Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

X grams of water were added to the 80 grams of a strong [#permalink]
18 Nov 2008, 10:16

4

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

36% (02:50) correct
64% (01:43) wrong based on 161 sessions

\(X\) 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 \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?

Re: X grams of water were added to the 80 grams of a strong [#permalink]
30 Oct 2014, 07:31

Expert's post

1

This post was BOOKMARKED

prasun84 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 \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?

(1) \(X = 80\)

(2) \(Y = 2\)

Statements (1) and (2) combined are insufficient. Denote the original concentration as \(C\). Construct an equation using S1 and S2: \(80C + 80*0 = \frac{160C}{2}\) \(80C = 80C\)

\(C\) cancels out, so we cannot determine the answer.

Thanks a lot for your reply. But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

Thanks a lot for your reply. But, I must admit my weakness on "Mixture" type of questions. I didn't find the MGMAT mathbook sufficient to face different types of Mixture problems.

Can you please explain any CORE FORMULA regarding this type of problem, or how to set the equations in such cases ?

Re: X grams of water were added to the 80 grams of a strong [#permalink]
01 Nov 2014, 07:05

Bunuel wrote:

prasun84 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 \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?

(1) \(X = 80\)

(2) \(Y = 2\)

Statements (1) and (2) combined are insufficient. Denote the original concentration as \(C\). Construct an equation using S1 and S2: \(80C + 80*0 = \frac{160C}{2}\) \(80C = 80C\)

\(C\) cancels out, so we cannot determine the answer.

Answer: E.

Hi Bunuel

Didn't quite get what you did up there.

This is what I did.

let in the original solution there be U unknown liquid and 80 gms of acid.

combining (1) & (2) --> {80/(80+u)}*(1/y) = {80/(80+80+u)} => u = 0 That's how i got C.

Re: X grams of water were added to the 80 grams of a strong [#permalink]
09 Dec 2014, 22:01

Expert's post

2

This post was BOOKMARKED

prasun84 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 \(\frac{1}{Y}\) times of the initial concentration, what was the concentration of acid in the original solution?

(1) \(X = 80\)

(2) \(Y = 2\)

Responding to a pm:

Quote:

I made this equation: A1/(80+x) = 1/Y *(A2/80). Is this equation right? Where A1 is the new concentration of acid, and A2 is the original one. I realize my initial wrror , which was to put both A1 and A2 as A (lack of concentration I suppose).

I don't know how you get this equation.

Amount of acid remains same after you add water so

Re: X grams of water were added to the 80 grams of a strong [#permalink]
26 Jun 2015, 22:56

1

This post received KUDOS

Can someone please tell me if my logic is correct?

According to the statements: if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives 1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

Re: X grams of water were added to the 80 grams of a strong [#permalink]
26 Jun 2015, 23:09

Expert's post

AjChakravarthy wrote:

Can someone please tell me if my logic is correct?

According to the statements: if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives 1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient Answer: OptionE _________________

Re: X grams of water were added to the 80 grams of a strong [#permalink]
27 Jun 2015, 00:15

GMATinsight wrote:

AjChakravarthy wrote:

Can someone please tell me if my logic is correct?

According to the statements: if a is the weight of acid and w is the weight of water then total weight is a + w = 80

Initial concentration of acid is a/(a+w) = a/80 >> this is what we need to find.

If x gms of water is added to the mixture then,

a/(80+x) = (1/y) a/80; which gives 1/80+x = 1/y80

Substituting value of x from statement 1 we get 1/160 = 1/160 >> Not Sufficient Substituting vale of y from statement 2 again gives the same fraction >> Not sufficient

Combining 1 & 2 does not give us any new info; hence E

This approach is correct?

This is absolutely fine approach.

Only the reasoning of highlighted part is not very clear.

We must be clear that, We only need to know the value of either a or W to answer the question

Statement 1: Gives us the value of y but no information about values of a or W hence Not Sufficient Statement 2: Gives us the value of y but no information about values of a or W hence Not Sufficient

Combining the two gives us the value of y only so NOT Sufficient Answer: OptionE

Thanks very much for the clarification

Regarding the highlighted part : just saying that wen we substitute the values and simplify, in each case we get 1/160 = 1/160....

gmatclubot

Re: X grams of water were added to the 80 grams of a strong
[#permalink]
27 Jun 2015, 00:15

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

In out-of-the-way places of the heart, Where your thoughts never think to wander, This beginning has been quietly forming, Waiting until you were ready to emerge. For a long...