document.write( "Question 1167264: Larry mitchel invested part of his $13,000 advance at 7% annual simple interest and the rest at 3% annual simple interest. If his total yearly interest from both accounts was $870, find the amount invested at each rate \n" ); document.write( "
Algebra.Com's Answer #791903 by greenestamps(13198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Here is a non-algebraic method for solving a \"mixture\" problem like this. \n" ); document.write( "(1) Find the amounts of interest if the whole $13,000 had been invested at each rate. \n" ); document.write( "$13,000 at 7% would yield $910 interest; $13,000 at 3% would yield $390 interest. \n" ); document.write( "(2) Where the actual interest lies between those two amounts exactly determines how much was invested at each rate. \n" ); document.write( "From 390 to 910 on a number line is a difference of 520; from 390 to 870 is a difference of 480. 480/520 = 48/52 = 12/13. \n" ); document.write( "$870 is 12/13 of the way from $390 to $910; that means 12/13 of the total was invested at the higher rate. \n" ); document.write( "ANSWERS: 12/13 of $13,000, or $12,000, was invested at 7%; the other $1000 was invested at 3%. \n" ); document.write( "CHECK: .07(12000)+.03(1000) = 840+30 = 870 \n" ); document.write( " \n" ); document.write( " |