document.write( "Question 1190296: Phyllis invested
\n" ); document.write( "56000
\n" ); document.write( " dollars, a portion earning a simple interest rate of
\n" ); document.write( "5
\n" ); document.write( " percent per year and the rest earning a rate of
\n" ); document.write( "6
\n" ); document.write( " percent per year. After one year the total interest earned on these investments was
\n" ); document.write( "3200
\n" ); document.write( " dollars. How much money did she invest at each rate?
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Algebra.Com's Answer #821928 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "A quick, informal solution method, if formal algebra is not required....

\n" ); document.write( "$56000 all invested at 5% would yield $2800 interest; all invested at 6% would yield $3360 interest. The ratio in which the total was split between the two investments is determined by where the actual interest of $3200 lies between $2800 and $3360.

\n" ); document.write( "(1) Look at the three interest amounts 2800, 3200, and 3360 on a number line and observe/calculate that 3200 is 400/560 = 5/7 of the way from 2800 to 3360.
\n" ); document.write( "(2) That means 5/7 of the total was invested at the higher rate.

\n" ); document.write( "ANSWER: 5/7 of the total $56,000, or $40,000, was invested at 6%; the other $16,000 at 5%.

\n" ); document.write( "CHECK: .06(40,000)+.05(16,000) = 2400+800 = 3200

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