Mixing Copper-Silver Alloys

Date: 06/09/99 at 02:46:21
From: Agnes
Subject: An alloy of copper and silver

Dear Doctor Math,

I have spent an hour staring at this problem but I still can't solve 
it. I would be VERY grateful if ouu could help me with it.

An alloy of copper and silver containing 60% pure silver is mixed with 
another alloy, also of copper and silver containing 65% pure silver. 
How much of each type is needed to produce 1.2 kg of alloy containing 
62% pure silver?

Agnes

Date: 06/09/99 at 11:47:10
From: Doctor Anthony
Subject: re: Re: An alloy of copper and silver

Let W = the weight of 60% alloy

We are told that the end mixture required is 1.2 kg of 62% pure 
silver. Therefore I assume that the weight of 60% alloy will be W kg 
(where W is to be found). If W is the weight of 60% alloy and if 1.2 
kg is the total weight of both alloys it follows that (1.2-W) kg is 
the weight of 65% alloy.

Then   1.2-W = the weight of 65% alloy

This ensures that TOTAL weight is 1.2 since  W + (1.2-W) = 1.2

Then the equation to satisfy is

   0.6W + 0.65(1.2 -W) = 0.62 x 1.2

This equation gives the weight of pure silver in alloy (1) and alloy 
(2) and equates it to the weight of pure silver in the final mixture.

   0.6W is the weight of pure silver in W kg of alloy (1)

   0.65(1.2-W) is the weight of pure silver in (1.2-W) kg of alloy (2)

   0.62 x 1.2 = 0.744 is the weight of pure silver in 1.2 kg of the 
      final mixture.

Simplifying the above equation we get:
   
   0.6W + 0.78 - 0.65W = 0.744   This gives the weights of pure silver
 
               - 0.05W = - 0.036   collecting terms

Divide both sides by -0.05 to find W

                     W = 0.72 kg

We require 0.72 kg of alloy (1) and then  1.2 - 0.72 = 0.48 kg of 
alloy (2).

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/