At what rate will the sum of 1200 amount to 2348.32 in 2yrs? if compounded quarterly

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Answer 1 · 2018-05-11T21:55:38

$35.018 %$ $color(blue)(35.018%)$

The formula for compound interest is given by:

$F V = P V {(1 + \frac{r}{n})}^{n t}$ $FV=PV(1+r/n)^(nt)$

Where:

$F V = future value$ $FV= "future value"$ .

$P V = present value$ $PV="present value"$

$r = interest rate given as a decimal$ $r="interest rate given as a decimal"$

$n = compounding period$ $n="compounding period"$

$t = time in years$ $t="time in years"$ .

From given information, we have:

$t = 2$ $t=2$

$n = quarterly = 4$ $n="quarterly"=4$

$P V = 1200$ $PV=1200$

$F V = 2348.32$ $FV=2348.32$

We are looking to find interest rate $r$ $r$ :

$1200 {(1 + \frac{r}{4})}^{8} = 2348.32$ $1200(1+r/4)^8=2348.32$

${(1 + \frac{r}{4})}^{8} = \frac{2348.32}{1200}$ $(1+r/4)^8=2348.32/1200$

$\frac{{(4 + r)}^{8}}{4^{8}} = \frac{2348.32}{1200}$ $(4+r)^8/4^8=2348.32/1200$

${(4 + r)}^{8} = 4^{8} (\frac{2348.32}{1200})$ $(4+r)^8=4^8(2348.32/1200)$

${(4 + r)}^{8} = 128249.5829$ $(4+r)^8=128249.5829$

$r = \sqrt[8]{128249.5829} - 4 = 0.350177674$ $r=root(8)(128249.5829)-4=0.350177674$

This is in decimal form, so:

$r = 35.018 %$ $r=35.018%$