Impossible Interest question.

• Jan 25th 2013, 01:00 AM
skg94
Impossible Interest question.
At the end of each quarter year, aaron makes a 625 payment into a mutual fund that earns an annual percentage rate of 6%, compounded quarterly. The future value, of aaron's investment is = FV= R[(1+i)^n -1] / i - where n is the number of equal periodic payments of R dollars, and i is the interest rate per compounding period expressed as a decimal. After how long will aaron's investment be worth 1 million?

I did 625 [(1.015^4x)-1] / .015 - wasnt right, tried all variations of 1.06 and 4x, x 1/4x , didnt get it.

Answer s 54.25 years, i got close tho at 54.05 years. But nope, not good enough, if ex. it was a numerical response i would get it wrong still.

Anyone know how to do this, and see what i did wrong?
• Jan 25th 2013, 09:37 AM
Soroban
Re: Impossible Interest question.
Hello, skg94!

You must have made some elementary errors.

Quote:

At the end of each quarter-year, Aaron makes a \$625 payment into a mutual fund
that earns an annual percentage rate of 6%, compounded quarterly.
The future value, of aaron's investment is: . $FV \:=\: R\frac{(1+i)^n -1}{i}$
. . where $n$ = number of equal periodic payments of $R$ dollars,
. . and $i$ = interest rate per period.
After how long will aaron's investment be worth 1 million dollars?

We have: . $FV \,=\, 1,\!000,\!000,\;\;R \,=\,625,\;\; i \,=\, \tfrac{6\%}{4} \,=\, 0.015$

Substitute: . $625\,\frac{1.015^n - 1}{0.015} \:=\:1,\!000,\!000$

. . . . . . . . . $1.015^n - 1 \:=\:24 \quad\Rightarrow\quad 1.015^n \:=\:25$

Take logs: . $\ln(1.015^n) \:=\:\ln(25) \quad\Rightarrow\quad n\!\cdot\!\ln(1.015) \:=\:\ln(25)$

Hence: . $n \;=\;\frac{\ln(25)}{\ln(1.015)} \;=\; 216.1971659$

$\text{Therefore, it will take }217\text{ quarters }=\: 54\tfrac{1}{4}\text{ years.}$ .**

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** . $\text{If }n = 216\text{, he will have "only" }\996,946.64$