document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1149106>Question 1149106</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A man  saved 3000 in a bank p, whose interest rate was X% per annum and 2000 in another bank Q whose interest rate was Y% per annum. His total interest in one year was 640. If he had saved 2000 in P and 3000 in Q for the same period,  he would have gained 20 as additional interest. Find <BR>\n" );
document.write( "(I) the value of x and Y. <BR>\n" );
document.write( "(II)  the investment return if the man saves 5000 in bank Q for 2years. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #770435 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=770435\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> interest in first part is 3000*p+2000*q=64000, because I multiplied the interest by 10 to make it an integer<BR>\n" );
document.write( "this is 3p+2q=64<BR>\n" );
document.write( "the second is 2000*p+3000q=66000<BR>\n" );
document.write( "2p+3q=66; multiply top by 2 and bottom by -3<BR>\n" );
document.write( "6p+4q=128<BR>\n" );
document.write( "-6p-9q=-198<BR>\n" );
document.write( "-5q=-70<BR>\n" );
document.write( "q=14%<BR>\n" );
document.write( "p=12% substituting\r<BR>\n" );
document.write( "\n" );
document.write( "5000*0.14*2=$1400 earned at a 14% rate.<BR>\n" );
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