Question 1125806
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Here is a method for solving "mixture" problems like this which, if you understand it, will get you to the answer much faster and with far less work than the traditional algebraic method.<br>
The key idea for this method is that the ratio in which the money was split between the two accounts is exactly determined by where the actual amount of interest lies between the amounts of interest that would have been earned at each of the two separate accounts.<br>
The explanation in words sounds confusing; but the actual calculations are quick and simple.  For this example:<br>
30,000 at 6% = 1800; 30,000 at 10% = 3000
Actual interest = 2360
2360 is (560/1200) of the way from 1800 to 3000.  (2360-1800 = 560; 3000-1800 = 1200).
That means 560/1200 of the money was invested at the higher rate.
Now convert that ratio into a ratio that is "convenient" for the total of $30,000 that was invested:  560/1200 = 56/120 = 14/30.
The amount invested at 10% was 14/30 of the $30,000, which is $14,000; the other $16,000 was invested at 6%.