<PRE><B>Question 1092605</B>
&lt;br&gt;(1) Traditional algebraic approach....&lt;br&gt;
You are mixing two ingredients to get the final mixture.  Find expressions for the amounts of copper in each.&lt;br&gt;
Ingredient A: x ounces of 20% copper
The amount of copper is 0.20(x)&lt;br&gt;
Ingredient B: 300 ounces of 80% copper
The amount of copper is 0.80(300) = 240&lt;br&gt;
Mixture: (x+300) ounces of 30% copper
The amount of copper in the mixture is 0.30(x+300)&lt;br&gt;
Write and solve the equation that says the amount of copper in the mixture is the sum of the amounts in the two ingredients:
{{{0.20x + 240 = 0.30(x+300)}}}&lt;br&gt;
I&#39;ll let you finish...&lt;br&gt;
Note: if you want to avoid calculations with decimals, multiply the whole equation by 10 before doing anything else.&lt;br&gt;&lt;br&gt;
But here is a faster way to solve mixture problems.&lt;br&gt;
The ratio in which the two ingredients need to be mixed is exactly determined by how close the percentage of the mixture is to the percentages of each ingredient.&lt;br&gt;
The mixture percentage, 30, is &quot;5 times as close to 20 as it is to 80&quot;.  That is, 80-30 = 50; 30-20 = 10; the 50 is 5 times the 10.&lt;br&gt;
That means you need 5 times as much of the 20% copper as the 80% copper.&lt;br&gt;
Given that there are 300 ounces of the 80% copper, you need 5*300 = 1500 ounces of the 20% copper.&lt;br&gt;
If you finished solving the problem by the algebraic method shown above, that should be the answer you got.</PRE>