document.write( "Question 1057528: If P20,000 invested at interest compounded annually amounts to P21,218 in two years, what is the rate of interest? \n" ); document.write( "
Algebra.Com's Answer #672572 by math_helper(2461)\"\" \"About 
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We'll need the compound interest formula:\r
\n" ); document.write( "\n" ); document.write( " \"+F+=+P+%281%2Br%2Fn%29%5E%28nt%29+\"
\n" ); document.write( "—
\n" ); document.write( "Where:
\n" ); document.write( " F = future value (or final value)
\n" ); document.write( " P = present value
\n" ); document.write( " n = number of compounding periods per year
\n" ); document.write( " r = rate, expressed as a decimal (so 4% is 0.04)
\n" ); document.write( " t = number of years
\n" ); document.write( "—\r
\n" ); document.write( "\n" ); document.write( "We know:
\n" ); document.write( " P = 20000
\n" ); document.write( " F = 21218
\n" ); document.write( " t = 2
\n" ); document.write( " n = 1 (compounds once per year)
\n" ); document.write( "We need to solve for r\r
\n" ); document.write( "\n" ); document.write( "Plugging in these values gives:\r
\n" ); document.write( "\n" ); document.write( " \"+21218+=+20000%281%2Br%2F1%29%5E%281%2A2%29+\"
\n" ); document.write( " \"+21218%2F20000+=+%281%2Br%29%5E2+\"
\n" ); document.write( " \"+1.0609+=+%281%2Br%29%5E2+\"
\n" ); document.write( " \"+1.03+=+1%2Br+\" ( sqrt of both sides )
\n" ); document.write( " \"+r+=+0.03+\"
\n" ); document.write( "—
\n" ); document.write( "Ans: r = 3%
\n" ); document.write( "—\r
\n" ); document.write( "\n" ); document.write( "Check: \"+20000%281%2B0.03%29%5E2++=+21218+\"
\n" ); document.write( "
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