Question

How much would you invest today to have $9500 in 8 years if the effective annual rate of interest is 4%?

Answer 1

Answer:

a) You should invest $6941.90 today.

b) The effective annual interest rate is 11%.

c) t is approximately 6.

Step-by-step explanation:

These are compound interest problems. The compound interest formula is given by:

$A = P(1 + \frac{r}{n})^{nt}$

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

a) How much would you invest today to have $9500 in 8 years if the effective annual rate of interest is 4%?

Here, we want to find the value of P when $A = 9500, t = 8, n = 1, r = 0.04$.

$9500 = P(1 + \frac{0.04}{1})^{8}$

$P = \frac{9500}{1.3685}$

$P = 6941.90$.

You should invest $6941.90 today.

b) Suppose that an investment of $5750 accumulates to $11533.20 at the end of 13 years, then the effective annual interest rate is i= ?

Here, we have that $A = 11533.20, P = 5750, t = 13, n = 1$, and we want to find the value of i, that is r on the formula above the solutions.

$11533.20 = 5750(1 + r)^{13}$

$\sqrt[13]{11533.20} = \sqrt[13]{5750(1 + r)^{13}}$

$2.05 = 1.94(1 + r)$

$r = 0.11$

The effective annual interest rate is 11%.

c) At an effective annual rate of interest of 5.3%, the present value of $7425.70 due in t years is $3250. Determine t.

Here, we have that $A = 7425.70 + 3250 = 10675.7, P = 7425.7, r = 0.053, n = 1$ and we have to find t. So

$10675.7 = 7425.7(1 + \frac{0.053}{1})^{t}$

$(1 + 0.053)^{t} = \frac{10675.7}{7425.7}$

$(1.0553)^{t} = 1.4377$

We have that:

log_{a}a^{n} = n

So

$log_{1.0553} (1.0553)^{t} = log_{1.0553} 1.4377$

$t = 5.95$

t is approximately 6.