document.write( "Question 1129650: Amy invests $1500 in one account and $1900 in an account paying 3 % higher interest. At the end of one year she had earned $261 in interest. At what rates did she invest?\r
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document.write( "1500 invested at -
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document.write( "1900 invested at- \n" );
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Algebra.Com's Answer #746283 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Here is a solution using some logical reasoning and some mental arithmetic, instead of formal algebra. \n" ); document.write( "First consider the \"extra\" 3% that is earned on the $1900 investment. The amount of interest there is .03($1900) = $57. \n" ); document.write( "The remaining amount of interest is $261-$57 = $204. That is the amount of interest on the whole investment of $3400. Since we have accounted for the interest from the extra 3% on the second investment, that $204 interest is now all at the lower rate. \n" ); document.write( " \n" ); document.write( "The $1500 was invested at 6%; the $1900 was invested at 6%+3% = 9%. \n" ); document.write( " |