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M03-04

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Re: M03-04 [#permalink]

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New post 16 Sep 2016, 09:11
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Re: M03-04 [#permalink]

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New post 23 Feb 2017, 01:57
wrong question.
The question should ask .......... what percent of original solution was replaced.
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Joined: 07 Nov 2016
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Concentration: Finance
Schools: Johnson '19 (WL)
GMAT 1: 650 Q43 V37
Re: M03-04 [#permalink]

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New post 13 Apr 2017, 17:57
I always get excited with problems likes these! They're weighted heavily and categorized as difficult, but it seems to just be a weighted average problem in disguise. So if you know how to work with weighted averages, these types of problems are like free chicken!

We're given 3 solution concentrations to work with.

Starting: 50% (S)
Added solution: 25% (A)
Resultant solution: 30% (R)

Draw them out on a number line in increasing order, with the resultant solution in the middle (we put the resultant in the middle since it's an average of the other two solutions)

A-----R--------------------S
25-----30--------------------50

Now count the percentages in between the two.

From A to R: 5
From R to S: 20

Now, set them into a ratio. 20:5. Now simplify to 4:1.

**This means that 4 parts of the 25% solution were replaced for every 1 part of the existing solution. In other words, \(\frac{4}{5}\) replaced. (part over whole of 4:1 ratio) and \(\frac{4}{5}\) expands to \(\frac{80}{100}\) or 80%

I used to get tripped up on setting the ratio up backwards, at first glance it looks like it would be 5:20, since the space between A-R is 5, and R-S is 20, but think about it logically. The resultant (or average) solution is much closer to the "added" solution, meaning the resultant solution contains more of the added solution than the original solution, which is what brought "R" closer to "A" on our number line.

Once you get the hang of setting these weighted averages problems up like this, it becomes super easy.

Ans: E
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New post 18 Jun 2017, 09:24
I think this question is wrong. because the question is asking how much alcohol
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Re: M03-04 [#permalink]

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New post 10 Jul 2017, 18:48
How to solve using "arrows"? I got 1 to 4 ratio. But apparently it means that not 20%, but 80% of the original solution was used.
>> !!!

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New post 31 Jul 2017, 16:37
Any takers?
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Re: M03-04 [#permalink]

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New post 28 Oct 2017, 00:13
Bunuel wrote:
Official Solution:

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. \(80\%\)


Initial solution is "half water/half alcohol mix" means that it's 50% 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 that 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, since finally 30% alcohol solution was obtained then the amount of alcohol in the end was \(0.3*1\).

So \(0.5(1-x)+0.25x=0.3\) \(\rightarrow\) \(x=0.8\), or 80%.


Answer: E



Hi Bunuel

I tried this question in a different way

Portion A Portion B
50% 25%


30%

5% 20%


Thus the portion removed is 1/5 which is 20 percent. Where did i go wrong??
Re: M03-04   [#permalink] 28 Oct 2017, 00:13

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