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 #770437 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=770437\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let x = the interest rate for investment P.<BR>\n" );
document.write( "let y = the interest rate for investment Q.\r<BR>\n" );
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document.write( "you get:\r<BR>\n" );
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document.write( "3000x + 2000y = 640<BR>\n" );
document.write( "2000x + 3000y = 660\r<BR>\n" );
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document.write( "multiply both sides of the first equation by 2 and multiply both sides of the second equation by 3 to get:<BR>\n" );
document.write( "6000x + 4000y = 1280<BR>\n" );
document.write( "6000x + 9000y = 1980<BR>\n" );
document.write( "subtract the first equation from the second to get:<BR>\n" );
document.write( "5000y = 700<BR>\n" );
document.write( "solve for y to get:<BR>\n" );
document.write( "y = .14<BR>\n" );
document.write( "replace y with .14 in the first original equation to get:<BR>\n" );
document.write( "3000x + 2000y = 640 becomes 3000x + 2000*.14 = 640 which becomes 3000x + 280 = 640.<BR>\n" );
document.write( "subtract 280 from both sides of this equation to get:<BR>\n" );
document.write( "3000x = 640 - 280 which becomes 3000x = 360.<BR>\n" );
document.write( "solve for x to get x = 360/3000 = .12<BR>\n" );
document.write( "you have x = .12 and y = .14<BR>\n" );
document.write( "go back to your two original equations and replace x with .12 and y with .14 to get:<BR>\n" );
document.write( "3000x + 2000y = 640 becomes 3000*.12 + 2000*.14 = 640 which becomes 360 + 280 = 640 which becomes 640 = 640 which is true.<BR>\n" );
document.write( "2000x + 3000y = 660 becomes 2000*.12 + 3000*.14 = 660 which becomes 240 + 420 = 660 which becomes 660 = 660 which is true.<BR>\n" );
document.write( "both original equations are true when x = .12 and y = .14.<BR>\n" );
document.write( "x% = x * 100 = 12%<BR>\n" );
document.write( "y% = y * 100 = 14%.<BR>\n" );
document.write( "since they asked for the value of x and y, and not for the value of x% and y%, your solution is x = .12 and y = .14.<BR>\n" );
document.write( "if the man has put all 5000 in bank Q, then he would have earned interest of 5000 * .14 = 700.\r<BR>\n" );
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