document.write( "Question 1120806: Rodrigo invests $4000, part of it at 10% annual interest and the rest at 12% annual interest. If he receives $460 in interest at the end of one year, how much did he invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #736679 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "The ratio in which the money must be split between the two investments is exactly determined by where the actual interest lies between the amounts of interest that would have been received if the money had been invested all at the lower rate or all at the higher rate.

\n" ); document.write( "$4000 at 10% would yield $400 interest
\n" ); document.write( "the actual interest was $460
\n" ); document.write( "$4000 at 12% would yield $480 interest

\n" ); document.write( "$460 is three-fourths of the way from $400 to $480; that means 3/4 of the money needs to be invested at the higher rate.

\n" ); document.write( "So 3/4 of the $4000, or $3000, was invested at 12%; the other $1000 was invested at 10%.

\n" ); document.write( "Check:
\n" ); document.write( "(.12)($3000) + (.10)($1000) = $360+$100 = $460 \n" ); document.write( "

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