How many gallons of a 30% alcohol solution must be mixed with 60 gallons of a 14%
solution to obtain a solution that is 20% alcohol
number of gallons of the 30% stuff --- x
.3x + .14(60) = .2(60+x)
30x + 14(60) = 20(60+x)
30x + 840 = 1200 +20x
10x = 360
x = 36
.3*V+.14*60=.2*(V+60)
solve for V
(2x2y2-3xy+4)(4xy2)
A mixture of alcohol and water contains a total of
52
52 oz of liquid. There are
13
13 oz of pure alcohol in the mixture. What percent of the mixture is water? What percent is alcohol?
A mixture of alcohol and water contains a total of
52 oz of liquid. There are
13 oz of pure alcohol in the mixture. What percent of the mixture is water? What percent is alcohol?
To determine the number of gallons of a 30% alcohol solution needed to mix with 60 gallons of a 14% alcohol solution to obtain a 20% alcohol solution, we can use the concept of a weighted average.
Let's consider that x gallons of the 30% alcohol solution need to be mixed with the 60 gallons of the 14% alcohol solution.
First, let's calculate the amount of alcohol in the 30% solution. Since it is a 30% alcohol solution, we have 30/100 * x gallons of alcohol.
Next, let's calculate the amount of alcohol in the 14% solution. Since it is a 14% alcohol solution, we have 14/100 * 60 = 8.4 gallons of alcohol.
Now, we can set up an equation to represent the overall alcohol content in the final solution:
(30/100) * x + 8.4 = (20/100) * (60 + x)
This equation represents the total amount of alcohol in the 30% solution plus the total amount in the 14% solution, which should equal the total amount of alcohol in the final solution.
Let's solve this equation to find the value of x:
(30/100) * x + 8.4 = (20/100) * (60 + x)
Multiply both sides of the equation by 100 to remove the percentages:
30x + 840 = 20(60 + x)
Distribute on the right side:
30x + 840 = 1200 + 20x
Combine like terms:
30x - 20x = 1200 - 840
10x = 360
Divide both sides by 10:
x = 36
Therefore, you would need to mix 36 gallons of the 30% alcohol solution with the 60 gallons of the 14% alcohol solution to obtain a solution that is 20% alcohol.