document.write( "Question 1185811: A bank loaned out ​$14,000​, part of it at the rate of 6% per year and the rest at 16% per year. If the interest received in one year totaled ​1500$​, how much was loaned at 6% \n" ); document.write( "
Algebra.Com's Answer #816664 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "A standard \"mixture\" problem....

\n" ); document.write( "First a standard setup for solving the problem using algebra.

\n" ); document.write( "x dollars at 6%, plus (14000-x) dollars at 16%, yields $1500 interest.

\n" ); document.write( "\".06%28x%29%2B.16%2814000-x%29=1500\"

\n" ); document.write( "Solve using basic algebra, although the numbers are a bit messy.

\n" ); document.write( "And now a quick and easy way to solve any 2-part mixture problem like this, if a formal algebraic solution is not required and your mental math skills are good.

\n" ); document.write( "$14,000 all invested at 6% would yield $840 interest; all at 16% would yield $2240 interest; the actual interest was $1500.

\n" ); document.write( "Picture the three interest amounts -- 840, 1500, and 2240 -- on a number line and observe/calculate that 1500 is 660/1400 of the way from 840 to 2240. That means 660/1400 of the total was loaned at the higher rate.

\n" ); document.write( "\"14000%28660%2F1400%29=6600\"

\n" ); document.write( "ANSWER: $6600 was loaned at 16%; the other $7400 at 6%.

\n" ); document.write( "CHECK: .16(6600)+.06(7400)=1056+444=1500

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