Question 1154741
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Here is a solution using the method I like best for "mixture" problems like this.<br>
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.<br>
(1) All $5500 invested at 6% would yield $330 interest; all at 12% would yield $660 interest.<br>
(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.<br>
(3) That means 32/55 of the total was invested at the higher rate.<br>
ANSWER: 32/55 of $5500, or $3200, at 12%; the other $2300 at 6%.<br>
CHECK: .12(3200)+.06(2300) = 384+138 = 522<br>