document.write( "Question 1153657: A total of $7000 is invested: part at 6% and the remainder at 10%. How much is invested at each rate if the annual interest is $470? \n" ); document.write( "
Algebra.Com's Answer #775959 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "A quick and easy way to solve this problem and a wide variety of similar \"mixture\" problems.... \n" ); document.write( "Consider the actual amount of interest and the amounts of interest if the whole amount were invested at each rate. $7000 at 6% gives $420 interest; $7000 at 10% gives $700 interest. Now consider those three amounts of interest on a number line. \n" ); document.write( "$470 is 50/280 = 5/28 of the way from $420 to $700. \n" ); document.write( "That means 5/28 of the total was invested at the higher rate. \n" ); document.write( "5/28 of $7000 = 5*$250 = $1250. \n" ); document.write( "ANSWER: $1250 at 10%; the other $5750 at 6%. \n" ); document.write( "CHECK: .10(1250)+.06(5750) = 125+345 = 470 \n" ); document.write( " \n" ); document.write( " |