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X grams of water were added to the 80 grams of a strong [#permalink]

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18 Nov 2008, 11:16

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

\(X\) 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]

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01 Nov 2014, 08:05

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.

\(X\) 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]

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26 Jun 2015, 23:56

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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

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
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Re: X grams of water were added to the 80 grams of a strong [#permalink]

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27 Jun 2015, 01:15

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....

Re: X grams of water were added to the 80 grams of a strong [#permalink]

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30 Jul 2016, 07:02

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Re: X grams of water were added to the 80 grams of a strong [#permalink]

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09 Aug 2017, 14:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: X grams of water were added to the 80 grams of a strong [#permalink]

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20 Aug 2017, 07:17

Hello Karishma,

I recently went through your article on solving weighted average and mixture problems. Please tell me if I am applying the approach correctly on this problem.

Let A1=a (conc. of acid in original solution) A2=100 (conc. of water added to the solution) A(avg.) = a/Y w1=80 w2=X Using Allegation : (80/X) = (100-(a/Y)) / ((a/Y)-a) We need to find "a".

(1) X=80 Cannot get the value of "a" from the above equation with only the value of X. Insufficient

(2) Y=2 Same as (1) Insufficient

(1) + (2) (80/80)=(100-(a/2)) / ((a/2)-a) => (a/2)-a = 100-(a/2) "a" gets cancelled out hence not solvable