+956 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.
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.
+129852
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
