document.write( "Question 1186718: A man borrows 7000 at the rate of 6% per annum compound interest.If he pays back 1300 at the end of each year,how much is he owing at the end of the third year? \n" ); document.write( "

Algebra.Com's Answer #817762 by Theo(13342) $\"\"$ $\"About$

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prinipal is equal to 7000
\n" ); document.write( "interest rate = 6% per year compounded annually.
\n" ); document.write( "he pays off the loan with 1300 deposits at the end of each year.\r
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\n" ); document.write( "\n" ); document.write( "here is what's happening.
\n" ); document.write( "all values are rounded to the nearest penny.\r
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\n" ); document.write( "\n" ); document.write( "time period 0 is the start of the loan.
\n" ); document.write( "time period 1 is the end of the first year.
\n" ); document.write( "time period 2 is the end of the second year.
\n" ); document.write( "etc.\r
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\n" ); document.write( "\n" ); document.write( "time period 0.\r
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\n" ); document.write( "\n" ); document.write( "loan starts.
\n" ); document.write( "remaining balance is 7000\r
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\n" ); document.write( "\n" ); document.write( "time period 1.\r
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\n" ); document.write( "\n" ); document.write( "interest charged on 7000 = 6% of 7000 = 420
\n" ); document.write( "remaining balance becomes 7420
\n" ); document.write( "payment of 1300 is posted.
\n" ); document.write( "remaining balance becomes 7420 minus 1300 = 6120.\r
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\n" ); document.write( "\n" ); document.write( "time period 2.\r
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\n" ); document.write( "\n" ); document.write( "interest charged on 6120 = 6% of 6120 = 367.2
\n" ); document.write( "remaining balance becomes 6487.2
\n" ); document.write( "payment of 1300 is posted.
\n" ); document.write( "remaining balance becomes 6487.2 minus 1300 = 5187.20.\r
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\n" ); document.write( "\n" ); document.write( "time period 3.\r
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\n" ); document.write( "\n" ); document.write( "interest charged on 5187.2 = 6% of 5187.2 = 311.23.
\n" ); document.write( "remaining balance becomes 5498.43.
\n" ); document.write( "payment of 1300 is posted.
\n" ); document.write( "remaining balance becomes 5498.43. minus 1300 = 4198.43.\r
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\n" ); document.write( "\n" ); document.write( "your solution is that he still owed 4198.43. at the end of the third year.\r
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\n" ); document.write( "\n" ); document.write( "shown below is the loan carried to the end.
\n" ); document.write( "1300 is paid at the end of each month except for the last month.
\n" ); document.write( "in the last month, a payment of $913.42 is all that is required to close out the loan.\r
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document.write( "	  rembal	  int	       pmt\r\n" );
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document.write( "0	$7,000.00		     $0.00\r\n" );
document.write( "1	$6,120.00	$420.00	     $1,300.00\r\n" );
document.write( "2	$5,187.20	$367.20	     $1,300.00\r\n" );
document.write( "3	$4,198.43	$311.23	     $1,300.00\r\n" );
document.write( "4	$3,150.34	$251.91	     $1,300.00\r\n" );
document.write( "5	$2,039.36	$189.02	     $1,300.00\r\n" );
document.write( "6	$861.72	        $122.36	     $1,300.00\r\n" );
document.write( "7	$0.00	        $51.70	     $913.42\r\n" );
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