Stacy invests some money in an account that earns a fixed rate of interest compounded annually. The amounts of the investment at the end of the first four years is shown in the table.

Year 1: Total Amount $2251.05

Year 2: Total Amount $2356.85

Year 3: Total Amount $2467.62

Year 4: Total Amount $2583.60

a) Determine the annual interest rate.

b) How much did Stacy invest?

c) How long will it take for Stacy’s investment to double?

Explain all your answers.

Julius
Jun 14, 2017

#1**+2 **

Year 4 / Year 1 =2,583.60/2,251.05 =1.147731...

1.147731^(1/3) =4.7% this the yearly compound rate.

2,251.05 / 1.047=$2,150 - the amount that Stacy invested.

2,150 x 1.047^n =4,300.

n = 15.09 years to double the money.

Guest Jun 14, 2017

#2**+2 **

Mmmmm....this one looks interesting !!!

We can let the amount after year 1 be represented as

P ( 1 + r) = 2251.05 (1)

Where P is the orginal amount invested and r is the interest rate

After year 2 we have

P (1 + r)^2 = 2356.85 (2)

Dividing (2) by (1) we have that

(1 + r) = 2356.85 / 2251.05 subtract 1 from both sides

r = 2356.85 / 2251.05 - 1 ≈ .047 ≈ 4.7 %

We can find the original amount invested as follows

P ( 1 + .047) = 2251.05 divide both sides by (1 + .047)

P = 2251.05 / ( 1 + .047 = $2150

We can find the time to double - the amount will be 2P - with this :

log [ (2P / P) ] / log ( 1 + interest rate )

log (2) / log (1+. 047) ≈ 15.09 years

CPhill
Jun 14, 2017