<PRE><B>Question 1163062</B>
&lt;br&gt;
A formal algebraic setup for solving the problem....&lt;br&gt;
x = amount at 4%
6000-x = amount at 1%&lt;br&gt;
The average return on the $6000 was 2%:&lt;br&gt;
{{{.04(x)+.01(6000-x) = .02(6000)}}}&lt;br&gt;
Solve using basic algebra (I leave that to you)....&lt;br&gt;
A quick and easy solution method, using logical reasoning and a bit of mental arithmetic....&lt;br&gt;
The average return of 2% is &quot;twice as close to 1% as it is to 4%&quot;.  That means the amount invested at 1% was twice the amount invested at 4%.&lt;br&gt;
ANSWER: $4000 at 1%; $2000 at 4%&lt;br&gt;
CHECK:
.01(4000)+.04(2000) = 40+80 = 120
.02(6000) = 120&lt;br&gt;
Or here is a variation of this easy method, thinking a bit differently....&lt;br&gt;
The average return of 2% is 1/3 of the way from 1% to 4% (picture the three percentages on a number line). Therefore, 1/3 of the total was invested at 4%.&lt;br&gt;
ANSWER: 1/3 of the total $6000, or $2000, at 4%; the rest at 1%.&lt;br&gt;
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