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Mixture Problem DS

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Mixture Problem DS

by Mo2men » Fri Oct 05, 2018 5:54 pm

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X grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became 1/y times of the initial concentration, what was the concentration of acid in the original solution?

1. x=80
2. y=2

OA: E

Dear GMATGuru,
Can you provide your insights on this question? Does the alligation method help in this question?
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Source: — Data Sufficiency |
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by GMATGuruNY » Sat Oct 06, 2018 3:05 am

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Mo2men wrote:X grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became 1/y times of the initial concentration, what was the concentration of acid in the original solution?

1. x=80
2. y=2
Statement 1 indicates that 80 grams of water are added to the 80 grams of original solution, DOUBLING the total volume from 80 grams to 160 grams
Statement 2 indicates that the added water reduces the acid concentration by 1/2.
Doubling the total volume will reduce the acid concentration by 1/2 for ANY nonzero concentration of acid.
To illustrate:

Case 1: acid concentration = 10%, implying 8 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 8/160 = 1/20 = 5%.
Case 2: acid concentration = 20%, implying 16 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 16/160 = 1/10 = 10%.
In each case, doubling the total volume reduces the acid concentration by 1/2.

Since the acid concentration can be any nonzero value, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Mo2men » Sat Oct 06, 2018 3:26 am

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GMATGuruNY wrote:
Mo2men wrote:X grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became 1/y times of the initial concentration, what was the concentration of acid in the original solution?

1. x=80
2. y=2
Statement 1 indicates that 80 grams of water are added to the 80 grams of original solution, DOUBLING the total volume from 80 grams to 160 grams
Statement 2 indicates that the added water reduces the acid concentration by 1/2.
Doubling the total volume will reduce the acid concentration by 1/2 for ANY nonzero concentration of acid.
To illustrate:

Case 1: acid concentration = 10%, implying 8 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 8/160 = 1/20 = 5%.
Case 2: acid concentration = 20%, implying 16 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 16/160 = 1/10 = 10%.
In each case, doubling the total volume reduces the acid concentration by 1/2.

Since the acid concentration can be any nonzero value, the two statements combined are INSUFFICIENT.

The correct answer is E.
Dear GMATGuru,

Thanks for your help. I have another logic to solve.

Statement 1: water = 80 gm that is equal to the same amount of the solution. It means that it will bring the concentration of the solution to HALF its original concentration But we do not have can't get the original as it could be any thing.

Statement 2: y= 2. It mean that water added is the same with same amount, which leads to same conclusion to the aboe.

Both statment reach same conlcusion

Answer: E

Is my reasoning valid?

Thanks
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by GMATGuruNY » Sat Oct 06, 2018 3:32 am

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Mo2men wrote:
GMATGuruNY wrote:
Mo2men wrote:X grams of water were added to 80 grams of a strong solution of acid. If as a result, the concentration of acid in the solution became 1/y times of the initial concentration, what was the concentration of acid in the original solution?

1. x=80
2. y=2
Statement 1 indicates that 80 grams of water are added to the 80 grams of original solution, DOUBLING the total volume from 80 grams to 160 grams
Statement 2 indicates that the added water reduces the acid concentration by 1/2.
Doubling the total volume will reduce the acid concentration by 1/2 for ANY nonzero concentration of acid.
To illustrate:

Case 1: acid concentration = 10%, implying 8 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 8/160 = 1/20 = 5%.
Case 2: acid concentration = 20%, implying 16 grams of acid in the original 80-gram solution
After 80 grams of water increases the total volume to 160 grams, the resulting acid concentration = 16/160 = 1/10 = 10%.
In each case, doubling the total volume reduces the acid concentration by 1/2.

Since the acid concentration can be any nonzero value, the two statements combined are INSUFFICIENT.

The correct answer is E.
Dear GMATGuru,

Thanks for your help. I have another logic to solve.

Statement 1: water = 80 gm that is equal to the same amount of the solution. It means that it will bring the concentration of the solution to HALF its original concentration But we do not have can't get the original as it could be any thing.

Statement 2: y= 2. It mean that water added is the same with same amount, which leads to same conclusion to the aboe.

Both statment reach same conlcusion

Answer: E

Is my reasoning valid?

Thanks
Yes -- each statement conveys the same information, since doubling the total volume is the equivalent of reducing the acid concentration by 1/2.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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