document.write( "Question 1108863: A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $385, how much was invested at each rate?
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document.write( "6 %+
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document.write( "11%= \n" );
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Algebra.Com's Answer #723861 by addingup(3677)![]() ![]() You can put this solution on YOUR website! Let \"part of it\" be x. So, x was invested at 6% or 0.06x \n" ); document.write( "The remainder is invested at 11%, and that amount = 3600 - x \n" ); document.write( "Put it all together: \n" ); document.write( "0.06x + 0.11(3600 - x) = 385 \n" ); document.write( "0.06x + 396 - 0.11x = 385 \n" ); document.write( "-0.05x = -11 divide, and remember that -/- = + so when you divide you can ignore the - sign, just divide 11/0.05 \n" ); document.write( "x = 220 this is how much is invested at 6% \n" ); document.write( "3600 - 220 = 3380 this is the amount invested at 11% \n" ); document.write( "------------------------------------------- \n" ); document.write( "Check: \n" ); document.write( "220 * 0.06 = 13.2 \n" ); document.write( "3380 * 0.11 = 371.80 \n" ); document.write( "13.2 + 371.80 = 385 Correct. \n" ); document.write( " \n" ); document.write( " |