document.write( "Question 1144404: Walt made an extra $9000 last year from a part-time job. He invested part of the money at 5% and the rest at 6%. He made a total of $480 in interest. How much was invested at 6%?
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Algebra.Com's Answer #765497 by greenestamps(13198)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "(1) A traditional algebraic solution.... \n" ); document.write( "let x be the amount invested at 5% \n" ); document.write( "then 9000-x is the amount invested at 6% \n" ); document.write( "The total interest is 5% of x plus 6% of (9000-x): \n" ); document.write( " \n" ); document.write( "Solve using basic algebra.... \n" ); document.write( "(2) A solution found by comparing the actual interest to the amounts of interest if the whole $9000 had been invested at one rate of the other.... \n" ); document.write( "$9000 all at 5% would yield $450 interest \n" ); document.write( "$9000 all at 6% would yield $540 interest \n" ); document.write( "$480 is 1/3 of the way from $450 to $540 (picture the three numbers 450, 480, and 540 on a number line...) \n" ); document.write( "Therefore, 1/3 of the total was invested at the higher rate. \n" ); document.write( "ANSWER: 1/3 of $9000, or $3000, at 6%; the other $6000 at 5%. \n" ); document.write( "CHECK: \n" ); document.write( ".05(6000)+.06(3000) = 300+180 = 480 \n" ); document.write( " |