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27 February, 08:00

How many ounces of iodine worth 30 cents an ounce must be mixed with 50 ounces of iodine worth 18 cents an ounce so that the mixture can be sold for 20 cents an ounce?

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Answers (2)
  1. 27 February, 11:11
    0
    10 ounces
  2. 27 February, 11:31
    0
    10 ounces

    Step-by-step explanation:

    The total amount of ounces will the the amount worth 30 cents an ounce (x) plus the 50 ounces worth 20 cents an ounce. This amount is x + 50. The total worth of the mixture should be the worth of the unknown quantity (30 cents an ounce * x ounces) plus the worth of the known quantity (18 cents an ounce * 50 ounces). The total worth is then 30x + 18 (50).

    Selling the mixture at 20 cents an ounce means the total worth divided by the total amount equals 20 cents per ounce:

    [30x + 18 (50) ] cents / [x + 50] ounces = 20 cents/ounce

    Multiplying both sides by (x + 50) ounces yields

    (30x + (18) (50) cents = 20 (x + 50) cents

    which solves to (30x + 900) cents = (20x + 1000) cents

    now we use algebra to solve for the value of x

    (30x + 900) - 900 = 20x + 1000 - 900 - -> 30x = 20x + 100 - ->

    30x - 20 x - -> 20x - 20 x + 100 - -> 10x = 100 - -> x = 10 ounces
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