Last visit was: 19 Nov 2025, 16:11 It is currently 19 Nov 2025, 16:11
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
655-705 Level|   Exponents|   Inequalities|   Roots|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,368
 [7]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,368
 [7]
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 21 Mar 2022, 05:23
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

52% (02:01) correct 48% (01:31) wrong based on 139 sessions

History

Date
Time
Result
Not Attempted Yet
Is \(x^3 > y\)?

(1) \(\sqrt[3]{x} > y\)

(2) \(x^2 > y\)
ShowHide Answer
Official Answer
E
User avatar
JerryAtDreamScore
User avatar
Dream Score Representative
Joined: 07 Oct 2021
Last visit: 02 Jul 2022
Posts: 380
Own Kudos:
417
Given Kudos: 2
Posts: 380
Kudos: 417
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 23 Mar 2022, 14:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Is \(x^3 > y\)?

(1) \(\sqrt[3]{x} > y\)

(2) \(x^2 > y\)
Show more

Breaking Down the Info:

We are allowed to cube both sides as that does not affect the signs of either side.

Statement 1 Alone:

Cube both sides to get \(x > y^3\). This is not enough to tell whether \(x^3 > y\). Insufficient.

Statement 2 Alone:

We don't know if x or y is positive or negative from this. Insufficient.

Both Statements Combined:

If x is positive, we can clearly have \(x^3 > y\).

If x is negative, then we can have \(x > y^3\), where x and y are both negative. E.g. \((-2) > (-2)^3\), this results in \(x^3 < y\), so combined it is still insufficient.

Answer: E
avatar
ND104
avatar
Joined: 23 Nov 2021
Last visit: 21 Nov 2022
Posts: 9
Own Kudos:
11
Given Kudos: 10
Location: India
Concentration: Entrepreneurship, Technology
Schools: IIMA PGPX'23 Sloan '23 Harvard '24 Ross '23 Stern '23 Stanford '23
GMAT 1: 590 Q44 V28
GPA: 3.49
WE:Web Development (Computer Software)
Schools: IIMA PGPX'23 Sloan '23 Harvard '24 Ross '23 Stern '23 Stanford '23
GMAT 1: 590 Q44 V28
Posts: 9
Kudos: 11
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 18 May 2022, 00:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Is \(x^3 > y\)?

(1) \(\sqrt[3]{x} > y\)

(2) \(x^2 > y\)
Show more

(I) \(\sqrt[3]{x} > y\) ,
case1- if x=1/8 and y=1/9, \(\sqrt[3]{1/8} > (1/9)\), is \((1/8)^3 > (1/9)\) -NO
case2- if x=8 and y=0, \(\sqrt[3]{8} > (0)\), is \((8)^3 > (0)\) -YES
statement 1 alone not sufficient

(II) \(x^2 > y\),
if x=-1/2 and y=0, \((-1/2)^2 > 0\), is \((-1/2)^3 > 0\) -NO
if x=8 and y=0, \((8)^2 > 0\), is \((8)^3 > (0)\) -YES
statement 2 alone not sufficient

Both Togather- \(\sqrt[3]{x} > y\) and \(x^2 > y\)
negative x not possible as sqrt[3]{x} needs to be 0 or positive, though Y negative is possible.
If x is positive and y is negative then cube root or square of positive number x will be positive, and always be grater than negative number y, resulting \(x^3 > y\) YES
but if x=1/8 and y=1/65 then statement1- (1/2) >(1/65) and statement2 - (1/64)>(1/65) but \(x^3 > y\) NO

ANSWER- E

Bunuel, KarishmaB, IanStewart, Archit3110
can anyone tell me, is there any better way to check if both statement togather sufficient or not? algebrically or otherwise?
User avatar
Archit3110
User avatar
Major Poster
Joined: 18 Aug 2017
Last visit: 19 Nov 2025
Posts: 8,422
Own Kudos:
4,982
 [1]
Given Kudos: 243
Status:You learn more from failure than from success.
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1: 545 Q79 V79 DI73
GMAT Focus 2: 645 Q83 V82 DI81
GPA: 4
WE:Marketing (Energy)
GMAT Focus 2: 645 Q83 V82 DI81
Posts: 8,422
Kudos: 4,982
 [1]
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 18 May 2022, 01:15
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
target is \(x^3 > y\)
best way is to use plug in numbers
-1<x<1

which you have done in both statements for fractions
when combining you can use x as an integer x as 8 and y as 1 and x as 1/8 and y as 1/65
still this remains insufficient to say that x^3>y as we get both yes and no to target

ND104

ND104
Bunuel
Is \(x^3 > y\)?

(1) \(\sqrt[3]{x} > y\)

(2) \(x^2 > y\)

(I) \(\sqrt[3]{x} > y\) ,
case1- if x=1/8 and y=1/9, \(\sqrt[3]{1/8} > (1/9)\), is \((1/8)^3 > (1/9)\) -NO
case2- if x=8 and y=0, \(\sqrt[3]{8} > (0)\), is \((8)^3 > (0)\) -YES
statement 1 alone not sufficient

(II) \(x^2 > y\),
if x=-1/2 and y=0, \((-1/2)^2 > 0\), is \((-1/2)^3 > 0\) -NO
if x=8 and y=0, \((8)^2 > 0\), is \((8)^3 > (0)\) -YES
statement 2 alone not sufficient

Both Togather- \(\sqrt[3]{x} > y\) and \(x^2 > y\)
negative x not possible as sqrt[3]{x} needs to be 0 or positive, though Y negative is possible.
If x is positive and y is negative then cube root or square of positive number x will be positive, and always be grater than negative number y, resulting \(x^3 > y\) YES
but if x=1/8 and y=1/65 then statement1- (1/2) >(1/65) and statement2 - (1/64)>(1/65) but \(x^3 > y\) NO

ANSWER- E

Bunuel, KarishmaB, IanStewart, Archit3110
can anyone tell me, is there any better way to check if both statement togather sufficient or not? algebrically or otherwise?
Show more
User avatar
IanStewart
User avatar
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 18 Nov 2025
Posts: 4,145
Own Kudos:
10,989
 [2]
Given Kudos: 99
Expert
Expert reply
Posts: 4,145
Kudos: 10,989
 [2]
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 18 May 2022, 03:54
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ND104

Both Togather- \(\sqrt[3]{x} > y\) and \(x^2 > y\)
negative x not possible as sqrt[3]{x} needs to be 0 or positive, though Y negative is possible.

can anyone tell me, is there any better way to check if both statement togather sufficient or not? algebrically or otherwise?
Show more

When information is not sufficient, there may not be any algebraic or conceptual way to prove that, so you should quickly fall back on testing scenarios when you don't see any alternative. I think you may have made that more challenging than it needs to be, though -- I wasn't able to follow how you reached the conclusion I highlighted above, that x cannot be negative, but that's not true here (we can't take square roots or any other even root of a negative number, but we can take cube roots or any other odd root of a negative number).

So when I use both Statements here, I notice that Statement 2 will be true no matter what if y is negative, because x^2 is positive (or zero). So if I make y negative, I don't need to even think about Statement 2, because it will be true. I only need to think about Statement 1. And because we want numbers that we can easily cube root, and that change when we cube root them (we want to avoid -1, 0 or 1 here), I'll make y = -8. Then we just want to quickly try to get a 'yes' and a 'no' answer to the question "Is x^3 > y?"

• If we make y = -8 and x = 1,000,000, then both Statements are true, and the answer to the question is 'yes'

• if we make y = -8 and x = -8, then both Statements are true (because the cube root of -8 is -2, which is greater than y here), and the answer to the question is 'no'

So the answer is E. I'd only test fractions as a last resort, but in certain inequality questions that resemble this one (where you are comparing powers of unknowns), testing fractions is genuinely necessary.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,589
Own Kudos:
1,079
Posts: 38,589
Kudos: 1,079
Is x^3 > y ? (1) x^(1/3) > y (2) x^2 > y 25 Jul 2023, 03:03
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.
Moderators:
Bunuel
Math Expert
105390 posts
kingbucky
496 posts