document.write( "Question 1133346: Two investments totaling $57,500 produce an annual income of $4245 . One investment yields 8% per year, while the other yields 7% per year. How much is invested at each rate?
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Algebra.Com's Answer #750534 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The traditional algebraic approach: 7% of amount x, plus 8% of amount ($57,500-x), equals $4245:

\n" ); document.write( " $\".07x%2B.80%2857500-x%29+=+4245\"$

\n" ); document.write( "Solve using basic algebra, although the decimals make for some ugly computations.

\n" ); document.write( "I leave it to you to finish the problem by that method.

\n" ); document.write( "An alternative method, based on finding where the actual interest of 4245 lies between the two amounts if all $57,500 had been invested in either one of the two accounts....

\n" ); document.write( "$57,500 at 7% = 4025
\n" ); document.write( "$57,500 at 8% = 4600

\n" ); document.write( "Find where the actual amount of interest lies between those two amounts:
\n" ); document.write( "The difference between those two amounts is $575.
\n" ); document.write( "The difference between the actual amount of interest and the amount of interest at 7% is $4245-$4025 = $220.

\n" ); document.write( "So the amount of actual interest is 220/575 of the way from 7% to 8%; that means 220/575 of the total was invested at the higher rate of 8%. Since the total amount was $57,500, the amount invested at 8% was $22,000; that means the amount invested at 7% was $57,500-$22,000 = $35,500.

\n" ); document.write( "ANSWER: $35,500 at 7%; $22,000 at 8%

\n" ); document.write( "Of course that is the answer you should get by solving the problem by the algebraic method.
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