document.write( "Question 1090115: Assuming an interest rate of 8% compounded annually. Answer the following question;
\n" ); document.write( "(a) how much money can be loaded now if $6,000 is to be repaid at the end of five years?
\n" ); document.write( "(b) how much money will be required in four years in order to repay a $15,000 loan borrowed now?
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Algebra.Com's Answer #704533 by EMStelley(208) $\"\"$ $\"About$

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The compound interest formula is:
\n" ); document.write( " $\"A+=+P%281+%2B+%28r%2Fn%29%29%5E%28nt%29\"$
\n" ); document.write( "Where A is the final amount, P is the principal amount, r is the rate in decimal form, n is the number of times per year the interest is compounded, and t is the number of years.
\n" ); document.write( "So in part a), we have r = 0.08, n = 1, t = 5, and we want A to be 6000 since that is the desired amount at the end of the 5 years. So we need to solve for P.
\n" ); document.write( " $\"6000+=+P%281+%2B+%280.08%2F1%29%29%5E%281%2A5%29\"$
\n" ); document.write( " $\"6000+=+P%281.08%29%5E5\"$
\n" ); document.write( " $\"P+=+4083.50\"$
\n" ); document.write( "Note that we round to two decimal places since this is money. So if you want $6,000 at the end of 5 years, you need to start with $4,083.50.
\n" ); document.write( "For part b), the $15,000 is our P, our initial amount. And t is 4. So we are solving for A.
\n" ); document.write( " $\"A+=+15000%281+%2B+%280.08%2F1%29%29%5E%281%2A4%29\"$
\n" ); document.write( " $\"A+=+15000%281.08%29%5E4\"$
\n" ); document.write( " $\"A+=+20407.33\"$
\n" ); document.write( "So at the end of 4 years, you will have to repay $20,407.33. \n" ); document.write( "

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