document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1193301>Question 1193301</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help me with the homework<BR>\n" );
document.write( "1. R2000 invested on the 6 march 2016 at simple interest rate of 15% p.a. accumulated to R2240. Calculate the date on which this transaction matured.\r<BR>\n" );
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document.write( "2. A certain amount of money was invested on the 3 march 2016 at a simple interest rate of 12%p.a. and accumulated to R612 on the 2 may 2016. How much was the money?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #825319 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=825319\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I=prt<BR>\n" );
document.write( "240=2000*0.15*t (in years)<BR>\n" );
document.write( "240=300t<BR>\n" );
document.write( "t=0.8 years or about 9.6 months or 292 days or 23 Dec 2016.<BR>\n" );
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document.write( "that is 2 months' time or 1/6 year<BR>\n" );
document.write( "I=prt<BR>\n" );
document.write( "612=p*0.12*(1/6)<BR>\n" );
document.write( "multiply by 6<BR>\n" );
document.write( "3672=0.12p<BR>\n" );
document.write( "p=3672/0.12=30,600R </SPAN>\n" );
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