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# M03 #04

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07 Jul 2009, 18:29
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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%

[Reveal] Spoiler: OA
E

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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Re: M03 #04 [#permalink]

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07 Jul 2009, 18:39
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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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Re: M03 #04 [#permalink]

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26 Sep 2009, 12:22
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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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Re: Percentage problem [#permalink]

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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%

-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]

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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.
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Re: Percentage problem [#permalink]

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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%

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Re: M03 #04 [#permalink]

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12 Feb 2010, 06:27
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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Re: M03 #04 [#permalink]

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12 Feb 2010, 07:38
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%

[Reveal] Spoiler: OA
E

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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Re: M03 #04 [#permalink]

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12 Feb 2010, 18:45
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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Re: M03 #04 [#permalink]

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17 Feb 2011, 05:32
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%.

Other solutions: mixture-problem-can-someone-explain-this-100271.html
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Re: M03 #04 [#permalink]

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17 Feb 2011, 12:36
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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: M03 #04 [#permalink]

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20 Feb 2011, 04:14
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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Re: M03 #04 [#permalink]

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20 Feb 2011, 10:26
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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Re: M03 #04 [#permalink]

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24 Sep 2011, 05:46
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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Re: M03 #04 [#permalink]

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24 Sep 2011, 06:17
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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Re: M03 #04 [#permalink]

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24 Sep 2011, 10:10
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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Re: M03 #04 [#permalink]

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24 Sep 2011, 10:45
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

(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]

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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.

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Re: M03 #04 [#permalink]

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28 May 2012, 23:23
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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Re: M03 #04 [#permalink]

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28 May 2012, 23:31
abhi398 wrote:
can anyone solve this by allIgation method ??

i generally solve this type of problem using the alligation technique .. I am getting 20% ..

Check this: mixture-problem-can-someone-explain-this-100271.html

Hope it helps.
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Re: M03 #04   [#permalink] 28 May 2012, 23:31

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# M03 #04

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