+0

# @CPhill Interest Question

0
406
2
+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?

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
+90023
+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
edited by CPhill  Jun 14, 2017