Last visit was: 20 Nov 2025, 00:36 It is currently 20 Nov 2025, 00:36
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
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,070
Own Kudos:
19,393
 [7]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,070
Kudos: 19,393
 [7]
1
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,070
Own Kudos:
19,393
 [2]
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,070
Kudos: 19,393
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
avatar
Shahaditya
Joined: 20 Nov 2018
Last visit: 04 Apr 2019
Posts: 1
Given Kudos: 3
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
tulullus
Joined: 23 May 2019
Last visit: 15 Jun 2019
Posts: 3
Posts: 3
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a poor-quality question. I do not understand how the answer to the question is B.
I know that given that the powers are even, the outcome will be positive. But, I do not know how to go from there.
avatar
Prospect2020
Joined: 23 Feb 2019
Last visit: 02 Mar 2020
Posts: 7
Own Kudos:
4
 [1]
Given Kudos: 63
Posts: 7
Kudos: 4
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
tulullus
I think this is a poor-quality question. I do not understand how the answer to the question is B.
I know that given that the powers are even, the outcome will be positive. But, I do not know how to go from there.

I will try to explain:

Given the stem, we know that either x or y has to be less than 0, but not BOTH, as if both are less than 0, then the expression would result in >0, which violates what we are given.

So we are trying to solve whether xyz>0. We know given the stem, that x and z have opposite signs, therefore, we are needing to determine if y is greater to or less than 0. We don't know the sign of y, due to its having a positive exponent in the stem (could be positive or negative)

Statement 1: x<0 --> this would imply that z>0, but gives no info on y and is therefore INSUFFICIENT

Statement 2: y<0 --> we already know that x and z have opposite signs and now know that y is negative. Therefore we have (pos*neg*neg) which is >0 OR (neg*neg*pos) which is also > 0 SUFFICIENT
User avatar
thelonghalloween
Joined: 14 Jun 2020
Last visit: 22 Oct 2021
Posts: 62
Own Kudos:
Given Kudos: 119
Location: India
GMAT 1: 680 Q49 V33 (Online)
GRE 1: Q170 V151
WE:Consulting (Consulting)
Products:
GMAT 1: 680 Q49 V33 (Online)
GRE 1: Q170 V151
Posts: 62
Kudos: 81
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I think this is a high-quality question and I agree with explanation.
User avatar
Showmeyaa
Joined: 24 Jun 2019
Last visit: 08 Sep 2023
Posts: 428
Own Kudos:
Given Kudos: 117
Location: India
Concentration: Marketing, Strategy
Products:
Posts: 428
Kudos: 517
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MathRevolution
If \(w^2 x^3 y^4 z^5 < 0\), is \(xyz > 0\)?

1) \(x < 0\)

2) \(y < 0\)

\(w^2 x^3 y^4 z^5 < 0\)
From the above, it is clear that:
xz < 0, which means that x and z are of opposite signs

1)x < 0, so z > 0(As x and z are of opposite signs)
Case1: y>0
xyz < 0
Case2:
y<0
xyz > 0 - Insufficient!
2)y < 0
x and z are of opposite signs
So, xyz will always be > 0 - Sufficient!

IMO, (B)!
User avatar
Anshika000
Joined: 07 Sep 2021
Last visit: 16 Jan 2025
Posts: 5
Given Kudos: 22
Location: Greece
Posts: 5
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
where can i get the theory of this concept.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 20 Nov 2025
Posts: 105,408
Own Kudos:
Given Kudos: 99,987
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,408
Kudos: 778,421
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Moderator:
Math Expert
105408 posts