<PRE><B>Question 1185811</B>
&lt;br&gt;
A standard &quot;mixture&quot; problem....&lt;br&gt;
First a standard setup for solving the problem using algebra.&lt;br&gt;
x dollars at 6%, plus (14000-x) dollars at 16%, yields $1500 interest.&lt;br&gt;
{{{.06(x)+.16(14000-x)=1500}}}&lt;br&gt;
Solve using basic algebra, although the numbers are a bit messy.&lt;br&gt;
And now a quick and easy way to solve any 2-part mixture problem like this, if a formal algebraic solution is not required and your mental math skills are good.&lt;br&gt;
$14,000 all invested at 6% would yield $840 interest; all at 16% would yield $2240 interest; the actual interest was $1500.&lt;br&gt;
Picture the three interest amounts -- 840, 1500, and 2240 -- on a number line and observe/calculate that 1500 is 660/1400 of the way from 840 to 2240.  That means 660/1400 of the total was loaned at the higher rate.&lt;br&gt;
{{{14000(660/1400)=6600}}}&lt;br&gt;
ANSWER: $6600 was loaned at 16%; the other $7400 at 6%.&lt;br&gt;
CHECK: .16(6600)+.06(7400)=1056+444=1500&lt;br&gt;
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