Question

aleksandra tomich invested $11,532, part at 9% simple interest and part at 3% simple interest for a period of 1 year. how much did she invest at each rate if each account earned the same interest? aleksandra invested $ at 9% and $ at 3%.

Explanation:

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

Then the amount invested at 3% is $11532 - x$.

Step2: Set up the interest - equal equation.

The simple - interest formula is $I=Prt$. Since $t = 1$ year, the interest from the 9% investment is $I_1=0.09x$, and the interest from the 3% investment is $I_2 = 0.03(11532 - x)$. Set $I_1=I_2$, so $0.09x=0.03(11532 - x)$.

Step3: Expand the right - hand side.

$0.09x=0.03\times11532-0.03x$.
$0.09x = 345.96-0.03x$.

Step4: Add $0.03x$ to both sides.

$0.09x + 0.03x=345.96$.
$0.12x=345.96$.

Step5: Solve for $x$.

$x=\frac{345.96}{0.12}=2883$.

Step6: Find the amount invested at 3%.

The amount invested at 3% is $11532 - 2883=8649$.

Answer:

Aleksandra invested $2883$ at 9% and $8649$ at 3%.