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%
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 me reduce that to an equation for you: 10x + 80(1-x) = 30; solve for x.
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.
1 percent alloy. The gold would be soft.
Pewter is 85% - 99% tin. Other constituents of the alloy are copper, antimony, bismuth and lead. There is no zinc in pewter
200 ounces.
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%
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.
Let me reduce that to an equation for you: 10x + 80(1-x) = 30; solve for x.
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.
Let ( x ) be the amount of the 40% copper alloy needed. The equation based on the copper content is: [ 0.40x + 0.70(500) = 0.50(x + 500) ] Solving this equation, we find: [ 0.40x + 350 = 0.50x + 250 ] [ 100 = 0.10x ] [ x = 1000 ] Thus, you need to mix 1000 ounces of the 40% copper alloy with 500 ounces of the 70% copper alloy to obtain an alloy that is 50% copper.
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.
The question can be rewritten as the following equation: (20%x + 60%100)/(100+x) = 30% Where x is the amount of 20% copper you need to add. This equation can be solved by first multiplying both sides by (100+x) to get: 20%x + 60%100 = 30%100 + 30%x Now 30%100 can be subtracted from both sides 20%x + 30%100 = 30%x Now 20%x can be subtracted from both sides 30%100 = 10%x Now both sides can be divided by 10% 300 = x Thus you need to add 300 ounces of 20% copper alloy in order to get an alloy that totals 30% copper.
Bronze.
In a brass alloy containing 75 percent copper and 25 percent zinc, copper is the solute. The solute is the component of a solution that is present in a lesser amount and is dissolved into the solvent, which is the component present in a larger amount. Since copper is being dissolved into zinc to form the brass alloy, copper is considered the solute.