document.write( "Question 1153657: A total of $7000 is invested: part at 6% and the remainder at 10%. How much is invested at each rate if the annual interest is $470? \n" ); document.write( "
Algebra.Com's Answer #775959 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "A quick and easy way to solve this problem and a wide variety of similar \"mixture\" problems....

\n" ); document.write( "Consider the actual amount of interest and the amounts of interest if the whole amount were invested at each rate. $7000 at 6% gives $420 interest; $7000 at 10% gives $700 interest. Now consider those three amounts of interest on a number line.

\n" ); document.write( "$470 is 50/280 = 5/28 of the way from $420 to $700.
\n" ); document.write( "That means 5/28 of the total was invested at the higher rate.

\n" ); document.write( "5/28 of $7000 = 5*$250 = $1250.

\n" ); document.write( "ANSWER: $1250 at 10%; the other $5750 at 6%.

\n" ); document.write( "CHECK: .10(1250)+.06(5750) = 125+345 = 470

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