8. earning interest isabel invested $6000, part at 9% per annum and the remainder at 4% per annum. after one…

8. earning interest isabel invested $6000, part at 9% per annum and the remainder at 4% per annum. after one year, the total interest from these investments was $420. how much did she invest at each rate?

8. earning interest isabel invested $6000, part at 9% per annum and the remainder at 4% per annum. after one year, the total interest from these investments was $420. how much did she invest at each rate?

Answer

Explanation:

Step1: Let the amount invested at 9% be $x$.

Then the amount invested at 4% is $(6000 - x)$.

Step2: Calculate the interest from each investment.

The interest from the 9% - investment after one - year is $0.09x$. The interest from the 4% - investment after one - year is $0.04(6000 - x)$.

Step3: Set up the equation based on the total interest.

The total interest is $420$, so $0.09x+0.04(6000 - x)=420$.

Step4: Expand and simplify the equation.

$0.09x + 240-0.04x=420$. Combine like - terms: $0.09x-0.04x=420 - 240$. So, $0.05x=180$.

Step5: Solve for $x$.

Divide both sides of the equation by 0.05: $x=\frac{180}{0.05}=3600$.

Step6: Find the amount invested at 4%.

The amount invested at 4% is $6000 - x=6000 - 3600 = 2400$.

Answer:

She invested $3600$ at 9% and $2400$ at 4%.