Last visit was: 12 Jul 2025, 15:23 It is currently 12 Jul 2025, 15:23
Close
GMAT Club Daily Prep
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
 [62]
2
Kudos
Add Kudos
60
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
740,759
 [7]
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
 [7]
4
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
avatar
sudd1
Joined: 30 Oct 2013
Last visit: 16 Dec 2016
Posts: 9
Own Kudos:
38
 [11]
Given Kudos: 29
Status:Brushing up rusted Verbal....
Location: India
Schools: AGSM '16
GMAT Date: 11-30-2014
GPA: 3.96
WE:Information Technology (Computer Software)
Schools: AGSM '16
Posts: 9
Kudos: 38
 [11]
11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
AlexGenkins1234
Joined: 18 Sep 2015
Last visit: 17 May 2023
Posts: 57
Own Kudos:
111
 [3]
Given Kudos: 611
GMAT 1: 610 Q47 V27
GMAT 2: 650 Q48 V31
GMAT 3: 700 Q49 V35
WE:Project Management (Healthcare/Pharmaceuticals)
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Bunuel
\(X\) 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) \(Y = 2\)

This procedure is according to here: mixture-problems-made-easy-49897.html
Attachments

File comment: Detailed Solution
123.png
123.png [ 48.09 KiB | Viewed 89939 times ]

avatar
xxx321
Joined: 04 Jun 2015
Last visit: 09 Sep 2020
Posts: 3
Given Kudos: 4
Location: United States
GMAT 1: 720 Q51 V35
GPA: 3.91
GMAT 1: 720 Q51 V35
Posts: 3
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I agree with explanation. Great question. Really tests your ability to solve for the correct variable. The question is asking for the concentration of Acid (i.e. what is Acid/80). Each statement gives you enough info to solve for X/Y, but have these 2 variables by themselves do not give you A/80 (try to find A/80; you can't with the given info).
avatar
sakshamjinsi
Joined: 05 Aug 2018
Last visit: 28 Jan 2019
Posts: 8
Own Kudos:
6
 [3]
Given Kudos: 13
GMAT 1: 650 Q46 V34
GMAT 1: 650 Q46 V34
Posts: 8
Kudos: 6
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
\(X\) 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) \(Y = 2\)

Simple Logical Explanation:

1) X = 80 G. This means that 80G of water was added to 80G of strong acid solution.
In this scenario, the proportion of acid in the final solution will reduce by half i.e. (If initially the proportion was 80% - it will become 40%).
However, Initial proportion cannot be determined.
Not Suff.

2) Y = 2. This means that the proportion of acid in the final solution reduces by half.
For this to happen exactly the same amount of water must be added i.e. 80G
However, Initial proportion cannot be determined.
Not Suff.

1 & 2 Together:
Not Sufficient. Just proves that the initial concentration was halved. Cannot determine initial concentration.

OA: E
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 11 Jul 2025
Posts: 16,101
Own Kudos:
74,289
 [2]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,289
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
\(X\) 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) \(Y = 2\)

Raxit85

Method 1: Think of the scale. You are adding solution 1 (pure water so 0% acid) with solution 2 (conc acid say c% acid) to give a mixture of weaker acid.
The weights are X and 80.

(1) \(X = 80\)

This means that both solutions are mixed equally so the average will be right in the middle of the two. So if concentration of original acid were c%, concentration of new solution would be c/2% (half of previous) because it will be in the middle of 0 and c%.
We don't know what c% is so we don't know what c/2% is.
Not sufficient.

(2) \(Y = 2\)

This tells us that the average concentration becomes half of previous concentration. So average lies exactly in the middle of the two. This means both solutions must be added in equal measure so water must have been 80 gms too. But again, we don't know c% so we cannot say what c/2% will be.

Using both statements, note that you get the same information from both of them. Since neither alone is sufficient, both together, they are not sufficient either. You don't have c% so you cannot find c/2%.

Method 2: Using the formula,

\(\frac{w_1}{w_2} = \frac{(A_2 - A_{avg})}{(A_{avg} - A_1)}\)

First sol is water (Acid = 0%) and second sol is concentrated acid (Acid = A2%). We need the value of A2.

\(\frac{X}{80} = \frac{(A_2 - A_2/Y)}{(A_2/Y - 0)}\)

Using both statements, you get
\(\frac{80}{80} = \frac{(A_2 - A_2/2)}{(A_2/2 - 0)}\)

1 = 1

Both statements are equivalent so you get that both sides are equal. You don't get the value of A2.

Answer (E)
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,755
Own Kudos:
34,062
 [10]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,755
Kudos: 34,062
 [10]
7
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Bunuel
\(x\) 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) \(y = 2\)

Given: x grams of water were added to 80 grams of a strong solution of acid. As a result, the concentration of acid in the solution became \(\frac{1}{y}\) times of the original concentration
Let A = the grams of ACID in the original strong solution
So, the ORIGINAL concentration of acid \(=\frac{A}{80}\)
After, adding x grams of water the NEW concentration of acid \(=\frac{A}{80+x}\)

We can now write the following equation: \(\frac{A}{80+x}=(\frac{1}{y})(\frac{A}{80})\)

Simplify to get: \(\frac{A}{80+x}=\frac{A}{80y}\)

Divide both sides of the equation by A to get: \(\frac{1}{80+x}=\frac{1}{80y}\)

Cross multiply to get: 80+x = 80y

Target question: What was the concentration of acid in the original solution?
In other words, what is the value of \(\frac{A}{80}\)?

Statement 1: x = 80
Aside: At this point all we know is that 80+x = 80y. As you can see, knowing the value of x won't help us, since we must determine the value of A to answer the target question
To see what I mean let's take our equation 80+x = 80y and replace x with 80 to get: 80+80 = 80y
When we solve this equation for y we get: y = 2, which doesn't help us determine the value of A
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y = 2
We can apply the same logic here that we applied for statement 1 to see that statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x = 80
Statement 2 tells us that y = 2
When we plug these values into the equation 80+x = 80y, we get: 80+80 = 80(2)
Once again we have no way to determine the value of A, which means we can't answer the target question.
The combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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?



(1) \(x = 80\)

(2) \(y = 2\)

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
User avatar
pudu
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 234
Own Kudos:
118
 [1]
Given Kudos: 16
Location: India
Posts: 234
Kudos: 118
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.

strong acid solution means only acid ions. then its concentration is 100%. Then it becomes C, right? How can we take it is of unknown concentration ?

(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

but bunuel, if it is 100 % acid... then adding 80 and taking 1/2 means 50%. then it is C... because strong acid solution contains only acid...i am not getting the logic
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pudu
Bunuel
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.

strong acid solution means only acid ions. then its concentration is 100%. Then it becomes C, right? How can we take it is of unknown concentration ?

(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

but bunuel, if it is 100 % acid... then adding 80 and taking 1/2 means 50%. then it is C... because strong acid solution contains only acid...i am not getting the logic

A strong acid solution does not mean that it's 100% acid. If you read the above solutions carefully you'd see that none of them assumes that. For example, check this official question: https://gmatclub.com/forum/if-12-ounces ... 97494.html
User avatar
pudu
Joined: 12 Mar 2023
Last visit: 06 Mar 2024
Posts: 234
Own Kudos:
Given Kudos: 16
Location: India
Posts: 234
Kudos: 118
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
pudu
Bunuel
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.

strong acid solution means only acid ions. then its concentration is 100%. Then it becomes C, right? How can we take it is of unknown concentration ?

(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

but bunuel, if it is 100 % acid... then adding 80 and taking 1/2 means 50%. then it is C... because strong acid solution contains only acid...i am not getting the logic

A strong acid solution does not mean that it's 1005% acid. If you read the above solutions carefully you'd see that none of them assumes that. For example, check this official question: https://gmatclub.com/forum/if-12-ounces ... 97494.html

got....thank you...i forgot the basic thing ...never bring outside knowledge in to the question...thank you so much...now it is clear
User avatar
BottomJee
User avatar
Retired Moderator
Joined: 05 May 2019
Last visit: 09 Jun 2025
Posts: 996
Own Kudos:
Given Kudos: 1,009
Affiliations: GMAT Club
Location: India
GMAT Focus 1: 645 Q82 V81 DI82
GMAT 1: 430 Q31 V19
GMAT 2: 570 Q44 V25
GMAT 3: 660 Q48 V33
GPA: 3.26
WE:Engineering (Manufacturing)
Products:
GMAT Focus 1: 645 Q82 V81 DI82
GMAT 3: 660 Q48 V33
Posts: 996
Kudos: 1,213
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I agree with explanation.
User avatar
deepak.brijendra
Joined: 02 May 2023
Last visit: 14 Aug 2024
Posts: 29
Given Kudos: 35
Location: India
GMAT 1: 540 Q57 V19
GMAT 1: 540 Q57 V19
Posts: 29
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I don't agree with the explanation. Adding both the statement
1) x=80 Gm
2) Y=2

Which means that the Acid concentration is half in the final Qty of 160Gm. Which is 80Gm.
Now since only water is added with 80Gm this means that the whole of 80Gm of Acid was present in the original solution and thus the Acid concentration was 100% in the original solution. Plz explain.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
740,759
 [1]
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
deepak.brijendra
Bunuel
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


I think this is a high-quality question and I don't agree with the explanation. Adding both the statement
1) x=80 Gm
2) Y=2

Which means that the Acid concentration is half in the final Qty of 160Gm. Which is 80Gm.
Now since only water is added with 80Gm this means that the whole of 80Gm of Acid was present in the original solution and thus the Acid concentration was 100% in the original solution. Plz explain.

I appreciate your efforts in analyzing the problem. However, I believe there's a misunderstanding in your approach. Let's look at it step by step:

When we dilute the 80-gram acid solution with 80 grams of water, the concentration is halved, but this doesn't imply that the original was 100% acid. For instance:

Example 1:
If the original 80 grams had 40 grams of acid (50% concentration) and we added 80 grams of water, the new solution would have 160 grams with 40 grams of acid, which is a 25% concentration. This is halving the concentration.

Example 2:
If the original 80 grams had 20 grams of acid (25% concentration) and we added 80 grams of water, the new solution would have 160 grams with 20 grams of acid, which is a 12.5% concentration. Again, this is halving the concentration.

From the statements, we know the concentration is halved, but we don't have a definitive starting concentration.

Hope it's clear.
User avatar
Hritgoel
Joined: 09 Sep 2023
Last visit: 23 Jan 2024
Posts: 30
Own Kudos:
Given Kudos: 6
Posts: 30
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi all, just one query I have. It is mentioned in the question, that it is weighing 80g initially. So, I thought, (acid+water = 80g), as acid solution contains both acid and water. How, 80g is just the weight of water?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 12 Jul 2025
Posts: 102,636
Own Kudos:
Given Kudos: 98,172
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,636
Kudos: 740,759
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hritgoel
Hi all, just one query I have. It is mentioned in the question, that it is weighing 80g initially. So, I thought, (acid+water = 80g), as acid solution contains both acid and water. How, 80g is just the weight of water?

The original acid solution was 80 grams (acid + water) and then it was diluted by x = 80 grams of water.
User avatar
Dbrunik
Joined: 13 Apr 2024
Last visit: 16 Feb 2025
Posts: 273
Own Kudos:
106
 [3]
Given Kudos: 267
Location: United States (MN)
Concentration: Finance, Technology
GMAT Focus 1: 625 Q84 V82 DI77
GMAT Focus 1: 625 Q84 V82 DI77
Posts: 273
Kudos: 106
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I got this right simply by recognizing a common trap. the Statements are both giving you the same information in different ways. if you are proficient in translating the wording into algebra, you can see this when trying each statement independently. once you see that one is insufficient, and the other provides effectively the same information, you know that combining them will not be sufficient.
User avatar
KuhuPyasi
Joined: 24 Jul 2024
Last visit: 12 Jul 2025
Posts: 2
Given Kudos: 81
Posts: 2
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I like the solution - it’s helpful.
User avatar
Premkungg
Joined: 19 Apr 2022
Last visit: 29 Mar 2025
Posts: 13
Given Kudos: 95
Posts: 13
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I like the solution - it’s helpful.
 1   2   
Moderators:
Math Expert
102636 posts
Founder
41096 posts