Question 435031
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This is essentially a mixture problem -- we are mixing money invested with a return of 6% and money invested at 8%.<br>
Here is a solution using a method that can be used in any 2-part mixture problem like this.<br>
All $2000 invested at 6% would yield a return of $120; all invested at 8% would yield a return of $160; the actual return is $144.<br>
The ratio in which the money is invested in the two places is exactly determined by where the actual return of $144 lies between $120 and $160.<br>
Use a number line if it helps to determine that $144 is three-fifths of the way from $120 to $160.<br>
(The difference between $160 and $120 is $40; the difference between $144 and $120 is $24.  $24 is 24/40 = 6/10 = 3/5 of $40)<br>
That means 3/5 of the total $2000 was invested at the higher rate.<br>
3/5 of $2000 is $1200, so $1200 of the total $2000 was invested at 8%.<br>
ANSWERS: $1200 was invested at 8%; $800 at 6%.<br>
CHECK: .08(1200)+.06(800) = 96+48 = 144<br>