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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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1
Water .......... Alcohol ......... Total

50 .............. 50 ....................... 100

Let "x" quantity is removed

50 - 0.5x ....... 50 - 0.5x ............... 100-x

Same "x" quantity Alcohol is added of 25% concerntration

50 - 0.5x......... 50 - 0.5x + 0.25x .......... 100 - x + x

Post addition, Alcohol concentration is 30%

$$50 - 0.25x = \frac{30}{100} * 100$$

0.25x = 20

x = 80%

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If a portion of a half water/half alcohol mix is replaced  [#permalink]

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PareshGmat wrote:
Water .......... Alcohol ......... Total

50 .............. 50 ....................... 100

Let "x" quantity is removed

50 - 0.5x ....... 50 - 0.5x ............... 100-x

Same "x" quantity Alcohol is added of 25% concentration

50 - 0.5x......... 50 - 0.5x + 0.25x .......... 100 - x + x

Post addition, Alcohol concentration is 30%

$$50 - 0.25x = \frac{30}{100} * 100$$

0.25x = 20

x = 80%

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.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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honchos wrote:
PareshGmat wrote:
Water .......... Alcohol ......... Total

50 .............. 50 ....................... 100

Let "x" quantity is removed

50 - 0.5x ....... 50 - 0.5x ............... 100-x

Same "x" quantity Alcohol is added of 25% concentration

50 - 0.5x......... 50 - 0.5x + 0.25x .......... 100 - x + x

Post addition, Alcohol concentration is 30%

$$50 - 0.25x = \frac{30}{100} * 100$$

0.25x = 20

x = 80%

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.

Can you elaborate as to how the highlighted value was computed?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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honchos 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%

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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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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I am thinking about that for over one hour now...
I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?

0.5(100−x)+0.25x=0.3*100 → x=80

First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.

Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters.
In my opinion 30 liters of alcohol we being replaced by water so 3/5.

Can somebody explain that to me comprehensibly?
Thanks a lot.
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Joined: 02 Aug 2009
Posts: 8756
Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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I am thinking about that for over one hour now...
I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?

0.5(100−x)+0.25x=0.3*100 → x=80

First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.

Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters.
In my opinion 30 liters of alcohol we being replaced by water so 3/5.

Can somebody explain that to me comprehensibly?
Thanks a lot.

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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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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I am thinking about that for over one hour now...
I don't understand why 80% can be the answer to the question what percentage of the original alcohol was replaced?

0.5(100−x)+0.25x=0.3*100 → x=80

First we take out 80 liters (x) of 100 liters solution. 20 Liters remain, of which 10 are alcohol. Then we add to that 20 liters 80 liters, which consist only of 25% alcohol (20 liters pure alcohol). Therefore the new solution logically consists of 70 (10+60) liters water and 30 (10+20) liters alcohol.

Now 80 liters or 80% of the of 100, is what we take out and add of the WHOLE solution. First we take 80 liters out than we add 80 liters.
In my opinion 30 liters of alcohol we being replaced by water so 3/5.

Can somebody explain that to me comprehensibly?
Thanks a lot.

Say, you have a 100 ml glass of 50% water + 50% alcohol solution.
So water is 50 ml in it and alcohol is 50 ml in it homogeneously mixed.

Now, of the 100 ml, say you take out, 20 ml solution (i.e. you take out 20% of the solution). How much water and how much alcohol did you take out? Since the solution has 50-50 alcohol and water, the 20 ml will have 10 ml water and 10 ml alcohol, right?

So of the 50 ml alcohol originally, you took out 10 ml alcohol. This means you took out 20% of the original alcohol.
Of the 50 ml water originally, you took out 10 ml water. This means you took out 20% of the original water.

So by taking out 20% of the original solution (which was 100 ml), you took out 20% of original water (which was 10 ml of 50 ml) and 20% of original alcohol (which was 10 ml of 50 ml).

Does this make sense?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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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..
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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let x=% of alcohol replaced
.5-.5x+.25x=.3
x=.2/.25=4/5=80%
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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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?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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kamkaul wrote:
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?

The question says:

"If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution"

So half water/half alcohol is the kind of original mix we have. We have a solution made of 50% alcohol and the other 50% is water.
We replace some part of it with a 25% alcohol solution (so it has 25% alcohol and 75% water)
How much of it do we replace, we do not know. If of the original, say 100 ml, we replace 20 ml, then we will take out 20 ml of 50% solution (so 10 ml alcohol 10 ml water) and instead put 20 ml of 25% solution (5 ml alcohol and 15 ml water).

Now we would have alcohol -> 40 ml + 5 ml = 45 ml
and we would have water -> 40 ml + 15 ml = 55 ml

So the resulting solution will be 45% alcohol solution.

Does this all make sense?

But our resulting solution is only 30% alcohol. We solve it using the introductory concept discussed here:
http://www.veritasprep.com/blog/2011/04 ... -mixtures/
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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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?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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kamkaul wrote:
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?

Did you check the post for which I provided the link above?

Solution 1 -> 50% alcohol
Solution 2 - > 25% alcohol
Final Solution -> 30% alcohol

w1/w2 = (30 - 25)/(50 - 30) = 1/4

So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2.
Hence, 4/5th of solution 1 was replaced. This is 80%.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

Nice explanation indeed. But I want to know what does 50% represents? Is it water or alcohol?
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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NaeemHasan wrote:
gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

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

25% and 30% are the concentrations of alcohol. So basically we are working with alcohol. Hence 50% is also the concentration of alcohol.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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VeritasPrepKarishma wrote:
NaeemHasan wrote:
gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

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

25% and 30% are the concentrations of alcohol. So basically we are working with alcohol. Hence 50% is also the concentration of alcohol.

Thank you. I did not understand the question at first. Thanks a ton.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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gmat1011 wrote:
Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

Loved this solution, so easy and quick! Thanks a lot.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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VeritasPrepKarishma wrote:
kamkaul wrote:
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?

Did you check the post for which I provided the link above?

Solution 1 -> 50% alcohol
Solution 2 - > 25% alcohol
Final Solution -> 30% alcohol

w1/w2 = (30 - 25)/(50 - 30) = 1/4

So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2.
Hence, 4/5th of solution 1 was replaced. This is 80%.

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,

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 Veritas Prep GMAT Instructor V
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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dave13 wrote:
VeritasPrepKarishma wrote:
kamkaul wrote:
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?

Did you check the post for which I provided the link above?

Solution 1 -> 50% alcohol
Solution 2 - > 25% alcohol
Final Solution -> 30% alcohol

w1/w2 = (30 - 25)/(50 - 30) = 1/4

So in the final solution, there will be only 1 part solution 1 for every 4 parts of solution 2. So how much of solution 1 was replaced by solution 2? Out of 5 parts of solution 1, 4 parts were removed and replaced by 4 parts of solution 2.
Hence, 4/5th of solution 1 was replaced. This is 80%.

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,

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 Dave, it is just a matter of nomenclature. You can take A1, w1 as the 50% alcohol solution and A2, w2 as 25% alcohol solution or the other way around. In any case, answer will be the same.

Check here: https://www.veritasprep.com/blog/2017/1 ... -averages/

This post deals with exactly this confusion.
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Re: If a portion of a half water/half alcohol mix is replaced  [#permalink]

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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%

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_________________ Re: If a portion of a half water/half alcohol mix is replaced   [#permalink] 25 Feb 2018, 15:21

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