x*(70/100)+y*(30/100)=100*(45/100) .7x+.3y=45 x+y=100 y=100-x .7x+.3(100-x)=45 .7x+30-.3x=45 .4x=15 x=15/.4=37.5 gallons y=100-x=100-37.5=62.5 gallons
50% alcohol Proof ÷ 2 = % alcohol
100% alcohol is more effective because of the more alcohol concentration in the drink.
You mix the pure alcohol with something else, for example, water.
Let a be the volume of 100% antifreeze then 55 - a is the volume of 10% antifreeze. 100a + 10(55 - a) = 20 x 55 = 1100 100a + 550 - 10a = 1100 90a = 550 a = 55/9 = 61/9 so 55 - a = 55 - 61/9 = 488/9. The mix is 61/9 gallons (6.111) 100% antifreeze and 488/9 gallons (48.889) 10% antifreeze.
Putting 100% "straight" antifreeze in your car's radiator will expose the cooling system to freezing at a much higher temperature than would the appropriate and recommended mix of roughly 50/50 antifreeze and water. Antifreeze works best WITH water.
No , antifreeze should be mixed with preferably distilled water at a 50 / 50 mix usually . If you live in a colder climate up to 60 % antifreeze is O.K.
They sell an 80 proof(40% alcohol) and a 100 proof(50% alcohol).
it means that for every 100 ml of the drink in question 18 ml of that 100 ml will be alcohol
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.
50% alcohol Proof ÷ 2 = % alcohol
You will never get it there. However much 50% you have, any amount of 20% in the mix will reduce the total below 50%. So, throw away the 20% and just use the 50% on its own.