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)\"\" \"About 
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\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( "\".05%28x%29%2B.06%289000-x%29+=+480\"

\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
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