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40 fl oz of the 16% solution and 24 of the other.

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7y ago

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How many ounces of a ten percent alcohol solution must be mixed with 6 ounces of a fifteen percent alcohol solution to make a thirteen percent alcohol solution?

4 ounces


A pharmacist needs to obtain a 70 percent alcohol solution How many ounces of a 30 percent alcohol solution must be mixed with 40 ounces of an 80 percent alcohol solution to obtain 70 percent alcohol?

(Note: This answer assumes that the "ounces" specified are avoirdupois or other weight ounces and that percentages are by weight; otherwise possible volume changes on dilution must by considered.) The weight of pure alcohol in each solution is the product of the percentage and the total weight of the solution. Therefore, designating the unknown weight of 30 % alcohol as w, from the problem statement 0.30w + 0.80(40) = 0.70(w + 40), or 0.30w + 32 = 0.70w + 28, or 32 - 28 = w(0.70 - 0.30) or w = 4/0.40 = 10 ounces of 30 % alcohol.


How many ounces of a 25 percent alcohol solution must be mixed with 9 ounces of a 30 percent alcohol solution to make a 28 percent alcohol solution?

Let a be the number of ounces of 25% alcohol required. Then, 25a + (30x9) = 28(9 + a) 25a + 270 = 252 + 28a 3a = 18 a = 6 Then 6 ounces of 25% alcohol + 9 ounces of 30% alcohol produces 15 ounces of 28% alcohol.


How many ounces of 15 percent percent alcohol must be mixed with 23 percent alcohol solution to make 100 ounces?

Let x be the ounces of 15% alcohol solution. The amount of alcohol in the 15% solution is 0.15x, and the amount of alcohol in the 23% solution is 0.23(100 - x). Setting up the equation 0.15x + 0.23(100 - x) = 0.15(100) solves for x, which is approximately 38.5 ounces of the 15% alcohol solution needed.


How many ounces of the 50 percent alcohol solution should be mixed with the 1 percent alcohol solution to make 8 oz of the 20 percent alcohol solution?

Let x = ounces of 50% solution, and y = ounces of 1% solution. So that we have: 0.5x + 0.01y = 8(0.2) which is a linear equation in two variables, meaning there are infinitely many choices of mixing those solutions.


How many ounces of a 35 percent alcohol solution must be mixed with 14 ounces of a 40 percent alcohol solution to make a 37 percent alcohol solution?

.35x+.40*14=0.37(x+14) .35x=.37x-5.6+5.18 .35x=.37x-.42 -.02x=.42 x=21 check 21*.35+14*.4=7.35+5.6=12.95 12.95/(14+21)=.37 so you would need 21 ounces of 0.35 mixed in with 14 ounces of 40 % to get a 37% solution of which you would have 35 ounces


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Do 1.5 ounces of liquor have more alcohol than 12 ounces of beer?

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Is four ounces of wine the same as four ounces of whiskey?

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