# compound interest

• Dec 22nd 2009, 08:30 AM
math321
compound interest
Kevin wishes to deposit $3,000 for a period of two years for a special purpose. After visiting several banks and investigating his options, he has selected three plans: A. 6.5% interest per year compounded daily B. 6.55% interest per year compounded monthly C. 6.6% interest per year compounded quarterly (a) Which plan will earn Kevin the most money? (b) How much will he lose if he chooses the worst plan? • Dec 22nd 2009, 08:49 AM masters Quote: Originally Posted by math321 Kevin wishes to deposit$3,000 for a period of two years for a special purpose. After
visiting several banks and investigating his options, he has selected three plans:
A. 6.5% interest per year compounded daily
B. 6.55% interest per year compounded monthly
C. 6.6% interest per year compounded quarterly
(a) Which plan will earn Kevin the most money?
(b) How much will he lose if he chooses the worst plan?

Hi math321,

Put it in the compound interest formula and see.

$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$

P = principal amount (the initial amount you deposit)
r = annual rate of interest (as a decimal)
t = number of years the amount is deposited.
A = amount of money accumulated after n years, including interest.
n = number of times the interest is compounded per year
• Dec 22nd 2009, 08:51 AM
math321
tanks i solved it

but im working on another one

11. Jim invests $22,000 of his retirement savings in CD that pays 6.85% annual interest compounded monthly. When will the CD be worth$35,000? Hint: Use logarithms.
• Dec 22nd 2009, 09:27 AM
masters
Quote:

Originally Posted by math321
tanks i solved it

but im working on another one

11. Jim invests $22,000 of his retirement savings in CD that pays 6.85% annual interest compounded monthly. When will the CD be worth$35,000? Hint: Use logarithms.

Same formula:

$\displaystyle A=P\left(1+\frac{r}{n}\right)^{nt}$

$\displaystyle 35000=22000\left(1+\frac{.0685}{12}\right)^{12t}$

Solve for t.

$\displaystyle \frac{35}{22}=\left(\frac{120685}{120000}\right)^{ 12t}$

$\displaystyle \log\frac{35}{22}=12t \log \frac{120685}{120000}$

$\displaystyle 12t=\frac{\log\frac{35}{22}}{\log\frac{120685}{120 000}}$

$\displaystyle t=\frac{\log\frac{35}{22}}{\log\frac{120685}{12000 0}} \div 12$