<PRE><B>Question 1160573</B>
&lt;br&gt;
Formal algebra....&lt;br&gt;
x amount at 8%; 10000-x at 11%; total interest 890:&lt;br&gt;
{{{.08(x)+.11(10000-x) = 890}}}&lt;br&gt;
Solve using basic algebra; I leave that to you.&lt;br&gt;
A much faster and easier informal solution method....&lt;br&gt;
(1) $10,000 all at 8% would yield $800 interest; all at 11% would yield $1100 interest.
(2) The actual interest amount, $890, is 3/10 of the way from $800 to $1100.  ($800 to $1100 is $300; $800 to $890 is $90; 90/300 = 3/10).&lt;br&gt;
(3) That means 3/10 of the $10,000 was invested at the higher rate.&lt;br&gt;
ANSWER: 3/10 of $10,000, or $3000, at 11%; the rest at 8%.&lt;br&gt;
CHECK: .11(3000)+.08(7000) = 330+560 = 890&lt;br&gt;
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