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M03-04

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Re: M03-04  [#permalink]

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New post 16 Sep 2016, 08:11
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Re: M03-04  [#permalink]

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New post 23 Feb 2017, 00:57
wrong question.
The question should ask .......... what percent of original solution was replaced.
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New post 13 Apr 2017, 16:57
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, 08:24
I think this question is wrong. because the question is asking how much alcohol
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New post 10 Jul 2017, 17:48
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, 15:37
Any takers?
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Re: M03-04  [#permalink]

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New post 27 Oct 2017, 23:13
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??
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New post 04 Jun 2018, 10:09
What if i assumed the alcohol portion to be 1-x instead of x?

Actual equation before replacement:
0.5x + 0.5 (1-x) = 1 litre, for example?

New equation after replacement:
0.5x (Water) + 0.25x (new alcohol solution) = 0.3*1 (new alcohol solution) ?

this gives me 20%

confused

could you please help me here Bunuel?
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New post 23 Oct 2018, 20:53
For all the people asking why are we taking 80% as the alcohol percent when we get 80% for the solution replaced?

The answer is, if you replace x% of a solution with something else, then any smaller portion of the whole (lets say 50ml of the total 100ml solution) will also be replaced with the same %.
For example:
Assume that one has Solution A: 100ml. Solution A is composed of 60ml Rum and 40ml Water.
One removes a portion of the Rum mixture, replaces it with something more potent. Regardless of how much replacement has been done, since the solution is always homogeneous (evenly mixed) the removal will affect all individual portions as the same.

So if 50% Rum mixture is replaced, then 50% of the water inside the mixture is also replaced.

Not an exact answer, but should get the wheels rolling for people trying to search for an answer as to why getting 80% for whole means the same to alcohol as well.
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New post 20 Dec 2018, 22:46
Let mixture volume be = 100 ml
Therefore : 50 ml alcohol + 50 ml water
X= total volume of alcohol
Yr = Volume of alcohol replaced
Ya = Volume of alcohol added
=> X - Yr + Ya = 30 ( resultant will be 30ml alcohol out of 100 ml)
50 - (50/100)Yr + (25/100)Ya = 30
= 20 = (25/100) Y (amount replaced and added are same, so denoting as Y)
Y = 80 ( This means you are replacing 80ml of the solution and adding 80 ml of 25% alcohol )
Hence: 80% (E)
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Re: M03-04 &nbs [#permalink] 20 Dec 2018, 22:46

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