<PRE><B>Question 1144404</B>
&lt;br&gt;
(1) A traditional algebraic solution....&lt;br&gt;
let x be the amount invested at 5%
then 9000-x is the amount invested at 6%&lt;br&gt;
The total interest is 5% of x plus 6% of (9000-x):&lt;br&gt;
{{{.05(x)+.06(9000-x) = 480}}}&lt;br&gt;
Solve using basic algebra....&lt;br&gt;
(2) A solution found by comparing the actual interest to the amounts of interest if the whole $9000 had been invested at one rate of the other....&lt;br&gt;
$9000 all at 5% would yield $450 interest
$9000 all at 6% would yield $540 interest&lt;br&gt;
$480 is 1/3 of the way from $450 to $540 (picture the three numbers 450, 480, and 540 on a number line...)&lt;br&gt;
Therefore, 1/3 of the total was invested at the higher rate.&lt;br&gt;
ANSWER: 1/3 of $9000, or $3000, at 6%; the other $6000 at 5%.&lt;br&gt;
CHECK:
.05(6000)+.06(3000) = 300+180 = 480</PRE>