please help on math?

Question

When one ounce of water is added to a mixture of alcohol and water, the new mixture is 20% alcohol. When one ounce of alcohol is added to the new mixture, the result is 33 1/3% alcohol. What was the percentage of alcohol in the mixture.

Answer 1

let original ratio be=x:y(i.e water:alch)
now,alch content : y/(x+y+1) =1/5
with adition of 1 ounce of alch. the content of alch. : y+1/(x+y+2)=1/3
solving the eq. we get y=1 & x=3
therefore alch content (original)=1/4=25%

Answer 2

Let a be the amount of alcohol in the initial mixture and m the amount of mixture at the beginning of the process.
Then:

"When one ounce of water is added to a mixture of alcohol and water, the new mixture is 20% alcohol" translates into the ecuation:

a = (20/100) (m + 1)

"When one ounce of alcohol is added to the new mixture, the result is 33 1/3% alcohol" translates into:

a + 1 = (33,33/100) (m + 2)

Solving the system, we have: m = 4, a = 1
So, initialy the mixture weighed 4 ounces and there was 1 ounce of alcohol therefore the percentage of alcohol was 25%

Edit: Before I had read that the first ounce added was of alcohol too, sorry, it was too early in the morning when I did it =). Now it's corrected

Answer 3

Let the quantity of alcohol and water in the original mixture be x oz and y oz respectively
therefore,by the first condition.
x/(y+1)=20/(100-20)
x/(y+1)=1/4
=>4x=y+1 [by cross-multiplication]
=> 4x-y=1..........(1)
from the second condition,we get
(x+1)/y=33 1/3/(100-33 1/3)
=>(x+1)/y=1/2
=>2(x+1)=y
=>2x+2=y
=>2x-y= -2........(2)
subtracting eqn 2 from eqn1
y=3
Putting the value of y in eqn 1 we get
4x=4
x=1
Therefore in the original mixture quantity of alcohol is 1 oz in (1+3) or 4 oz of mixture
therefore Percentage of alcohol
=1/4*100
=25%