document.write( "Question 1147775: jeff invests $5500 in two different accounts. the first account paid 11%, thr second account paid 9% in intrest. at the end of the first year he had earned $525 in intrest. how much was in each account? \n" ); document.write( "
Algebra.Com's Answer #769138 by greenestamps(13215)\"\" \"About 
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\n" ); document.write( "By algebra....

\n" ); document.write( "The two rates are 11% (0.11) and 9% (0.09)
\n" ); document.write( "The two amounts are, respectively, x and 5500-x
\n" ); document.write( "The two amounts of interest are 0.11(x) and 0.09(5500-x)
\n" ); document.write( "The total interest is $525:

\n" ); document.write( "\"0.11%28x%29%2B0.09%285500-x%29+=+525\"

\n" ); document.write( "Solve using basic algebra (I leave that to you)

\n" ); document.write( "A much easier and faster way to solve \"mixture\" problems like this:

\n" ); document.write( "(1) Determine the amounts of interest if the whole amount was invested at each rate.
\n" ); document.write( "(2) Determine where the actual amount of interest lies between those two amounts. That exactly determines how the total must be split between the two rates.

\n" ); document.write( "$5500 all at 11% yields $605 interest
\n" ); document.write( "$5500 all at 9% yields $495 interest

\n" ); document.write( "$525 is 3/11 of the way from $495 to $605 (525-495 = 30; 605-495 = 110; 30/110 = 3/11)

\n" ); document.write( "That means 3/11 of the total was at the higher rate.

\n" ); document.write( "ANSWER: 3/11 of the $5500, or $1500, at 11%; the other $4000 at 9%

\n" ); document.write( "CHECK:
\n" ); document.write( ".11(1500) + .09(4000) = 165+360 = 525\r
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