Question

How many gallons each of 20​% alcohol and 10​% alcohol should be mixed to obtain 10gal of 11​% ​alcohol?

Answer 1

To obtain a 10-gallon mixture with 11% alcohol, you should mix 1 gallon of 20% alcohol and 9 gallons of 10% alcohol.

Step-by-step explanation:

To solve this problem, we can set up a system of equations.

Let x represent the volume of the 20% alcohol and y represent the volume of the 10% alcohol.

We know that the total volume of the mixture is 10 gallons, so we have the equation x + y = 10.

We also know that the percentage of alcohol in the mixture is 11%, so we can write the equation (0.20x + 0.10y) / 10 = 0.11.

Simplifying this equation, we get 0.20x + 0.10y = 1.1.

We now have a system of equations: x + y = 10 and 0.20x + 0.10y = 1.1.

We can solve this system using substitution or elimination. Let's use the elimination method.

Multiply the second equation by 10 to eliminate the decimal: 2x + y = 11.

Subtract the first equation from the second equation to eliminate x: 2x + y - (x + y) = 11 - 10.

Simplify: x = 1.

Substitute x = 1 into the first equation to find y: 1 + y = 10.

Solve for y: y = 9.

Therefore, you should mix 1 gallon of the 20% alcohol with 9 gallons of the 10% alcohol to obtain 10 gallons of an 11% alcohol mixture.

Answer 2

20​% alcohol = 1 gal

10​% alcohol = 9 gal

Step-by-step explanation:

Let

x = 20​% alcohol

y = 10​% alcohol

x + y = 10 (1)

0.20x + 0.10y = 10*0.11

0.20x + 0.10y = 1.1 (2)

From (1)

x = 10 - y

Substitute x = 10 - y into (2)

0.20x + 0.10y = 1.1

0.20(10 - y) + 0.10y = 1.1

2 - 0.20y + 0.10y = 1.1

- 0.20y + 0.10y = 1.1 - 2

-0.10y = -0.9

y = -0.9/-0.10

y = 9 gal

substitute y = 9 into (1)

x + y = 10

x + 9 = 10

x = 10 - 9

x = 1 gal

20​% alcohol = 1 gal

10​% alcohol = 9 gal