document.write( "Question 87268: Mike invested $6000 for one year. He invested part of it at 9% and the rest at 11%. At the end of the year he earned $624 in interest. How much did he invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #63207 by malakumar_kos@yahoo.com(315)\"\" \"About 
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\n" ); document.write( " let the amount invested at 9% be = x interest =9x/100
\n" ); document.write( " amount invested at 11%=(6000-x) interest =(6000-x).11/100
\n" ); document.write( " total interest = 624 hence 9x/100+(6000-x).11/100=624
\n" ); document.write( " 9x/100-11x/100+660=624 -2x/100=624-660
\n" ); document.write( " -2x/100=-36 or x=1800 6000-1800=4200
\n" ); document.write( " amount invested at 9% interest=Rs1800
\n" ); document.write( " amount invested at 11%interest=Rs 4200
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