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?

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- Apr 20th 2013, 09:40 PMhighlife2323Interest 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 20th 2013, 11:06 PMBradynsRe: Interest Rate
IIRC

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

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,

$\displaystyle n= (5.25)(4)$

$\displaystyle n= 21$

$\displaystyle C = 20,000$

$\displaystyle P = 16,000$

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

Original Equation

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

Multiply out by P

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

Take the 21st root of both sides.

$\displaystyle \sqrt[21]{\frac{C}{P}} = 1+\frac{p}{21}$

multiply both sides by 21, then subtract 21

$\displaystyle p = 21\left ( \sqrt[21]{\frac{C}{P}} \right )-21 $

Plug in the values

$\displaystyle 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, 07:28 PMBradynsRe: Interest Rate
Sorry, I got the equation wrong.

This is the correct one:

$\displaystyle C = P\left ( 1 + \frac{p}{n} \right )^{nt}$

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..

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

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

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

p = 0.0427302 or 4.3%

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