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)![]() ![]() You can put this solution on YOUR website! \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( " \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( " \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 \n" ); document.write( " \n" ); document.write( " |