document.write( "Question 1149612: Phyllis invested 27000 dollars, a portion earning a simple interest rate of 5 percent per year and the rest earning a rate of 7 percent per year. After one year the total interest earned on these investments was 1570 dollars. How much money did she invest at each rate? \n" ); document.write( "
Algebra.Com's Answer #770953 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "By the standard algebraic solution method....

\n" ); document.write( "Let x be the amount at 5%; then the amount at 7% is 27000-x. Then the total interest is 1570 dollars:

\n" ); document.write( "\".05%28x%29%2B.07%2827000-x%29+=+1570\"
\n" ); document.write( "\".05x%2B1890-.07x+=+1570\"
\n" ); document.write( "...

\n" ); document.write( "You can finish the solution by that method.

\n" ); document.write( "Here is a non-algebraic method for finding the answer to \"mixture\" problems like this that I find much easier and faster than that algebraic approach.

\n" ); document.write( "(1a) $27000 all at 5% would yield $1350 interest
\n" ); document.write( "(1b) the actual interest was $1570
\n" ); document.write( "(1c) $27000 all at 7% would yield $1890 interest

\n" ); document.write( "(2) The actual interest of $1570 is 220/540 = 11/27 of the way from $1350 to $1890. (If it helps, picture the three interest amounts on a number line. $1350 to $1570 is a difference of $220; $1350 to $1890 is a difference of $540. So $1570 is 220/540 of the way from $1350 to $1890.)

\n" ); document.write( "That means 11/27 of the investment was at the higher rate.

\n" ); document.write( "ANSWER: 11/27 of $27,000, or $11,000, at 7%; the other $16,000 at 5%.

\n" ); document.write( "CHECK: .07(11000)+.05(16000) = 770+800 = 1570

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