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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% Source: GMAT Club Tests - hardest GMAT questions 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
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for such problems, when total quantity is not given, always assume it as 1 unit (litre, number, km, etc)
say 1 liter initial 50% solution was there,, ie concentration is 50%
say x liter is taken out (if you take total as 1, then x will be a fraction)
so 50%(1-x) alcohol remains in the original solution. To this x liter of 25% solution is added, ie, 25%(x) liter alcohol is added
this gives 30% of 1liter solution again
so 50(1-x) + 25x = 30*1
solve for x, which would be 4/5, or 80% of 1 liter
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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%
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E. Those interested in learning the methods to solve such problems, they are very effectively explained here: http://www.onlinemathlearning.com/mixtu ... s.html#mix
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Re: Percentage problem [#permalink]
 11 Jan 2010, 11:17
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
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Re: Percentage problem [#permalink]
11 Jan 2010, 14:05
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.
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Re: Percentage problem [#permalink]
11 Jan 2010, 20:52
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%
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I solved it using just a small variation of the methods explained above.
x = original quantity of solution
y = quantity of solution replaced
Therefore, the equation for the alcohol part of the solution is:
x(50%) - y(50%) + y(25%) = x(30%)
~ (x/2 )- (y/2) + (y/4) = 3x/10
Multiply throughout by 20,
10x - 10y + 5y = 6x
~4x = 5y
~y = 4x/5
~y = 80%x
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ThrillRide 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% Source: GMAT Club Tests - hardest GMAT questions 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.
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1/2(x-y) + 1/4 y = 3/10 (x-y+y)
multiplying by 20
10 (x-y) + 5 y = 6x
4x=5y
y=4/5 x= 80% x
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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. So 0.5(1-x)+0.25x=0.3 --> x=0.8, or 80%. Answer: E. Other solutions: mixture-problem-can-someone-explain-this-100271.html
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Let there be total 100 ml of mix 50 ml W + 50 ml A Let x ml be the amount of new mix - 0.25x Alcohol 30 ml of Alcohol now in 100 ml 0.30 * 100 = 0.25x + 0.50(100-x) 30 = 0.25x + 50 - 0.50x => 0.25x = 20 => x = 80 So 80 ml of original mix was replaced. => Alcohol replaced in original mix = 80 * 50/100 = 40 ml => 40/50 * 100 = 80% So answer is E.
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Assume that there is 1 litter of mix hence 0.5(1-X)+.25X = 0.3 where X is portion of alcohol 0.5-0.5X + 0.25X = 0.3 0.2 = .25X X= 80%
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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 .
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ravijackson wrote: 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
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fluke wrote: ravijackson wrote: 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%. Again , open to discussion .
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ravijackson wrote: 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.
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Re: Percentage problem [#permalink]
24 May 2012, 16:24
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.
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can anyone solve this by allIgation method ?? i generally solve this type of problem using the alligation technique .. I am getting 20% ..
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close
