document.write( "Question 1161595: Amanda made an extra $70,00 last year from a part time job. She invested part of the money at 10% and the rest at 7%. She made a total of $6400 in interest. How much was invested at 7% \n" ); document.write( "
Algebra.Com's Answer #785195 by greenestamps(13215) You can put this solution on YOUR website! \n" ); document.write( "(1) A traditional algebraic setup for solving.... \n" ); document.write( "x = amount at 10% \n" ); document.write( "70000-x = amount at 7% \n" ); document.write( "10% of x, plus 7% of (70000-x), equals 6400: \n" ); document.write( " \n" ); document.write( "Solve using basic algebra.... \n" ); document.write( "-------------------------------------------- \n" ); document.write( "(2) A quick and easy way to solve a 2-part \"mixture\" problem like this, if an algebraic solution is not required.... \n" ); document.write( "(1) All $70,000 invested at 10% would yield $7000 interest; all at 7% would yield $4900 interest. \n" ); document.write( "(2) The actual interest, $6400, is 5/7 of the way from $4900 to $7000. ($4900 to $7000 is a difference of $2100; $4900 to $6400 is a difference of $1500; 1500/2100 = 5/7) \n" ); document.write( "(3) That means 5/7 of the total was invested at the higher rate. \n" ); document.write( "ANSWER: 5/7 of $70,000, or $50,000, at 10%; the other $20,000 at 7%. \n" ); document.write( "CHECK: \n" ); document.write( ".10(50000) + .07(20000) = 5000+1400 = 6400 \n" ); document.write( " \n" ); document.write( " |