Question

What is the percent of alcohol, in a mixture made by mixing 5 liters of 15%, 1 liter of 40% and 1 liter of 85% alcohol?

Answer 1

Answer:

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Answer 2

Answer:

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = y

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% y

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% ySubstitute y by x + 40 in the last equation to obtain.

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% ySubstitute y by x + 40 in the last equation to obtain.Solve for x.

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% ySubstitute y by x + 40 in the last equation to obtain.Solve for x.x = 80 liters

Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hencex + 40 = yWe shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.20% x + 50% * 40 = 30% ySubstitute y by x + 40 in the last equation to obtain.Solve for x.x = 80 liters80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.

Explanation:

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