what you are trying to explain?

You have calculated X. This is the question-

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%


Even I was saying this the question didn't asked what %age of total solution was replaced, but the question asked what %age of original alcohol was replaced. X here is the amount of total solution replaced.

50- 0.5x = The amount of alcohol replaced, which is equal to 40 and is also 80% coincidentally because alcohol was 1/2 the Original mixture but neither yous nor the original solution addressed this.

Honchos, it is not 80% coincidentally. It has to be 80% because 80% of the solution was replaced. Look at my explanation given in the post above:

"we find the value of x, which is the fraction of the 50% SOLUTION that was replaced. We found that 4/5th of the original solution was replaced. Note that we assume solutions to be homogeneous. This means that if 4/5th of the solution was replaced, 4/5th of the original alcohol was replaced and 4/5th of the original water was replaced.
Say you had 100 ml solution with 50 ml each of water and alcohol. You removed 4/5th of this solution i.e. 80 ml. So you removed 40 ml of each -water and alcohol i.e. you removed 4/5th of alcohol and 4/5th of water."
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hi madmax1000,
the Q asks us how much original alcohol is replaced...
as found by you and taken by you....
original alcohol =50l and after removing 80 l, 40 l of original alcohol has been replaced..
what is later added is not the replaced(original) alcohol but the replacement alcohol and what has been asked is the original alcohol, which has come down from 50 to 10...
so replaced alcohol=40/50..
hope it helped..
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yes thanks i was deducting the amount of alcohol that was taken out of the alcohol and replaced, because i thought that doesnt count..

Hi
Can anyone explain the meaning of the question?? I've been trying to comprehend..say the solution is 100 ml
50 ml water and 50 ml alcohol..
now it says half water/half alcohol is replaced... as in each component is 25ml each and so 25+25=50? I dint understand this.can anyone help please?

Hi Karishma,

Thanks so much, I did understand the question now.. Original composition = 50% water + 50% alcohol ; after replacement 25% alcohol + 75 % water = 30% alcohol..
Now if I apply allegation:
old alcohol new alcohol
50 25
30
5 20
old alcohol/new alcohol = 1/4
How can I proceed after this...? Also I am finding it difficult to construct the same through equations..could you please help?

Nice explanation indeed. But I want to know what does 50% represents? Is it water or alcohol?

Thank you. I did not understand the question at first. Thanks a ton.

Hi VeritasPrepKarishma

now I got totally confused

why here you write this solution w1/w2 = (30 - 25)/(50 - 30) = 1/4 here you follow weighted average formula, ok

A1 = 50
A2 = 25
Aavg = 30

in this case i understand, since 25 is closer to 30 we take A2 = 25 meaning it 25 has more volume,

But in this link
https://gmatclub.com/forum/weighted-ave ... l#p2003243

you write in this way
\(\frac{w_1}{w_2} = \frac{(A_2 – A_{avg})}{(A_{avg} – A_1)} = \frac{(50 – 30)}{(30 – 25)} = \frac{4}{1}\)

here you take A1 = 25 , A2 = 50 and Aavg = 30 - Why ?

can you explain the difference between the first and the second solution?

thank you

Hi All,

We're told that we're going to mix a certain amount of a 50% alcohol solution with a certain amount of a 25% alcohol solution and end up with a 30% alcohol solution. This is an example of a 'Weighted Average' question. Here's how we can set up that calculation using Algebra:

A = number of ounces of 50% solution
B = number of ounces of 25% solution
A+B = total ounces of the mixed solution

(.5A + .25B)/(A+B) = .3

.5A + .25B = .3A + .3B
.2A = .05B
20A = 5B
4A = B
A/B = 1/4

This means that for every 1 ounce of solution A, we have 4 ounces of solution B.

To answer the question that's asked, imagine that you have 5 ounces of solution A. We're told to REPLACE 4 ounces of it with 4 ounces of solution B (thus, we'll end up with the 30% mixture that we're after).

Since 4 ounces of the 5 ounces were replaced, 4/5 of the original alcohol and 4/5 of the original water was replaced.

4/5 = 80%

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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