Question

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

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

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

Answer 1

a) You should invest $6941.90 today.

b) The effective annual interest rate is 11%.

c) t is approximately 6.

Explanation:

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

`A = P(1 + (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 + (0.04)/(1))^(8)`

`P = (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 + (0.053)/(1))^(t)`

`(1 + 0.053)^(t) = (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.