document.write( "Question 1205484: Ethan invested $19,000 for 1 year, part at 11% simple interest and part at 13% simple interest. If he earned a total of $2,290 in interest, how much was invested at each rate?
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Algebra.Com's Answer #842327 by greenestamps(13198) $\"\"$ $\"About$

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\n" ); document.write( "Here is a solution by a method completely different from the standard algebraic solutions shown by the other tutors. This method can be use to solve any 2-part \"mixture\" problem like this one.

\n" ); document.write( "All $19,000 invested at 11% would yield $2090 interest; all invested at 13% would yield $2470 interest.

\n" ); document.write( "The amounts invested at the two rates to yield $2290 in interest is exactly determined by where the actual interest of $2290 lies between $2090 and $2470.

\n" ); document.write( "Use a number line, if it helps, to observe/calculate that $2290 is $2000/$3800 = 20/38 = 10/19 of the way from $2090 to $2470. That means 10/19 of the total was invested at the higher rate.

\n" ); document.write( "10/19 of $19,000 is, trivially, $10,000.

\n" ); document.write( "ANSWER: $10,000 was invested at 13%, the other $9000 at 11%

\n" ); document.write( "CHECK: .13(10000)+.11(9000) = 1300+990 = 2290

\n" ); document.write( "For me, at least, that solution is much easier than a formal algebraic solution using either one or two variables.

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