document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1038087>Question 1038087</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Question 1. <BR>\n" );
document.write( "How long will it take for an investment to double at 7%p.a. compounded monthly?\r<BR>\n" );
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document.write( "Question 2.<BR>\n" );
document.write( "How long will it take for regular monthly deposit of $200 to accrue to $15000 if interest is paid at the rate of 7.25%p.a. a compounded quarterly?\r<BR>\n" );
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document.write( "Question 3.<BR>\n" );
document.write( "What is the present value of an annuity where $160 is deposited at the end of each quarter for 3.5 years into an account bearing 14% p.a. interested compounded monthly?\r<BR>\n" );
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document.write( "Question 4.<BR>\n" );
document.write( "At the end of each quarter $700 is invested in an account. The account pays interest at 8% p.a compounded monthly. How much is in the account after two years? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #652830 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=652830\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Can do the first two here.<BR>\n" );
document.write( "A=Ao(1+.07/12)^12t<BR>\n" );
document.write( "A=2 Ao<BR>\n" );
document.write( "2=(1+.0058333)^12t<BR>\n" );
document.write( "ln of both sides<BR>\n" );
document.write( "ln2=12t*ln (1.00583333)<BR>\n" );
document.write( "without rounding, divide both sides by ln(1.0583333)<BR>\n" );
document.write( "then divide both by 12<BR>\n" );
document.write( "t=9.93 years<BR>\n" );
document.write( "The rule of 72 would say about 10 years.  This is reasonable.<BR>\n" );
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document.write( "15000=200(1+(.0725)/4)^4t<BR>\n" );
document.write( "75=[1+(.0725)/4)]^4t<BR>\n" );
document.write( "ln 75=4t ln (1+(0.0725)/4))<BR>\n" );
document.write( "t=60.09 years\r<BR>\n" );
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document.write( " </SPAN>\n" );
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