document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160573>Question 1160573</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A woman has a total of $10,000 to invest. She invests part of the money in an account that pays 8% per year and the rest in an account that pays 11% per year. If the interest earned in the first year is $890, how much did she invest in each account? </SPAN>\n" );
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document.write( "Formal algebra....<br><BR>\n" );
document.write( "x amount at 8%; 10000-x at 11%; total interest 890:<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=.08%28x%29%2B.11%2810000-x%29+=+890 ALIGN=MIDDLE  ALT=\".08%28x%29%2B.11%2810000-x%29+=+890\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Solve using basic algebra; I leave that to you.<br><BR>\n" );
document.write( "A much faster and easier informal solution method....<br><BR>\n" );
document.write( "(1) $10,000 all at 8% would yield $800 interest; all at 11% would yield $1100 interest.<BR>\n" );
document.write( "(2) The actual interest amount, $890, is 3/10 of the way from $800 to $1100.  ($800 to $1100 is $300; $800 to $890 is $90; 90/300 = 3/10).<br><BR>\n" );
document.write( "(3) That means 3/10 of the $10,000 was invested at the higher rate.<br><BR>\n" );
document.write( "ANSWER: 3/10 of $10,000, or $3000, at 11%; the rest at 8%.<br><BR>\n" );
document.write( "CHECK: .11(3000)+.08(7000) = 330+560 = 890<br><BR>\n" );
document.write( " </SPAN>\n" );
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