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Two investments totaling $49,500
$
⁢
49,500
produce an annual income of $1770
$
⁢
1770
. One investment yields 2%
2
%
per year, while the other yields 4%
4
%
per year. How much is invested at each rate?

Sagot :

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:

[tex]I(t) = Prt[/tex]

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 [tex]P_1 + P_2 = 49500[/tex].
  • One investment yields 2% and the other yields 4%, hence the rates are [tex]r_1 = 0.02, r_2 = 0.04[/tex].
  • The annual income is of $1770, hence [tex]t = 1, I_1 + I_2 = 1770[/tex].

Considering the rates, we have that:

[tex]I_1 = 0.02P_1[/tex]

[tex]I_2 = 0.04P_2[/tex]

Considering that [tex]I_1 + I_2 = 1770 \rightarrow I_2 = 1770 - I_1[/tex]:

[tex]I_1 = 0.02P_1[/tex]

[tex]1770 - I_1 = 0.04P_2[/tex]

Adding the equations:

[tex]0.02P_1 + 0.04P_2 = 1770[/tex]

Considering that [tex]P_1 + P_2 = 49500 \rightarrow P_2 = 49500 - P_1[/tex], we have that:

[tex]0.02P_1 + 0.04P_2 = 1770[/tex]

[tex]0.02P_1 + 0.04(49500 - P_1) = 1770[/tex]

[tex]0.02P_1 = 210[/tex]

[tex]P_1 = \frac{210}{0.02}[/tex]

[tex]P_1 = 10500[/tex]

[tex]P_2 = 49500 - P-1 = 49500 - 10500 = 39000[/tex]

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

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