Please help me with this problem:
A bourbon that is 51 proof is 25.5% alcohol by volume while one that is 82 proof is 41% alcohol. How many liters of 51 proof bourbon must be mixed with 1.0 L of 82 proof bourbon to produce a 66 proof bourbon? Give answer to 3 decimal places.
Thanks so much! :)
amount of final alcohol=sum of alcohol added.
V51+V82=V66
But
.255*V51+.41*V82=.33*V66
Now it is given that V82 is 1 liter.
You have two equations, with two unknowns (V51, and V66)
I would put the expression for V66 from the first equation (V51+V82) or V51+1 into the second equation for V66, then solve for V51
I am not understanding how to do this problem. Basically, whst I need to know is how many liters of 51 proof bourbon must be mixed with 1.0L of 82 proof bourbon to produce a 66 proof bourbon. This is am extra credit assignment, and I really need to know the answer. Please help!
You know that if you use x liters of 51 proof bourbon, then you wind up with x+1 liters of mixture. The amount of alcohol in each part must add up to the amount of alcohol in the final mixture.
Naturally, the amount of alcohol in each part is "proof"/2 * volume, since 1% alcohol is 2 proof.
So, expressing % as decimals,
.255 x + .41 * 1 = .33(x+1)
x = 16/15
Note that since the same scaling factor (2proof/1%) was applied to all terms, we could have just used the proof value directly:
51x + 82 = 66(x+1)
x = 16/15
To solve this problem, let's denote:
x = the number of liters of 51 proof bourbon to be mixed
Now, let's break down the problem step by step:
Step 1: Calculate the total alcohol in each bourbon
Alcohol in 51 proof bourbon = 25.5% of x liters = 0.255x
Alcohol in 82 proof bourbon = 41% of 1 liter = 0.41 * 1 = 0.41
Step 2: Determine the total volume of the mixture
The total volume of the mixture is given by adding the volumes of the two bourbons: x + 1 liter.
Step 3: Calculate the alcohol content of the mixture
To find the alcohol content of the mixture, divide the total alcohol by the total volume and multiply by 100 to express it as a percentage:
Alcohol content of the mixture = (0.255x + 0.41) / (x + 1) * 100
Step 4: Set up an equation
We want the resulting bourbon to have a proof of 66. Since proof is 2 times the alcohol content, we have:
Proof of the mixture = 2 * (Alcohol content of the mixture) = 66
So, our equation is:
66 = 2 * [(0.255x + 0.41) / (x + 1) * 100]
Now, let's solve for x:
66 = 2 * [(0.255x + 0.41) / (x + 1) * 100]
Divide both sides by 2:
33 = (0.255x + 0.41) / (x + 1) * 100
Multiply both sides by (x + 1) / 100:
33(x + 1) = 0.255x + 0.41
Expand and simplify:
33x + 33 = 0.255x + 0.41
Subtract 0.255x and 33 from both sides:
32.745x = -32.59
Divide both sides by 32.745:
x = -32.59 / 32.745
Calculating this value gives us x ≈ -0.995
Since it doesn't make sense to have a negative volume, there is no solution to this problem.
Therefore, it is not possible to mix the bourbons to produce a 66 proof bourbon.