• Nov 25th 2006, 08:34 PM
xxpink__saltxx
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 PM
Quick
Quote:

Originally Posted by xxpink__saltxx
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

$\displaystyle 999=250\times i^x$ 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 AM
Soroban
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 AM Quick Quote: Originally Posted by Soroban Therefore, the annual interest rate is about$\displaystyle 46.2\%.\$

Yes, but the question asks for daily interest rate :eek:
• Nov 30th 2006, 12:11 AM
ardnas
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.