document.write( "Question 1147279: Mikalya invests $1600 in one account and $1000 in an account paying 2% higher interest. At the end of one year she had earned $72 in interest.At what rates did she invest?? \n" ); document.write( "

Algebra.Com's Answer #768616 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The numbers in the problem make it easy to find a solution by trial and error.

\n" ); document.write( "The total amount invested was $2600; the total interest was $72. That's an interest rate a bit below 3%.

\n" ); document.write( "If the two interest rates differ by 2%, they are probably 2% and 4%. So try those:

\n" ); document.write( "$1600 at 2% = $32 interest
\n" ); document.write( "$1000 at 4$ = $40 interest

\n" ); document.write( "The total interest is $72, as required.

\n" ); document.write( "ANSWER: $1600 at 2%; $1000 at 4%.

\n" ); document.write( "Now see if you can get that result using algebra.

\n" ); document.write( "The interest on $1600 at x percent is $\"1600%2A%28x%2F100%29\"$
\n" ); document.write( "The interest on $1000 at (x+2) percent is $\"1000%2A%28%28x%2B2%29%2F100%29\"$
\n" ); document.write( " $\"1600%2A%28x%2F100%29%2B1000%2A%28%28x%2B2%29%2F100%29+=+72\"$
\n" ); document.write( " $\"1600x+%2B+1000%28x%2B2%29+=+7200\"$

\n" ); document.write( "I'll leave it to you to do the little remaining work to verify the answer of 2% and 4%. \n" ); document.write( "

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