document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1198536>Question 1198536</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What sum of money would Max need to invest to accumulate a total of $100000 at the end of 7 years at 8% p.a. interest compounding biannually? Answer to the nearest cent. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #832154 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=832154\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Compounding interest formula<BR>\n" );
document.write( "A = P*(1+r/n)^(n*t)\r<BR>\n" );
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document.write( "A = final amount after t years<BR>\n" );
document.write( "P = deposit, aka starting amount<BR>\n" );
document.write( "r = annual interest rate in decimal form<BR>\n" );
document.write( "n = compounding frequency<BR>\n" );
document.write( "t = number of years\r<BR>\n" );
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document.write( "biannual = once every six months (aka twice a year)<BR>\n" );
document.write( "So we have n = 2\r<BR>\n" );
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document.write( "The goal is to have A = 100000 dollars at the end of t = 7 years, where the money is compounded n = 2 times a year. <BR>\n" );
document.write( "The interest rate in decimal form is r = 0.08<BR>\n" );
document.write( "Your teacher asks you to find the deposit P.\r<BR>\n" );
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document.write( "Summary of values<BR>\n" );
document.write( "A = 100000 <BR>\n" );
document.write( "P = unknown<BR>\n" );
document.write( "r = 0.08<BR>\n" );
document.write( "n = 2<BR>\n" );
document.write( "t = 7\r<BR>\n" );
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document.write( "Let's solve for P.<BR>\n" );
document.write( "A = P*(1+r/n)^(n*t)<BR>\n" );
document.write( "100000 = P*(1+0.08/2)^(2*7)<BR>\n" );
document.write( "100000 = P*(1+0.04)^(14)<BR>\n" );
document.write( "100000 = P*(1.04)^(14)<BR>\n" );
document.write( "100000 = P*1.7316764476028<BR>\n" );
document.write( "100000 = 1.7316764476028P<BR>\n" );
document.write( "1.7316764476028P = 100000<BR>\n" );
document.write( "P = 100000/1.7316764476028<BR>\n" );
document.write( "P = 57747.5082821806<BR>\n" );
document.write( "P = 57747.51\r<BR>\n" );
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document.write( "Answer: Max needs to invest <font color=red>$57,747.51</font><BR>\n" );
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