X grams of water were added to the 80 grams of a strong

Responding to a pm:


I don't know how you get this equation.

Amount of acid remains same after you add water so

Amount of acid = C1*V1 (initial) = C2*V2 (final)
C1*80 = C1/Y * (80+x)
80 = (80+x)/Y

The equation is independent of C1.

Look at the statements now:
(1) \(X = 80\)
If X = 80, Y = 2
But you don't get the value of C1.

(2) \(Y = 2\)
If Y = 2, you get x = 80
But you don't get the value of C1.

Answer (E)
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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

This approach is correct?

While statement (2) might seem new and exciting, it actually provides identical information to (1).

To the explanation!

From (1), we know that we doubled the volume of the solution. There's a handy formula to remember for mixture questions:

(concentration 1)(volume 1) = (concentration 2)(volume 2)

or:

(C1)(V1) = (C2)(V2)

for short.

We know that the original volume is 80 grams (we're measuring in grams, so technically it's weight, not volume, but that doesn't matter) and that there's some acid in there as well. "Strong" has no specific meaning, so we don't know the quantity of acid.

So, based on (1), we know that:

(C1)(80) = (C2)(80 + 80)

(C1)(80) = (C2)(160)

(C1) = (C2)(160/80)
C1 = 2(C2)

so, from the original statement, we know that C1 is twice as strong as C2. Of course, this doesn't give us the actual value of C1, so it's insufficient.

From (2), we know that the new solution is 1/Y times as strong as the original. Since Y = 2, we know that the new solution is 1/2 times as strong. Also nothing about numbers, insufficient.

Now let's look at the statements together:

(1) C1 is twice as much as C2

(2) C2 is half as much as C1

When we stack them together like that, we quickly realize that each statement gives us identical information! Since neither statement was good enough by itself, they certainly won't be sufficient together: choose (E).

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.
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Bunuel

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 in advance

Mixture problems made easy: mixture-problems-made-easy-49897.html

All DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
All PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope this helps.
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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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Consider Variables A1 and A2 for the acid
Initial concentration is A1/80 and final concentration is nothing but A2/(80+x)
It is given that A2/(80+x) = 1/Y*(A1/80)
We have to find A1/80
We need the values of A2, X, Y
Combining both the statements we have only X and Y
Answer is E

If such a question appears on the GMAT, the best approach, instead of panicking and trying to write equations, is to take a deep breath and just go through this slowly and thoroughly.

It's a VERY LOGICAL question.

If the same amount of water is added to some amount of acid to dilute it (in this case, 80 grams of water is added to 80 grams of a strong solution of acid), NO MATTER what the concentration of acid in the original solution, it will ALWAYS be halved.

Hence, the concentration of acid in the original solution can have any (infinite) number of values.

HIT KUDOS IF YOU LIKE THE SOLUTION

If 80 gms of solution had C gms of acid

Hence from statement 1

we get C/80 - C/160 = y/100

since there are 2 variables in equation and it can't be solved , hence 1 is insufficient

According to statement 2,

C/80 - C/(80+x) = 50/100 = 1/2

Hence, 2 is insufficient

If we combine both statements,

80C + 80*0 (since water has 0 concentration of the acid) = (160C)/2

This becomes 80C=80C

This equation can't be solved.

Hence, answer = E

Hi Bunuel!

You expect us to know what C stands for. Can you please elaborate what C means in your solution for this question.

Thanx.

C is the concentration of acid in the solution, so (acid in grams)/(solution in grams). For example, if acid in the solution is 50 grams and the solution is 80 grams, then C = 50/80 = 5/8.
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This question puzzles me, if (1) you add 80g of water to 80g solution that results in (2) ½ =50% drop of concentration, this should only happen if initial solution was 100% acid in fact not a solution.
What am I missing? ?

It's not 50% concentration -- it's a 50% drop in concentration.
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Hi,

I have a doubt. Can we solve this problem by alligation method?

Regards.
Niwedita B

Bunuel Can we solve this problem by alligation method?

Official Solution:


A strong acid solution weighing 80 grams was diluted by adding \(x\) grams of water. After dilution, the concentration of the acid in the solution became \(\frac{1}{y}\) times its initial concentration. What was the concentration of the acid in the original solution?

The question looks intimidating and one might jump right into constructing equation. But take a moment, step back and use logic.

(1) \(x = 80\).

We can see that the initial solution was diluted with an equal amount of water (80 grams of solution was diluted with 80 grams of water). Whatever the concentration of the acid was in the original solution, this operation, would halved it in the resulting solution. For example, if the initial concentration of acid was 10%, so if there were 8 grams of acid in the original solution, the concentration of acid in the resulting solution would be 5% (8 grams of acid over 160 grams). However, while we can infer that the concentration of acid was halved, we don't have any information on what it actually was. Therefore, statement (1) alone is not sufficient to determine the original concentration of acid.

(2) \(y = 2\)

This statement also tells us that the concentration of acid was halved in the resulting solution, but we have no idea what the original concentration actually was. Therefore, statement (2) is also not sufficient.

(1)+(2) Notice that both statements convey the same information: the concentration of the acid in the original solution was halved in the resulting solution, but we still lack information about its actual value. Therefore, combining the statements is not sufficient to determine the original concentration of the acid.


Answer: E
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one part that I find a bit difficult to understand, after getting the equation A(89)+80= 1/2(A) (80+80)
why should I multiply the 80g of water (ibolded) by zero?