Question 1205484
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Here is a solution by a method completely different from the standard algebraic solutions shown by the other tutors.  This method can be use to solve any 2-part "mixture" problem like this one.<br>
All $19,000 invested at 11% would yield $2090 interest; all invested at 13% would yield $2470 interest.<br>
The amounts invested at the two rates to yield $2290 in interest is exactly determined by where the actual interest of $2290 lies between $2090 and $2470.<br>
Use a number line, if it helps, to observe/calculate that $2290 is $2000/$3800 = 20/38 = 10/19 of the way from $2090 to $2470.  That means 10/19 of the total was invested at the higher rate.<br>
10/19 of $19,000 is, trivially, $10,000.<br>
ANSWER: $10,000 was invested at 13%, the other $9000 at 11%<br>
CHECK: .13(10000)+.11(9000) = 1300+990 = 2290<br>
For me, at least, that solution is much easier than a formal algebraic solution using either one or two variables.<br>