SOLUTION: A metallurgist needs to make 12 grams of an alloy containing 60% gold. He is going to melt and combine one metal that is 90% gold with another metal that is 40% gold. How much of e
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-> SOLUTION: A metallurgist needs to make 12 grams of an alloy containing 60% gold. He is going to melt and combine one metal that is 90% gold with another metal that is 40% gold. How much of e
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Question 1165355: A metallurgist needs to make 12 grams of an alloy containing 60% gold. He is going to melt and combine one metal that is 90% gold with another metal that is 40% gold. How much of each should he use? Found 2 solutions by josgarithmetic, greenestamps:Answer by josgarithmetic(39620) (Show Source):
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,...make 12 grams of an alloy containing 60% gold. He is going to melt and combine one metal that is 90% gold with another metal that is 40% gold. How much of each s....
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H=90
L=40
T=60
M=12
v=unknown amount of the H% alloy
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This shows a way to find the FRACTION of the alloy-to-prepare which is the H% alloy.
Substitute the given values and evaluate v and M-v.
That other tutor loves that formula with all those variables representing the numbers in the problem.
I dislike memorizing formulas with lots of variables that I can derive in a couple of seconds....
A traditional algebraic approach would be something like this:
x grams of 40%, plus (12-x) grams of 90%, equals 12 grams of 60%:
The equation is easily solved using basic algebra.
If a formal algebraic solution is not required, here is a much faster path to the answer (for ANY 2-part mixture problem like this).
The target 60% is 2/5 of the way from 40% to 90%. (Picture the three percentages on a number line, if it helps -- 60 is 20/50 = 2/5 of the way from 40 to 90....)
Therefore, 2/5 of the mixture needs to be the higher percentage metal.
ANSWER: 2/5 of 12 grams, or 4.8 grams, of 90%; the other 7.2 grams of 40%.
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