Question 1152349
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If an algebraic solution is required, then the solution provided by the other tutor is a good presentation of the usual method for solving the problem.<br>
This kind of problem -- as well as a great deal of similar types of problems -- is very common on competitive middle school or high school math contests.  Here is a quick way to solve this kind of problem with a few quick calculations -- much faster than with the equation used in the formal algebraic method.<br>
(1) Find out how the actual interest of $820 compares to the interest that would have been earned if the whole amount were invested at each of the two rates:
$12,000 at 6% --> $720 interest
actual -- $820 interest
$12,000 at 8% --> $960 interest<br>
From $720 to $960 is a difference of $240; from $720 to $820 is a difference of $100.  So the actual interest is "100/240 = 5/12 of the way" from $720 to $960.<br>
Since the actual interest if 5/12 of the way from the amount it would have been if all the money had been invested at 6% to the amount it would have been if it had all been invested at 8%, that means 5/12 of the total was invested at 8%.<br>
ANSWER: 5/12 of the $12,000, or $5000, at 8%; the other $7000 at 6%.<br>
CHECK: .08(5000)+.06(7000) = 400+420 =820<br>
All the words of explanation might make this seem like a lengthy process; but it is not.  Being good with mental arithmetic, I solve this kind of problem typically in 7-8 seconds once I have read the given information.<br>
Without all the words, here are the needed calculations:
.06(12000) = 720; .08(12000) = 960
960-720 = 240; 820-720 = 100; 100/240 = 5/12
(5/12)*12000 = 5000<br>