document.write( "Question 1116363: susan invest 2 times as much money at 9% as she does at 6%. If her total interest after 1 year is $1440, how much does she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #731251 by greenestamps(13198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "...or here is a way to solve the problem with logical analysis instead of formal algebra \n" ); document.write( "The amount invested at 9% is 2 times the amount invested at 6%; and the 9% interest rate is 1.5 times the 6% interest rate. Together those two things mean the interest earned at 9% is 2*1.5 = 3 times the interest earned at 6%. \n" ); document.write( "So 1/4 of the interest earned is from the 6% investment and 3/4 is from the 9% investment. \n" ); document.write( "1/4 of the total interest of $1440 is $360, so the amount of interest from the 6% investment is $360. \n" ); document.write( "And then the amount invested at 6% is $360/.06 = $6000. \n" ); document.write( "Then the amount invested at 9% is 2*$6000 = $12000. \n" ); document.write( " |