# Interest Rate

• Apr 20th 2013, 10:40 PM
highlife2323
Interest Rate
Need help!

At what interest rate, to the nearest hundredth of a percent, will $16,000 grow to$20,000 if invested for 5.25 years and interest rate compounded quarterly?
• Apr 21st 2013, 12:06 AM
Re: Interest Rate
IIRC

$C = P(1+\frac{p}{n})^n$

Where,
C is the compounded amount
P is the principle (starting) amount
p is the rate (what you're after)
n is time.

Given that it is done quarterly for 5.25 years,
$n= (5.25)(4)$
$n= 21$

$C = 20,000$
$P = 16,000$

Let's do some algebra with our n value as 21.

Original Equation
$C = P(1+\frac{p}{21})^{(21)}$

Multiply out by P
$\frac{C}{P} = (1+\frac{p}{21})^{(21)}$

Take the 21st root of both sides.
$\sqrt[21]{\frac{C}{P}} = 1+\frac{p}{21}$

multiply both sides by 21, then subtract 21
$p = 21\left ( \sqrt[21]{\frac{C}{P}} \right )-21$

Plug in the values
$p = 21\left ( \sqrt[21]{\frac{20,000}{16,000}} \right )-21$

That should find you your rate.
x = 21((1.25)^(1/21)) -21 - Wolfram|Alpha

p = 22.43%
• Apr 21st 2013, 08:28 PM
Re: Interest Rate
Sorry, I got the equation wrong.

This is the correct one:
$C = P\left ( 1 + \frac{p}{n} \right )^{nt}$

C is the compounded amount
P is the principle (starting) amount
p is the rate (what you're after)
n is times compounded annually.
t is the number of years.

Doing the same sort of stuff..

$(20,000) = (16,000)\left ( 1 + \frac{p}{4} \right )^{21}$

$\sqrt[21]{\frac{20,000}{16,000}} = 1 + \frac{p}{4}$

$p = (4)\sqrt[21]{\frac{20,000}{16,000}}-(4)$

p = 0.0427302 or 4.3%

x = 4((1.25)^(1/21)) -4 - Wolfram|Alpha