document.write( "Question 1188420: a) A total of $40,000 was invested, part of it at 12% interest and the remainder at 15%.
\n" ); document.write( "If the total yearly interest from both investments was $5,800. How much was invested at each rate
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Algebra.Com's Answer #819513 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Bad data, requiring answers that are not whole numbers of dollars, and not even whole numbers of cents....

\n" ); document.write( "x = amount invested at 12%
\n" ); document.write( "40000-x = amount invested at 15%

\n" ); document.write( "The total interest was 5800:

\n" ); document.write( "\".12%28x%29%2B.15%2840000-x%29=5800\"

\n" ); document.write( "You can do the algebra to try to find the answer.

\n" ); document.write( "Here is a quick informal way to find the answer.

\n" ); document.write( "All $40,000 invested at 12% would yield $4800 interest; all at 15% would yield $6000.
\n" ); document.write( "Look at the three numbers 4800, 5800, and 6000 on a number line and observe/calculate that 5800 is 1000/1200 = 5/6 of the way from 4800 to 6000.
\n" ); document.write( "That means 5/6 of the total was invested at the higher rate.

\n" ); document.write( "5/6 of $40,000 is not a whole number....

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