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/
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