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M03 #04

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Re: M03 #04 [#permalink] New post 14 Jun 2012, 00:59
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

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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

Hi,

Using the alligation method:
we have A:B = 1:4 (refer attached diagram), so final solution must have 1 part of 50% solution & 4 parts of 25% solution,

thus, 4 parts of 50% solution should be removed and replaced by 25% solution
so, answer is 80% (E)

Regards,
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all.jpg
all.jpg [ 6.09 KiB | Viewed 1161 times ]


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Re: M03 #04 [#permalink] New post 21 Feb 2013, 06:21
abhi398 wrote:
can anyone solve this by allIgation method ??

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


As mentioned in the other thread, you need to reverse the ratios. If you do this, you will the right answer, 80%. Reversing the ratios is the key step in the Alligation Method.

Please read the document in the link below for further information on how to apply the Alligation Method.

www.gmatclub.com/forum/demystifying-the ... 40276.html

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Re: M03 #04 [#permalink] New post 21 Feb 2013, 06:23
abhi398 wrote:
can anyone solve this by allIgation method ??

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


Hi,

You need to reverse the ratios. That's the key step in the Alligation Method.

www.gmatclub.com/forum/demystifying-the ... 40276.html

Cheers,
Der alte Fritz
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Re: M03 #04 [#permalink] New post 21 Feb 2013, 12:50
lets consider we ve 1 litre of soln with us..
and we replaced X litre of the originl 50% with X lite 25% resulting in 30% solution (1 litre)..

so equating the water part

0.5 - 0.5X + 0.75X = 0.7
Solving for X we get .80 or 80 %
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Re: M03 #04 [#permalink] New post 22 Feb 2013, 12:12
abhi398 wrote:
can anyone solve this by allIgation method ??

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


Abhi .. by using the allegation technique in the diagram you will get the final ratio of alcohol : water in the mixture = 5:20 i.e. 1:4

that means that at the moment 5 parts of solution 1 part is alcohol, so the alcohol removed is 4 parts i.e. 4/5* 100 = 80%

other way let x be the part removed from the solution and let 1 L be the original solution so 50(1-x) + 25x = 30 (1) (Both sides are having percentages)

Solving for 25x = 20 4:5 or 80%
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Re: M03 #04 [#permalink] New post 23 Mar 2013, 07:24
solved this but still I feel I missed something ..0.5x+0.25y=0.3(x+y).....x/y =4/1 ie x=4/5=80%.But I do not know why I still feel I should have subtracted this 80% from 100% to be 20% ... anybody can help please?
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Re: M03 #04 [#permalink] New post 23 Mar 2013, 07:53
Expert's post
TheNona wrote:
solved this but still I feel I missed something ..0.5x+0.25y=0.3(x+y).....x/y =4/1 ie x=4/5=80%.But I do not know why I still feel I should have subtracted this 80% from 100% to be 20% ... anybody can help please?


Check here: m03-80634.html#p873581

Hope it helps.
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Re: M03 #04 [#permalink] New post 07 Feb 2014, 11:36
Let's say total of 10 liters; 5 water; 5 alcohol.
Total x removed; water = alcohol = 5 -(x/2)
25% of x added back = (0.25x).
Final volume of alcohol = 3.
So, [5-(x/2)] + 0.25x = 3
Solving x = 8 or 80%.
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Re: M03 #04 [#permalink] New post 21 Apr 2014, 05:34
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The percentage replaced is x -> (1-x)*0.5+0.25x=0.3 -> 0.25x=0.2->x=80%
Re: M03 #04   [#permalink] 21 Apr 2014, 05:34
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