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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%

OA explanation: Find the difference between the original and final amounts of alcohol. Suppose we had 1 liter or 1 gallon of the initial solution, and has been replaced. We can build an equation using this information:

How do you solve this? I don't get how the OA came up with the equation below. thanks

Here is the how I solved the problem Consider the quantity of solution as 1 litter. It is given that 50% of the solution was alcohol. Let X be the amount of alcohol removed from the solution.

The equation to remove X amount of alcohol from the 50% alcohol solution is 0.5 litters alcohol - 0.5 * X litters alcohol

The X litters removed from the solution is replaced by 25% alcohol solution. 0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol

The resulting solution contains 30% alcohol. Therefore the final equation is 0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol = 0.30 litters of alcohol.

If you solve for the above equation, you will get X as 0.8, which is 80%

Re: Percentage problem [#permalink]
11 Jan 2010, 10:17

amitanand wrote:

Hi Experts,

Can anyone help me in the below question:

Q: 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%

Thanks in Advance -Amit

b) 20% solution1 is 50% alcohol and solution2 is 25% alcohol. to get 30% alcohol solution we need solution1/ solution2 in ratio 1:4

Re: Percentage problem [#permalink]
11 Jan 2010, 13:05

Interesting question. My answer is E.

Let x be removed out of the first solution of 50% alcohol and x be added from the other solution of 25% alcohol to make a wholesome (a peg ??) 30% alcohol solution ( I don't take it with water though!!!)

Anyways, in equation form it is 50(1-x) + 25x = 30, which gives x = 4/5. so 80% quantity of the first solution is replaced with 80% quantity of second solution to get a wholesome 30% solution.
_________________

Re: Percentage problem [#permalink]
11 Jan 2010, 19:52

amitanand wrote:

Hi, Thanks for the reply. But the OA is E(80%). Could you please justify? -Amit

ohk...read the q wrong...since we have 100% of original solution [50% alcohol] and after mixing it is only 20% so the 1st solution is replaced by 100 - 20 = 80%

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%

OA explanation: Find the difference between the original and final amounts of alcohol. Suppose we had 1 liter or 1 gallon of the initial solution, and has been replaced. We can build an equation using this information:

How do you solve this? I don't get how the OA came up with the equation below. thanks

Let x - amount of total mixture y - amount of mixture replaced with 25% alcohol

Now, 0.5(x-y) + 0.25y = 0.3x 0.5(x-y) represents 50% alcohol quantity in new mixture 0.25y represents 25% alcohol quantity in new mixture 0.3x represents total alcohol quantity in new mixture

solving y = (4/5)x hence y is 80% of x that means E.

This question was also posted in PS subforum. Below is my solution from there.

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%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.

Let the portion replaced be x and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of a portion will be 0.5(1-x) and the amount of alcohol added will be 0.25x, so total amount of alcohol will be 0.5(1-x)+0.25x. On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was 0.3*1.

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

0.8*0.5=0.4 g of alcohol replaced.

0.4 is 80% of 0.5

0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

Again , open to discussion .

I'm going to do this backwards:

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?

1 G mix 0.5 g Original Alcohol 0.5 g Original Water ***********

Answer: 80% of original alcohol was replaced.

0.4g of alcohol from 0.5g of original alcohol was replaced, (80% of original alcohol replaced- removed AND something else was added), with 0.1g Alcohol+0.3g Water.

Also, 0.4g of water from 0.5g of original water was replaced, (80% of original water replaced- removed AND something else was added), again with 0.1g Alcohol+0.3g Water

So, A+W 0.5+0.5

Removed: 0.4A+0.4W

Remaining 0.1A+0.1W

Added (Placed back for alcohol)+(Placed back for water) (0.1A+0.3W)+(0.1A+ 0.3W)

Total Content: 0.1A+0.1W+0.1A+0.3W+ 0.1A+ 0.3W 0.3A+0.7W

I don't see any problem with the wordings.
_________________

Re: Percentage problem [#permalink]
24 May 2012, 15:24

ravijackson wrote:

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced?

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistakes in my logic .

I also arrived at the same answer. I think the question should as "what % of original mixture was replaced" or 40% should be one of the choices.