document.write( "Question 1156044: You invested $ 4000 between two accounts paying 2 % and 7 % annual interest, respectively. If the total interest earned for the year was $ 230, how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #778767 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "A traditional algebraic approach.... \n" ); document.write( "x = amount at 2% \n" ); document.write( "4000-x = amount at 7% \n" ); document.write( " \n" ); document.write( "Solve using basic algebra (straightforward, but it takes time).... \n" ); document.write( "Here is how I prefer to solve two-part \"mixture\" problems like this. It uses only logical reasoning and a few simple mental calculations -- as opposed to writing and solving an algebraic equation. \n" ); document.write( "$4000 all at 2% would yield $80 interest \n" ); document.write( "$4000 all at 7% would yield $280 interest \n" ); document.write( "The actual interest of $230 is 3/4 of the way from $80 to $280 (80 to 280 is 200; 80 to 230 is 150; 150/200 = 3/4) \n" ); document.write( "Therefore 3/4 of the $4000 was invested at the higher rate. \n" ); document.write( "ANSWER: 3/4 of the $4000, or $3000, was invested at 7%; the remaining $1000 was invested at 2%. \n" ); document.write( "CHECK: \n" ); document.write( ".07(3000)+.02(1000) = 210+20 = 230 \n" ); document.write( " \n" ); document.write( " |