SOLUTION: This is a story problem: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelfths pure gold. How many
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-> SOLUTION: This is a story problem: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelfths pure gold. How many
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Question 170296This question is from textbook Introductory Algebra
: This is a story problem: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelfths pure gold. How many ounces of each should be melted and mixed in order to obtain a 6-oz mixture that is two-thirds pure?
Could you please lead me in the right direction for solving this problem? This question is from textbook Introductory Algebra
You can put this solution on YOUR website! This is a story problem: A goldsmith has two alloys that are different purities of gold. The first is three-fourths pure gold and the second is five-twelfths pure gold. How many ounces of each should be melted and mixed in order to obtain a 6-oz mixture that is two-thirds pure?
Could you please lead me in the right direction for solving this problem?
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In the first alloy, the gold is (3/4) of the alloy.
In the 2nd alloy, the gold is 5/12 of the alloy.
In the resulting alloy, 1/3 of 6 oz is gold, which is 4 oz.
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He can use x ounces of the 1st, and 6-x of the 2nd alloy.
So, x*(3/4) + (6-x)*(5/12) = 4
3x/4 + (6-x)*(5/12) = 4
Multiply by 12 to eliminate fractions
9x + 5*(6-x) = 48
9x + 30 -5x = 48
4x = 18
x = 4.5 (oz of the 1st alloy)
6-4.5 = 1.5 (oz of the 2nd alloy)
