document.write( "Question 1148298: Jerry invested $10,000, part at 8% simple interest and the rest at 5% simple interest for a period of 1 year. If he received a total annual interest of $575 from both investments, how much did he invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #769666 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "Using the traditional algebraic solution method....

\n" ); document.write( "8% of x, plus 5% of ($10,000-x), equals $575:

\n" ); document.write( " $\".08%28x%29%2B.05%2810000-x%29+=+575\"$
\n" ); document.write( " $\".08x%2B500-.05x+=+575\"$
\n" ); document.write( " $\".03x+=+75\"$
\n" ); document.write( " $\"x+=+75%2F.03+=+2500\"$

\n" ); document.write( "ANSWER: $2500 at 8%; the other $7500 at 5%.

\n" ); document.write( "Here is a non-algebraic method that will get you to the solution to \"mixture\" problems like this much faster and with far less work, if you understand how to use it.

\n" ); document.write( "(1) The $10,000 all earning 8% interest would yield $800 interest; all at 5% would yield $500 interest.
\n" ); document.write( "(2) The actual interest, $575, is 1/4 of the way from $500 to $800. (Picture the three numbers on a number line -- 500, 575, and 800; 575 is 1/4 of the way from 500 to 800.)
\n" ); document.write( "(3) That means 1/4 of the total was invested at the higher rate.

\n" ); document.write( "ANSWER: 1/4 of $10,000, or $2500, at 8%; the other $7500 at 5%.

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