SOLUTION: A merchant blends tea that sells for $3.50 an ounce with tea that sells for $2.70 an ounce to produce 80 oz of a mixture that sells for $3.00 an ounce. How many ounces of each type

Question 1092137: A merchant blends tea that sells for $3.50 an ounce with tea that sells for $2.70 an ounce to produce 80 oz of a mixture that sells for $3.00 an ounce. How many ounces of each type of tea does the merchant use in the blend?

______oz (at $3.50/oz)

______oz (at $2.70/oz)

Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
30____oz (at $3.50/oz)

50____oz (at $2.70/oz)

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You can start using x for how much of the cheaper tea, and 80-x is how much of the more expensive tea.
, accounting for costs.
Solve for x. Results shown above.

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The traditional algebraic method for solving mixture problems is of course fine. An answer has already been posted getting you started on that method for solving your problem. But for a much faster and easier method, try this:

(1) Where the cost per ounce of the mixture lies relative to the costs per ounce of the two ingredients exactly determines the ratio in which the two ingredients are to be mixed.

(2) The $3.00 per ounce cost of the mixture is 3/8 of the way from the $2.70 per ounce cost of the lower priced tea to the $3.50 per ounce price of the more expensive tea. That is,

(3) Since the $3 per ounce cost of the mixture is 3/8 of the way from $2.70 to $3.50, 3/8 of the mixture must be the more expensive tea.

3/8 of the total 80 ounces is 30 ounces, so you need 30 ounces of the more expensive tea and 80-30=50 ounces of the less expensive tea.
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