SOLUTION: combine an alloy that costs $4.30 and ounce with an alloy that cost $1.80 an ounce. Hou many ounces of each were used to make a mixture of 200 ounces costing $2.50 per ounce?

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Question 416360: combine an alloy that costs $4.30 and ounce with an alloy that cost $1.80 an ounce. Hou many ounces of each were used to make a mixture of 200 ounces costing $2.50 per ounce?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = ounces of $4.30/oz alloy needed
Let b = ounces of $1.80/ oz alloy needed
In words:
(cost of 1st alloy + cost of 2nd alloy) / (total ounces of both) = cost/ounce
(1) +%284.3a+%2B+1.8b%29%2F200+=+2.5+
(2) a+%2B+b+=+200
-----------------
(1) +%284.3a+%2B+1.8b%29%2F200+=+2.5+
(1) +4.3a+%2B+1.8b+=+500+
(1) 43a+%2B+18b+=+5000
-----------------
Multiply both sides of (2) by 18 and
subtract from (1)
(1) 43a+%2B+18b+=+5000
(2) -18a+-+18b+=+-3600
+25a+=+1400
+a+=+56
and,
b+=+200+-+56
b+=+144
56 ounces of $4.30/oz alloy are needed
144 ounces of $1.80/ oz alloy are needed
check answer:
(1) +%284.3a+%2B+1.8b%29%2F200+=+2.5+
(1) +%284.3%2A56+%2B+1.8%2A144%29%2F200+=+2.5+
(1) +%284.3%2A56+%2B+1.8%2A144%29%2F200+=+2.5+
+240.8+%2B+259.2+=+500+
+500+=+500+
OK