<PRE><B>Question 1195060</B>
&lt;br&gt;
The solution shown by the other tutor is the standard formal algebraic method for solving &quot;mixture&quot; problems like this.&lt;br&gt;
If a formal algebraic solution is not required, any problem like this can be solved by seeing where the actual interest lies between the two amounts of interest that would have been earned if the whole amount had been invested at each rate.&lt;br&gt;
$20,000 all at 11% --&gt; $2200 interest
actual interest: $1210
$20,000 all at 2% --&gt; $400 interest&lt;br&gt;
Look at those three interest amounts on a number line and determine where the actual interest lies between the maximum and minimum amounts.&lt;br&gt;
$2200-$400 = $1800
$1210-$400 = $810&lt;br&gt;
810/1800 = 9/20&lt;br&gt;
Those simple calculations show that the actual interest is 9/20 of the way from the minimum of $400 to the maximum of $2200.  That fraction 9/20 means 9/20 of the total was invested at the higher rate.&lt;br&gt;
9/20 of $20,000 is $9000, so&lt;br&gt;
ANSWER: $9000 was invested at 11%; the other $11000 at 2%&lt;br&gt;
CHECK: 0.11(9000)+0.02(11000) = 990+220=1210&lt;br&gt;
Here are some words that summarize the solution by this method:&lt;br&gt;
&quot;The actual amount of interest was 9/20 of the way from the smallest amount to the largest amount; so 9/20 of the total was invested at the higher rate.&quot;&lt;br&gt;
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