SOLUTION: How much of an alloy that is 20% copper should be mixed with 300 ounces of an alloy that is 80% copper in order to get an alloy that is 30% ​copper?
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Question 1092605: How much of an alloy that is 20% copper should be mixed with 300 ounces of an alloy that is 80% copper in order to get an alloy that is 30% copper? Found 2 solutions by jorel1380, greenestamps:Answer by jorel1380(3719) (Show Source):
You can put this solution on YOUR website! Let n be the amount of 20% alloy. Then:
0.2n+0.8(300)=0.3(300+n)
0.2n+240=0.3n+90
0.1n=150
n=1500 oz.
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You are mixing two ingredients to get the final mixture. Find expressions for the amounts of copper in each.
Ingredient A: x ounces of 20% copper
The amount of copper is 0.20(x)
Ingredient B: 300 ounces of 80% copper
The amount of copper is 0.80(300) = 240
Mixture: (x+300) ounces of 30% copper
The amount of copper in the mixture is 0.30(x+300)
Write and solve the equation that says the amount of copper in the mixture is the sum of the amounts in the two ingredients:
I'll let you finish...
Note: if you want to avoid calculations with decimals, multiply the whole equation by 10 before doing anything else.
But here is a faster way to solve mixture problems.
The ratio in which the two ingredients need to be mixed is exactly determined by how close the percentage of the mixture is to the percentages of each ingredient.
The mixture percentage, 30, is "5 times as close to 20 as it is to 80". That is, 80-30 = 50; 30-20 = 10; the 50 is 5 times the 10.
That means you need 5 times as much of the 20% copper as the 80% copper.
Given that there are 300 ounces of the 80% copper, you need 5*300 = 1500 ounces of the 20% copper.
If you finished solving the problem by the algebraic method shown above, that should be the answer you got.
