document.write( "Question 1149229: You invested $ 4000 between two accounts paying 7 % and 9 % annual interest, respectively. If the total interest earned for the year was $ 300 comma how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #770574 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Using the standard formal algebraic solution.... \n" ); document.write( "7% of x, plus 9% of (4000-x) equals 300: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "ANSWER: $3000 at 7%; $1000 at 9%. \n" ); document.write( "CHECK: \n" ); document.write( ".07(3000)+.09(1000) = 210+90 = 300 \n" ); document.write( "Here is a much faster path to solving a mixture problem like this if an algebraic solution is not required. \n" ); document.write( "(1) $4000 all at 7% would yield $280 interest; all at 9% would yield $360 interest. \n" ); document.write( "(2) The actual interest, $300, is 1/4 of the way from $280 to $360. (Picture the three numbers 280, 300, and 360 on a number line. 280 to 360 is a difference of 80; 280 to 300 is a difference of 20. 20 is 1/4 of 80.) \n" ); document.write( "(3) Therefore, 1/4 of the total was invested at the higher rate. \n" ); document.write( "ANSWER: 1/4 of $4000, or $1000, at 9%; the other $3000 at 7%. \n" ); document.write( "Obviously the same answer as before.... \n" ); document.write( " \n" ); document.write( " |