Question

Two investments totaling $49,500

Answer 1

Using simple interest, it is found that:

• $10,500 was invested at 2%.

• $39,000 was invested at 4%.

Simple Interest

Simple interest is used when there is a single compounding per time period.

The interest after t years in is modeled by:

$I(t) = Prt$

In which:

• P is the initial amount.

• r is the interest rate, as a decimal.

In this problem, considering that there are two investments:

• A total of $49,500 was invested, hence $P_1 + P_2 = 49500$.

• One investment yields 2% and the other yields 4%, hence the rates are $r_1 = 0.02, r_2 = 0.04$.

• The annual income is of $1770, hence $t = 1, I_1 + I_2 = 1770$.

Considering the rates, we have that:

$I_1 = 0.02P_1$

$I_2 = 0.04P_2$

Considering that $I_1 + I_2 = 1770 \rightarrow I_2 = 1770 - I_1$:

$I_1 = 0.02P_1$

$1770 - I_1 = 0.04P_2$

Adding the equations:

$0.02P_1 + 0.04P_2 = 1770$

Considering that $P_1 + P_2 = 49500 \rightarrow P_2 = 49500 - P_1$, we have that:

$0.02P_1 + 0.04P_2 = 1770$

$0.02P_1 + 0.04(49500 - P_1) = 1770$

$0.02P_1 = 210$

$P_1 = \frac{210}{0.02}$

$P_1 = 10500$

$P_2 = 49500 - P-1 = 49500 - 10500 = 39000$

Hence:

• $10,500 was invested at 2%.

• $39,000 was invested at 4%.

You can learn more about simple interest at brainly.com/question/25296782