Question

You have $4000 to invest. If you'd like to earn exactly $638
in interest each year, how much should you invest in each of
two accounts, one that earns 8% per year and one that
earns 21% per year. (Round your answer to the nearest
dollar)
8%: $
21%: $

Answer 1

Answer:

8%: $1554

21%: $2446

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

$E = P*I*t$

In which E are the earnings, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

$T = E + P$.

In this question:

Two earnings, that i am going to call A(8% per year) and B(21% per year).

Two principals, for A i am going to call P and for B it is the rest, so 4000 - P.

A:

One year, so $t = 1$

8% interest, so $r = 0.08$

Earnings A.

$A = P*I*t$

$A = 0.08P$

B:

21% interest, so $r = 0.21$

Principal (4000 - P).

$B = P*I*t$

$B = 0.21*(4000 - P)$

You'd like to earn exactly $638 in interest each year.

This means that $A + B = 638$

Then

$B = 638 - A$

Now we have to solve the following system:

$A = 0.08P$

$638 - A = 0.21*(4000 - P)$

So, on the second equation:

$A = 638 - 0.21*(4000 - P)$

Replacing on the first:

$A = 0.08P$

$638 - 0.21*(4000 - P) = 0.08P$

$638 - 840 + 0.21P = 0.08P$

$0.21P - 0.08P = 840 - 638$

$0.13P = 202$

$P = \frac{202}{0.13}$

$P = 1553.8$

Rounding up to the nearest integer

P = 1554.

So on A, the 8% interest, you invest $1554.

On B, the 21% interest, you invest 4000 - 1554 = $2446.