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present value

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\\var{period[2]}\\left(\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1\\right)

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Suppose $\\var{years}$ years ago you deposited $\\$\\var{P}$ into a bank account and today the balance accumulated is $\\$\\var{S}$. If the interest was compounded {period[0]} what was the rate of compound interest per annum?

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[[0]] $\\%$ (to two decimal places)

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If you are unsure of how to do a question, click on Reveal answers to see the full working. Then, once you understand how to do the question, click on Try another question like this one to start again. Do each question repeatedly to ensure you have mastered it.

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We are asked to find the interest rate per annum using compound interest. Therefore we will use the compound interest equation

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$S=P(1+i)^n$,

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where $S$ is the future value, $P$ is the present value, $i$ is the interest rate per time period and $n$ is the number of time periods.

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In our situation we have,

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$S=\\var{S}$

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$P=\\var{P}$,

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$n=\\var{years}\\times\\var{period[1]}=\\var{n}$, 

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and therefore we have

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$\\var{S}=\\var{P}\\left(1+r\\right)^\\var{n}$,

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which we need to rearrange to solve for $i$.

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We want to get $i$ by itself. We start by dividing both sides by $\\var{P}$ (to remove the multiplication by $\\var{P}$)

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$\\displaystyle \\frac{\\var{S}}{\\var{P}}=\\left(1+i\\right)^\\var{n}$.

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Next we remove the power on the right hand side by raising both sides to the power of $\\frac{1}{\\var{n}}$ (this is equivalent to applying $\\sqrt[\\var{n}]{\\phantom{XX}}$ to both sides)

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}=\\left(\\left(1+i\\right)^\\var{n}\\right)^{\\frac{1}{\\var{n}}}$.

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Which simplifies (by an index law) to 

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}=1+i$.

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Next to get $i$ by itself we subtract $1$ from both sides to get 

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1=i$.

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Recall that $i$ is the interest rate per time period but we were asked for the interest rate per annum so we multiply by $\\var{period[1]}$ to get 

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$\\begin{array}\\displaystyle \\text{interest rate pa}&=\\var{period[1]}\\left(\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1\\right)\\\\&\\approx\\var{ipadec}\\\\&=\\var{iparounded}\\%\\quad \\text{(2 decimal places)}\\end{array}$

", "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}]}], "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}]}