Question: Metal alloy A contains 20% copper and metal alloy B contains45% copper

Question

Question: Metal alloy A contains 20% copper and metal alloy B contains45% copper. How many pounds of each alloy should be combined toproduce 100 pounds of a new alloy that contains 35% copper?
I need to know the equations, and how/why you put themtogether. I don't just want to answer, I can solve the equations onmy own! Metal alloy A contains 20% copper and metal alloy B contains45% copper. How many pounds of each alloy should be combined toproduce 100 pounds of a new alloy that contains 35% copper?
I need to know the equations, and how/why you put themtogether. I don't just want to answer, I can solve the equations onmy own!

Explanation / Answer

Let x = amount of alloy A in the final mix Let y = amount of alloy B in the final mix
We know that x + y = 100; from this we can say x = 100 -y
We also know that .2 x + .45 y = .35 * 100 Here Iconverted all the percents into decimals. In words, this saysthat the amount of copper in A is .2 times its weight. Theamount of copper in B is .45 times its weight. The end resultis the sum of these two and is .35 times its weight of 100 pounds. Lets multiply everything here by 100 so we haveintegers:
20x + 45 y = 3500; Now we substitute the equality for x(above) and get one equation in y alone:
20 (100 - y) + 45 y = 35; 2000 -20y +45y = 3500; 25y =1500; y = 60 pounds
x = 100 - y = 40 pounds,

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