Last visit was: 12 Jul 2025, 10:08 It is currently 12 Jul 2025, 10:08
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 July 2025
Posts: 102,636
Own Kudos:
740,709
 [5]
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: 740,709
 [5]
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
User avatar
Nidzo
Joined: 26 Nov 2019
Last visit: 31 May 2025
Posts: 960
Own Kudos:
Given Kudos: 59
Location: South Africa
Posts: 960
Kudos: 1,322
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Crytiocanalyst
Joined: 16 Jun 2021
Last visit: 27 May 2023
Posts: 951
Own Kudos:
Given Kudos: 309
Posts: 951
Kudos: 202
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
starredastro
Joined: 23 Jul 2021
Last visit: 03 Oct 2022
Posts: 49
Own Kudos:
Given Kudos: 34
Location: India
Schools: ISB '23 (S)
GPA: 3.88
Schools: ISB '23 (S)
Posts: 49
Kudos: 14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Nidzo
1) Looking at the top line alone: 9ab = 18 tells us that either one of a or b is 2 and the other is 1.
Looking at the bottom line: 3+a = 5 tells us that a = 2, and therefore b = 1

SUFFICIENT

2) Only gives us the value of b, and we cannot deduce a from this.

INSUFFICIENT

Answer A

Your answer is correct, but your approach is incorrect.
Why are you assuming that a and b are integers?
If you simplify the equation using the (a^2-b^2) = (a-b)(a+b) formula in the denominator, after applying the LCM you will get the exact expression mentioned in statement (1).
Hence, option A is sufficient alone to answer the target question!
User avatar
Namangupta1997
Joined: 23 Oct 2020
Last visit: 05 Apr 2025
Posts: 146
Own Kudos:
Given Kudos: 63
GMAT 1: 710 Q49 V38
GMAT 1: 710 Q49 V38
Posts: 146
Kudos: 6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi chetan2u Bunuel

Why are we not considering the fact that "a" cannot be equal to +-3 as (9-a^2) is in denominator ? If we use b=1, it ensures that the a =2, and hence removes the possibility of "a" being +-3. Why isn't the correct answer C ? Using the equation in statement 1, we just get x = 18(3-a)/5(3-a) which would be equal to 18/5 IF it was mentioned a is not equal to 3.

What am I missing ? Please clarify.
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 12 Jul 2025
Posts: 11,295
Own Kudos:
Given Kudos: 333
Status:Math and DI Expert
Products:
Expert
Expert reply
Posts: 11,295
Kudos: 41,698
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If \(x = \frac{9b - 3ab}{\frac{3}{a} - \frac{a}{3}}\), what is x?


(1) \(\frac{9ab}{3+a} = \frac{18}{5}\)

(2) \(b = 1\)


Statement I should tell you that the initial equation requires to be simplified to check statement I.

\(x = \frac{9b - 3ab}{\frac{3}{a} - \frac{a}{3}}\)

\(x = \frac{9b - 3ab}{\frac{9-a^2}{3a}}\)

\(x=\frac{3a(9b-3ab)}{9-a^2}\)

\(x=\frac{3a*3b(3-a)}{(3-a)(3+a)}=\frac{9ab}{3+a}\)

Statement I gives this value as 18/5.
Sufficient

Statement II gives b=1
So, \(x=\frac{9*a*1}{3+a}=\frac{9a}{3+a}\)
x depends on a.
If a is 1, x is 9/4
If a is 2, x is 18/5
If a is 3, x is 27/6 or 9/2 and so on
Insufficient

A
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,373
Own Kudos:
Posts: 37,373
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
102636 posts