mmslf75 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 ?


St 1 X = 80

St 2 Y = 2


OA is E , as stated on GMATCLub..

Please explain

this question involves 3 variables, concentration of acid in original solutions, %(weight) of water and Y

stmt 1 and 2 give only two variables hence E

mmslf75 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 ?


St 1 X = 80

St 2 Y = 2

Please explain

This is a very strange question.

First, I've never heard of a liquid solution measured in grams; in the real world, we'd measure a liquid solution with a liquid measurement, e.g. litres or millilitres or ounces.

Second, our mixture is referred to as a "strong solution of acid". Does that mean that it's more than 50% acid? The GMAT doesn't expect us to be chemists, so that description has to be misleading, since we don't need to know any scientific terminology.

Ignoring those issues, the answer is indeed E.

We can think of this in terms of equations and unknowns, as does Satish - that's a great approach to a large number of DS questions.

We can also pick numbers to show that even combined there's more than 1 possible answer:

if we start with all acid, the original concentration is 100%.

When we add 80 grams of water, the new concentration is 80/160 = 50%.

Since 1/Y = 1/2, these numbers work.

if we start with half acid and half water, the original concentration is 50%.

When we add 80 grams of water, the new concentration is 40/160 = 25%.

Since 1/Y = 1/2, these numbers also work.

So, the original concentration could be either 100% or 50%: insufficient.

* * *

A large number of the GMATClub questions are unimpressive, including this one.

Stuart Kovinsky wrote:

mmslf75 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 ?


St 1 X = 80

St 2 Y = 2

Please explain

This is a very strange question.

First, I've never heard of a liquid solution measured in grams; in the real world, we'd measure a liquid solution with a liquid measurement, e.g. litres or millilitres or ounces.

Second, our mixture is referred to as a "strong solution of acid". Does that mean that it's more than 50% acid? The GMAT doesn't expect us to be chemists, so that description has to be misleading, since we don't need to know any scientific terminology.

Ignoring those issues, the answer is indeed E.

We can think of this in terms of equations and unknowns, as does Satish - that's a great approach to a large number of DS questions.

We can also pick numbers to show that even combined there's more than 1 possible answer:

if we start with all acid, the original concentration is 100%.

When we add 80 grams of water, the new concentration is 80/160 = 50%.

Since 1/Y = 1/2, these numbers work.

if we start with half acid and half water, the original concentration is 50%.

When we add 80 grams of water, the new concentration is 40/160 = 25%.

Since 1/Y = 1/2, these numbers also work.

So, the original concentration could be either 100% or 50%: insufficient.

* * *

A large number of the GMATClub questions are unimpressive, including this one.

thanks stuart