[tex]\[ -0.04x = -280 \][/tex]
- Divide by -0.04 to solve for \( x \):
[tex]\[ x = \frac{-280}{-0.04} \][/tex]
[tex]\[ x = 7000 \][/tex]
5. Determine the Remaining Investment:
- Since \( x = 7000 \), the amount invested at 6% is $7000.
- The amount invested at 10% is therefore:
[tex]\[ 11000 - x = 11000 - 7000 = 4000 \][/tex]
6. Conclusion:
- $7000 should be invested at 6%.
- $4000 should be invested at 10%.
Therefore, to earn a total interest of [tex]$820 in one year, $[/tex]11,000 should be invested with [tex]$7000 at 6% and $[/tex]4000 at 10%.
1. Define Variables:
- Let's denote the amount invested at 6% as \( x \) dollars.
- The remaining amount, which is invested at 10%, will then be \( 11000 - x \) dollars.
2. Set Up the Interest Equations:
- The interest earned from the amount invested at 6% will be \( 0.06x \).
- The interest earned from the amount invested at 10% will be \( 0.10 (11000 - x) \).
3. Total Interest Equation:
- The total interest earned from both investments should equal $820. Therefore, we can set up the following equation:
[tex]\[ 0.06x + 0.10(11000 - x) = 820 \][/tex]
4. Solve the Equation:
- First, distribute the 0.10 in the second term:
[tex]\[ 0.06x + 0.10 \cdot 11000 - 0.10x = 820 \][/tex]
[tex]\[ 0.06x + 1100 - 0.10x = 820 \][/tex]
- Combine like terms:
[tex]\[ 0.06x - 0.10x + 1100 = 820 \][/tex]
[tex]\[ -0.04x + 1100 = 820 \][/tex]
- Isolate the variable term:
[tex]\[ -0.04x = 820 - 1100 \][/tex]
[tex]\[ -0.04x = -280 \][/tex]
- Divide by -0.04 to solve for \( x \):
[tex]\[ x = \frac{-280}{-0.04} \][/tex]
[tex]\[ x = 7000 \][/tex]
5. Determine the Remaining Investment:
- Since \( x = 7000 \), the amount invested at 6% is $7000.
- The amount invested at 10% is therefore:
[tex]\[ 11000 - x = 11000 - 7000 = 4000 \][/tex]
6. Conclusion:
- $7000 should be invested at 6%.
- $4000 should be invested at 10%.
Therefore, to earn a total interest of [tex]$820 in one year, $[/tex]11,000 should be invested with [tex]$7000 at 6% and $[/tex]4000 at 10%.
