document.write( "Question 1154741: In investing &5500 of a couples money, a financial planner put some of it into savings paying 6% annual simple interest. The rest was invested in a riskier minimall development plan paying 12% annual simple interest. The combined interest earned the first year was $522. How much money was invested at each rate \n" ); document.write( "
Algebra.Com's Answer #777238 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Here is a solution using the method I like best for \"mixture\" problems like this. \n" ); document.write( "If you understand the method, and if an algebraic solution is not required, you will get to the answer faster and with much less work than using algebra. \n" ); document.write( "(1) All $5500 invested at 6% would yield $330 interest; all at 12% would yield $660 interest. \n" ); document.write( "(2) The actual interest of $522 is 32/55 of the way from $330 to $660. ($330 to $660 is a difference of $330; $330 to $522 is a difference of $192; 192/330 = 32/55. \n" ); document.write( "(3) That means 32/55 of the total was invested at the higher rate. \n" ); document.write( "ANSWER: 32/55 of $5500, or $3200, at 12%; the other $2300 at 6%. \n" ); document.write( "CHECK: .12(3200)+.06(2300) = 384+138 = 522 \n" ); document.write( " \n" ); document.write( " |