Question

paul and donna petrie invested $36,000, part at 6% simple interest and the rest at 4% simple interest for a period of 1 year. how much did they invest at each rate if their total annual interest from both investments was $1700? the amount invested at 6% is $. the amount invested at 4% is $

Explanation:

Step1: Let the amount invested at 6% be $x$.

Then the amount invested at 4% is $(36000 - x)$.

Step2: Calculate the interest from each investment.

The interest from the 6% - investment is $0.06x$ (since simple - interest formula is $I = Prt$ and $t = 1$ year). The interest from the 4% - investment is $0.04(36000 - x)$.

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

The total annual interest is $1700$, so $0.06x+0.04(36000 - x)=1700$.

Step4: Expand and simplify the equation.

$0.06x + 1440-0.04x=1700$. Combine like - terms: $(0.06x - 0.04x)+1440 = 1700$, which gives $0.02x+1440 = 1700$.

Step5: Solve for $x$.

Subtract 1440 from both sides: $0.02x=1700 - 1440=260$. Then divide both sides by 0.02: $x=\frac{260}{0.02}=13000$.

Step6: Find the amount invested at 4%.

The amount invested at 4% is $36000 - x=36000 - 13000 = 23000$.

Answer:

The amount invested at 6% is $13000.
The amount invested at 4% is $23000.