document.write( "Question 1149612: Phyllis invested 27000 dollars, a portion earning a simple interest rate of 5 percent per year and the rest earning a rate of 7 percent per year. After one year the total interest earned on these investments was 1570 dollars. How much money did she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #770953 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "By the standard algebraic solution method.... \n" ); document.write( "Let x be the amount at 5%; then the amount at 7% is 27000-x. Then the total interest is 1570 dollars: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "... \n" ); document.write( "You can finish the solution by that method. \n" ); document.write( "Here is a non-algebraic method for finding the answer to \"mixture\" problems like this that I find much easier and faster than that algebraic approach. \n" ); document.write( "(1a) $27000 all at 5% would yield $1350 interest \n" ); document.write( "(1b) the actual interest was $1570 \n" ); document.write( "(1c) $27000 all at 7% would yield $1890 interest \n" ); document.write( "(2) The actual interest of $1570 is 220/540 = 11/27 of the way from $1350 to $1890. (If it helps, picture the three interest amounts on a number line. $1350 to $1570 is a difference of $220; $1350 to $1890 is a difference of $540. So $1570 is 220/540 of the way from $1350 to $1890.) \n" ); document.write( "That means 11/27 of the investment was at the higher rate. \n" ); document.write( "ANSWER: 11/27 of $27,000, or $11,000, at 7%; the other $16,000 at 5%. \n" ); document.write( "CHECK: .07(11000)+.05(16000) = 770+800 = 1570 \n" ); document.write( " \n" ); document.write( " |