Math
posted by Elizabeth .
Aaron has two types do iodine in his lab. He has brand 213 iodine, which costs 36 cents per ounce. He also has brand 856 iodine, which costs 1.58$ per ounce. How many ounces of brand 213 should be used if he has 63 ounces of brand 856 and wants to make a mixture that costs approximately 0.53$ an ounce

the value of the parts must add up to the value of the whole.
If there are x oz. of brand 213, then
.36x + 1.58*63 = .53(x+63)
x = 389.12
Seems reasonable, since the cost of the mixture is much closer to that of the cheap brand; he'd need more of the cheap stuff than the costly one.
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