SOLUTION: i have no idea how to solve this, can someone please help me? a silversmith has two alloys. the first alloy is 28% silver and the second is 60% silver. how many grams of each sho
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Question 113028: i have no idea how to solve this, can someone please help me? a silversmith has two alloys. the first alloy is 28% silver and the second is 60% silver. how many grams of each should be mixed to produce 80 grams of an alloy that is 52% silver? So i should have came up with 2 answers right? stumped! Found 2 solutions by Earlsdon, josmiceli:Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Let x = the number of grams of 28% Ag alloy. (Note: Ag = chemical symbol for silver)
and (80-x) = the number of grams of 60% Ag alloy.
The sum of these two amounts equals the 80 grams of 52% Ag alloy.
Basically, we will add the amounts of silver in the two alloys to come up with the amount of silver in the 80 grams of silver alloy.
The amount of silver in the x grams of the 28% Ag alloy can be written:
x(0.28)
The amount of silver in the (80-x) grams of 60% Ag alloy can be written:
(80-x)(0.6), so if we add these two we should get 80 grams of 52% Ag alloy which contains 80(0.52) grams of silver.
x(0.28)+(80-x)(0.6) = 80(0.52) Simplify and solve for x.
0.28x + 48 - 0.6x = 41.6 Combine like-terms.
-0.32x = -6.4
x = 20 and (80-x) = 60
So, the silversmith would need 20 grams of the 28% Ag alloy and 60 grams of the 60% Ag alloy to obtain 80 grams of 52% Ag alloy.
Check: Add the number of grams of silver in the 28% Ag alloy to the number of grams of silver in the 60% Ag alloy to get the number of grams of silver in the 80 grams of 52% Ag alloy.
20(0.28)+60(0.6) = 5.6+36 = 41.6
80(0.52) = 41.6
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