SOLUTION: You have a vast quantity of yellow brass - 67% copper and 33% zinc - and red brass - 85% copper and 15% zinc - on hand. Someone wants an order of 500 tons of brass containing 80% c
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Question 262332: You have a vast quantity of yellow brass - 67% copper and 33% zinc - and red brass - 85% copper and 15% zinc - on hand. Someone wants an order of 500 tons of brass containing 80% copper and 20% zinc. How many of each kind of brass is needed to be melted together to fill this person's order? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Hi---
This offers a little different twist to the mixture problems. I started working on this problem a couple of days ago and got diverted. Thought I would go ahead and work it since it has not been worked:
Let x=amount of yellow brass needed
And let y=amount of red brass needed
Now we know that the amount of copper in the yellow brass (0.67x) plus the amount of copper in the red brass (0.85y) has to equal the amount of copper in the final mixture (0.80*500=400). The same argument holds for the zinc, so our two equations are:
0.67x+0.85y=400---------------------eq1
0.33x+0.15y=100----------------------eq2
I divide each term in eq1 by 0.85 and get:
0.788x+y=470.6--------------------------------eq1a
I divide each term of eq2 by 0.15 and get:
2.2x+y=666.7----------------------------------eq2a
subtract eq 1a from eq2a and get:
1.412x=196.1
x=138.9 tons---------------------amount of yellow brass needed
substitute into eq2a
2.2*138.9+y=666.7
y=361.12 tons-------amount of red brass needed
CK
138.9+361.12=500
500.02~~~500
