SOLUTION: Marsha and Jan both invested money March 1, 2007. Marsha invested $9,000 at Bank A where the interest was compounded quarterly. Jan invested $6,000 at Bank B where he interest was

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Marsha and Jan both invested money March 1, 2007. Marsha invested $9,000 at Bank A where the interest was compounded quarterly. Jan invested $6,000 at Bank B where he interest was       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 907782: Marsha and Jan both invested money March 1, 2007. Marsha invested $9,000 at Bank A where the interest was compounded quarterly. Jan invested $6,000 at Bank B where he interest was compounded continuously. On March 1, 2010, Marsha had a balance of $10,678.42 while Jan had a balance of $6,943.92. What was the interest rate at each bank?
(Round to the nearest tenth of a percent)

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Marsha and Jan both invested money March 1, 2007.
Marsha invested $9,000 at Bank A where the interest was compounded quarterly. Jan invested $6,000 at Bank B where he interest was compounded continuously.

On March 1, 2010, Marsha had a balance of $10,678.42 while Jan had a balance of $6,943.92. What was the interest rate at each bank?
Marsha Info::
A(3) = 9000(1+(i/4))^(4*3) = 10,678.42
(1+(i/4))^(12) = 1.1865
(1+(i/4) = 1.0144
i/4 = 0.0144
interest = 0.0576
rate = 5.8%
-------------------------
B(3) = 6000*e^(3i) = 6943.92
e^(3i) = 1.1573
3i = ln(1.1573)
i = 0.0487
rate = 4.9%
-------------------
Cheers,
Stan H.
--------------
(Round to the nearest tenth of a percent)