Wiki User
∙ 11y agoLet me reduce that to an equation for you:
10x + 80(1-x) = 30; solve for x.
Wiki User
∙ 11y agoSince the percentages of copper in the two components to be mixed are symmetric about the desired result, the answer is that the same amounts should be used. 600 ounces of the 30% copper.
900 ounces. Since this contains 20% copper, The copper content will be 180 ounces. The original 300 ounces contain 30% copper which is also 180 ounces. Hence in the resulting mixture of 1200 ounces (300+900), the total copper is 360 ounces (180+180). Hence the copper content of resulting mixture is 360/1200 which is 30%
Designate the weight in ounces of the first alloy, containing 40 percent copper, as w. Then, from the problem statement and the fact that percentages can be converted to decimals by dividing by 100, 0.40w + (0.80)(400) = 0.60(400 + w). Applying the usual methods of algebra, multiplying out results in: 0.40 w + 320 = 240 + 0.60w; transposing like terms with sign change and collecting results in: (0.40 - 0.60)w = 240 -320; or -0.20 w = -80, or w = 400.
the answer is alloy
Circulation-strike half dollars dated 1971 and later contain about 0.95 gm of nickel. The rest is copper, either in the pure copper core or mixed in the alloy used for the outer cladding. Older halves don't contain any nickel. Those dated 1965-70 are made of 40% silver and 60% copper, and halves (as well as dimes and quarters) dated 1964 and earlier are 90% silver alloy.
How much of an alloy that is 10% copper should be mixed with 400 ounces of an alloy that is 70% copper in order to get an alloy that is 20%
200 ounces.
Since the percentages of copper in the two components to be mixed are symmetric about the desired result, the answer is that the same amounts should be used. 600 ounces of the 30% copper.
Let x be the amount of 20% copper alloy needed. The equation is: 0.20x + 0.50(200) = 0.30(x + 200) Solving for x, we find that 100 ounces of the 20% copper alloy should be mixed with 200 ounces of the 50% copper alloy to get the desired 30% copper alloy.
900 ounces. Since this contains 20% copper, The copper content will be 180 ounces. The original 300 ounces contain 30% copper which is also 180 ounces. Hence in the resulting mixture of 1200 ounces (300+900), the total copper is 360 ounces (180+180). Hence the copper content of resulting mixture is 360/1200 which is 30%
Designate the weight in ounces of the first alloy, containing 40 percent copper, as w. Then, from the problem statement and the fact that percentages can be converted to decimals by dividing by 100, 0.40w + (0.80)(400) = 0.60(400 + w). Applying the usual methods of algebra, multiplying out results in: 0.40 w + 320 = 240 + 0.60w; transposing like terms with sign change and collecting results in: (0.40 - 0.60)w = 240 -320; or -0.20 w = -80, or w = 400.
Let x be the amount of the 14% copper alloy. Then, the amount of the 23% copper alloy is 90 - x. Setting up the equation (0.14x + 0.23(90 - x) = 0.18*90) and solving for x gives x = 30. Therefore, you need 30 ounces of the 14% copper alloy.
To create an alloy with 40% copper, you would need to mix equal parts of the 20% alloy with the 70% alloy. This means you need to mix 50 ounces of the 20% alloy with 50 ounces of the 70% alloy to achieve the desired 40% copper content in the final alloy.
The industrial name for gold mixed with copper is called "rose gold." It gets its pinkish hue from the copper alloy mixed with gold.
Copper is typically mixed with gold to create a red hue. The higher the copper content in the alloy, the redder the gold will appear.
Zinc is mixed with copper to make brass. The proportions of copper and zinc can vary depending on the type of brass being produced.
If aluminum is mixed with copper, they can form an alloy called aluminum bronze. This alloy exhibits improved strength and corrosion resistance compared to pure aluminum or copper. The properties of the alloy can be tailored by adjusting the ratio of aluminum to copper.