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# Solution Problem

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Senior Manager
Joined: 05 Oct 2008
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Solution Problem [#permalink]

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25 Jan 2009, 11:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution went down by one-third. How much of the 24%-solution was used?

* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams
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Joined: 24 Jan 2009
Posts: 18
Re: Solution Problem [#permalink]

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25 Jan 2009, 12:12
study wrote:
After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution went down by one-third. How much of the 24%-solution was used?

* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams

Let X be amount of alcohol and Y the solution:

$$\frac{x}{y}=\frac{24}{100}$$
$$\frac{x}{y+200}=\frac{16}{100}$$

Cross-multiplying both equations you'll have someting like:
$$25x=6y$$
$$25x=4y+800$$

Then, $$2y = 800$$, $$y = 400$$

Answer E
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Re: Solution Problem [#permalink]

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25 Jan 2009, 12:31
study wrote:
After 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution went down by one-third. How much of the 24%-solution was used?

* 180 grams
* 220 grams
* 250 grams
* 350 grams
* 400 grams

solution = s
alcohal = 24% s
water = 76% s
added water = 200 grms
weight after addition = 200+s

24% of s = 16% of (200+s)
1.5s = 200+s
s = 400
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Re: Solution Problem [#permalink]

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25 Jan 2009, 15:00
Or, if you're familiar with weighted averages:

we're mixing two components, one at 24%, one at 0%, to get a 16% result. They must then be in a 2 to 1 ratio (their ratio must equal the ratio of the distances to the overall average, i.e. 16-0 to 24-16). So we must have twice as much of the 24% mixture as we have of water. 400.
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Re: Solution Problem   [#permalink] 25 Jan 2009, 15:00
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# Solution Problem

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