Thread: Compound interest : Present value

Sally invests some money at 6%/a compounded annually. After 5 years, she takes the prinicpal interest and reinvests it all at 7.2%/a compounded quarterly for 6 more years.At the end of this time, her investment is worth $14784.56. How much did sally originally invest ? 2. Re: Compound interest : Present value Hello, darkangel06! I must assume that you know the Compound Interest Formula. . .$\displaystyle A \;=\:P(1 + i)^n\;\;\text{ where: }\:\begin{Bmatrix}P &=& \text{principal invested} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods}\end{Bmatrix}$Sally invests some money at 6%/a compounded annually. After 5 years, she takes the prinicpal and interest and reinvests it all . . at 7.2%/a compounded quarterly for 6 more years. At the end of this time, her investment is worth$14784.56.
How much did Sally originally invest?

Let $\displaystyle P$ = her original investment.

She earns $\displaystyle i = 6\% = 0.06$ annual interest for $\displaystyle n = 5$ years.

At the end of five years, she will have:.$\displaystyle P(1.06)^5$ dollars.

She invests this amount at. $\displaystyle \frac{7.2\%}{4} = 1.8\% = 0.018$ quarterly for 24 quarters.

At the end of the six years, she will have:.$\displaystyle P(1.06)^5\cdot(1.018)^{24}$ dollars,
. . which, we are told, will equal $\displaystyle \$14,\!784.56.$There is our equation!. . . . .$\displaystyle P(1.06)^5(1.018)^{24} \;=\;14,\!784.56\displaystyle \text{Solve for }P\!:\;\;P \;=\;\frac{14,784.56}{(1.06)^5(1.018)^{24})} \;=\; 7199.9989497$Therefore, Sally originally invested$\displaystyle \$7,\!200.$

3. Re: Compound interest : Present value

ohh okay thank you I get it now

4. Re: Compound interest : Present value

And how do we do this :
Tia is investing $2500 that she would like to grow to$6000 in 10 years. At what annual interest rate, compounded quarterly, must Tia invest her money?
So i know that A=$6000 and Present value=$2500 and n=40 but how do i get the interest rate ?

5. Re: Compound interest : Present value

Hi,
As Soroban points out, $\displaystyle A=P(1+i/4)^n$, where i is the annual interest rate and n = 40.
So you need to solve $\displaystyle 6000=2500(1+i/4)^{40}$ for i.