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How many gallons each of 25% alcohol and 10% alcohol should be mixed to obtain 15 gal of 21% alcohol?Gallons ofPure AlcoholGallons ofSolutionХy15Percent(as a decimal)25% = 0.2510% = 0.121% =How many gallons of 25% alcohol should be in the mixture? gal

Sagot :

Answer

11 gallons of the 25% alcohol is required for the mixture.

4 gallons of the 10% alcohol is required for the mixture.

Explanation

Let the number of gallons of 25% alcohol required be x

Let the number of gallons of 10% alcohol required be y

The total amount of gallons required is 15 gallons. In mathematical terms,

x + y = 15

The alcohol content of the 15 gallons is to be 21%.

21% of 15 gallons = 3.15 gallons

From the first statements,

Let the number of gallons of 25% alcohol required be x

Let the number of gallons of 10% alcohol required be y

25% of x gallons = 0.25x gallons

10% of y gallons = 0.10y gallons

0.25x + 0.10y = 3.15

We can then bring these two equations together to solve simultaneously

x + y = 15

0.25x + 0.10y = 3.15

Solving this simultaneously with the calculator, we obtain

x = 11 gallons

y = 4 gallons

Hope this Helps!!!