Question 1139798
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Here is a solution by a very different method that can be used on any 2-part mixture problem like this.  Especially if the numbers are "nice", this method is much faster than the standard algebraic solution method.<br>
$10,500 all invested at 6% would yield $630 interest; all at 15% would yield $1575 interest.  The actual interest amount was $990.<br>
Look at those three interest amounts (on a number line, if it helps) and observe/calculate that $990 is $360/$945 = 72/189 = 8/21 of the way from $630 to $1575.<br>
That means 8/21 of the total was invested at the higher rate.<br>
8/21 of the total $10,500 is 8*$500 = $4000.<br>
ANSWER: $4000 was invested at 15%; the other $6500 at 6%.<br>
CHECK: .15(4000)+.06(6500) = 600+390 = 990<br>