can anyone help me with a formula for this?

What interest rate is required to earn $250 interest on a $999 investment over 3 years compounding daily

Thank You

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- Nov 25th 2006, 08:34 PMxxpink__saltxxcompound interest? help please
can anyone help me with a formula for this?

What interest rate is required to earn $250 interest on a $999 investment over 3 years compounding daily

Thank You - Nov 25th 2006, 08:47 PMQuick
$\displaystyle 999=250\times i^x$ where $\displaystyle i$ is interest (plus 1) and $\displaystyle x$ is the number of time periods.

So how many days are in 3 years, let's assume there's no leap-year. The answer is 1095 days

So: $\displaystyle 999=250\times i^{1095}$

Thus: $\displaystyle \frac{999}{250}=i^{1095}$

Then: $\displaystyle \sqrt[1095]{\frac{999}{250}}=i\approx 1.00126591$

So interest is: $\displaystyle \approx .126591\%$ - Nov 26th 2006, 05:01 AMSoroban
Hello, xxpink__saltxx!

Quote:

What interest rate is required to earn $250 interest on a $999 investment

over 3 years compounding daily?

The compound interest formula is: .$\displaystyle A \;= \;P(1 + i)^n$

where: $\displaystyle A$ = final amount, $\displaystyle P$ = principal, $\displaystyle i$ = periodic interest rate, $\displaystyle n$ = number of periods.

We are given: .$\displaystyle A = 999,\;P = 250$

Since the interest is compounded daily, the interest rate is: $\displaystyle \frac{I}{365}$

. . and the number of periods is: $\displaystyle 3 \times 365 \,=\,1095$

So we have: .$\displaystyle 999 \;=\;250\left(1 + \frac{I}{365}\right)^{1095}$

. . and we must solve for $\displaystyle I$, the__annual__interest rate.

We have: .$\displaystyle \left(1 + \frac{I}{365}\right)^{1095} \:=\:\frac{999}{250} \:=\:3.996$

Raise both sides to the $\displaystyle \frac{1}{1095}$ power:

. . $\displaystyle \left[\left(1 + \frac{I}{365}\right)^{1095}\right]^{\frac{1}{1095}} \;=\;(3.996)^{\frac{1}{1095}} \quad\Rightarrow\quad 1 + \frac{I}{365}\:=\:1.001265909$

. . Then: .$\displaystyle \frac{I}{365}\:=\:0.001265909\quad\Rightarrow\quad I \:=\;0.462056836$

Therefore, the annual interest rate is about $\displaystyle 46.2\%.$

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Check

$\displaystyle A \;= \;\$250\left(1 + \frac{0.462}{365}\right)^{1095} \:=\:\$998.8298956 \:\approx\:\$999$*. . . Yes!*

- Nov 26th 2006, 09:11 AMQuick
- Nov 30th 2006, 12:11 AMardnas
Soroban was right. The solutions posted were correct.

Plus, the question does not ask for the "daily" interest rate. It only asks for the "interest rate", and in reality, all banks state annual interest rates.