sanju09 wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

[spoiler]Source: gmatclub.com[/spoiler]

Let the original volume be 100
water =50, alcohol =50
x units is replaced with the new solution.
of this x units, x/2 is alcohol and x/2 is water.
Now, this is replaced with x/4 alcohol and 3x/4 water.
Resulting total alcohol = 50-x/2+x/4 = 30
x=80%

Hope this helps!!

sanju09 wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
(A) 3%
(B) 20%
(C) 66%
(D) 75%
(E) 80%

[spoiler]Source: gmatclub.com[/spoiler]

Another approach for mixture problems is called alligation.

If we have two solutions of X% (higher) and Y% (lower), and we want to mix them to get Z%:

Proportion needed of X = Z-Y
Proportion needed of Y = X-Z

So if we have 50% alcohol and 25% alcohol, and we want a mixture of 30% alcohol:

Proportion needed of 50% alcohol = 30-25 = 5
Proportion needed of 25% alcohol = 50-30 = 20
So (25% alcohol)/total = 20/(5+20) = 20/25 = 80/100 = 80%.

Thus, 80% of the original alcohol is being replaced.

The attached .pdf illustrates the alligation method.

The correct answer is E.