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
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! :)
3 answers
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
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
0 answers