Question

A business has a savings account that earns a 3% annual interest rate. At the end of 1996, the business had $4,000 in the account. The formula
F = p (1+r/100)
t is used to determine the amount in the savings account.
F is the final amount,
p is the initial investment amount,
r is the annual interest rate, and
t is the time in years.
To the nearest dollar, how much did the business initially invest in 1991?
A. $3,450

B. $3,455

C. $4,631

D. $4,637

Answer 1

Answer:

The amount the business initially invested in 1991 is approximately;

A. $3,450

Step-by-step explanation:

The question relates to investment and finance

The parameters given are;

The annual interest rate on the business savings account = 3%

The amount the business had in the account at the end of 1996 = $4,000

The given formula used to determine the final amount in the savings account, $F = p \cdot \left(1 + \dfrac{r}{100} \right )^t$

Where;

F = The final amount

p = The initial (principal) investment

r = The annual interest rate

t = The time in years

Therefore, we have;

The required information = The initial amount, 'p', the business invested in 1991

The time in years, 't', from 1991 to 1996 = 1996 - 1991 = 5

∴ The time in years from 1991 to 1996, t = 5 years

The final amount in 1996, F = $4,000

r = 3%

Plugging in the values in the given equation gives;

$4000 = p\cdot \left(1 + \dfrac{3}{100} \right )^5$

Therefore;

$p = \dfrac{4000}{\left(1 + \dfrac{3}{100} \right )^5} = 3,450.43513754$

To the nearest dollar, the initial amount the business invested in 1991, p ≈ $3,450.