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)![]() ![]() ![]() You can put this solution on YOUR website! \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 \n" ); document.write( " |