document.write( "Question 1157106: A man invested part of $15,000 at 12% and the remainder at 8%. If his annual
\n" ); document.write( "income from the two investments is $1456, how much does he have invested at
\n" ); document.write( "each rate?
\n" ); document.write( "

Algebra.Com's Answer #779933 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "Here is a way to solve two-part mixture problems like this quickly without formal algebra. If you understand the method, it will get you to the answer much faster and with much less effort (and easier calculations) than the formal algebraic method.

\n" ); document.write( "(1) Determine what the interest would be if the whole amount were invested at each rate
\n" ); document.write( "$15,000 at 12% --> $1800 interest
\n" ); document.write( "$15,000 at 8% --> $1200 interest

\n" ); document.write( "(2) Determine where the actual interest lies between those two amounts (imagine the three percentages on a number line: 1200, 1456, 1800)
\n" ); document.write( "1200 to 1800 is a difference of 600
\n" ); document.write( "1200 to 1456 is a difference of 256

\n" ); document.write( "The actual interest is 256/600 = 64/150 of the way from 1200 to 1800

\n" ); document.write( "(3) The fraction of the total invested at the higher interest rate is that same fraction.
\n" ); document.write( "64/150 of $15,000 is $6400

\n" ); document.write( "ANSWER: $6400 invested at 12%; the other $8600 at 8%

\n" ); document.write( "CHECK:
\n" ); document.write( ".12(6400)+.08(8600) = 768+688 = 1456
\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );