4 ounces
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.
6%
There are 3 litres of alcohol in your starting mixture of 4 litres. If you add 6 litres of water you will have 3 litres of alcohol in a total of 10 litres. This is the required strength.
The question is best solved using basic algebra. You need 20 gallons of 32% alcohol. This will contain 0.32*20 = 6.4 gallons of pure alcohol. Now suppose you have X gallons of 25% alcohol in the mixture. That contains 0.25X gallons of pure alcohol. Also, since you have 20 gallons in total, you must have 20-X gallons of the 35% alcohol. This will contain 0.35*(20-X) = 7 - 0.35X gallons of pure alcohol. Then, the total amount of pure alcohol is 0.25X + 7 - 0.35X = 7 - 0.1X gallons. So you have 7 - 0.1X = 6.4 or 0.6 = 0.1X or X = 6. So the answer is 6 gallons of 25% alcohol and 14 gallons of the stronger stuff!
There is no way to calculate that with the information provided. We need to know the volume of each.
4 ounces
It is usually a mixture of alcohol and water in which 40%, by volume is alcohol.
16%
No.
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.
Standard.4.7%/5.0%
do you mean alcohol by volume - then Molson Canadian in 5%
I believe it's 14ml in a liter.
It could mean blood alcohol content is 0.1%.
6%
2.8 percent alcohol