Last visit was: 23 May 2024, 04:49 It is currently 23 May 2024, 04:49
Toolkit
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

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.

# If x > y > 0, which of the following must be true:

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626019 [18]
Given Kudos: 81940
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626019 [0]
Given Kudos: 81940
General Discussion
Manager
Joined: 04 Oct 2013
Posts: 141
Own Kudos [?]: 592 [2]
Given Kudos: 29
GMAT 1: 590 Q40 V30
GMAT 2: 730 Q49 V40
WE:Project Management (Entertainment and Sports)
Manager
Joined: 24 Jun 2014
Posts: 53
Own Kudos [?]: 49 [2]
Given Kudos: 105
Concentration: Social Entrepreneurship, Nonprofit
Re: If x > y > 0, which of the following must be true: [#permalink]
2
Kudos
Assuming x and y to be integers
X=3,Y=2
X^2=3^2=9
Y^2=2^2=4
So I is true

Now lets look at option III because if option 3 is true E is the answer or else A
As per question X>Y>0 Hence |X| >Y

So OA=E
Manager
Joined: 14 Jul 2014
Posts: 67
Own Kudos [?]: 95 [1]
Given Kudos: 49
Re: If x > y > 0, which of the following must be true: [#permalink]
1
Kudos
Ans - E

Since given both X & Y are +ve,

Stmnt 1 - Quickly plugged in decimals ( x = 0.2, y = 0.1) and (x = 1.2 and y = 1.1) and integers

Stmnt 2 - No need for any computation
Infact , irrespective of the sign , whenever given that X > Y, any odd power will always hold true
i.e. x^3 > y^3, x^5 > y^5, x^7 > y^7,
x^1/3 > y^1/3 (cube root)...etc

Stmnt 3 - No need for any computation - Modulas does not matter as the constraint in the Q is that both X & Y are +ve
Intern
Joined: 22 Aug 2014
Posts: 30
Own Kudos [?]: 37 [1]
Given Kudos: 6
Re: If x > y > 0, which of the following must be true: [#permalink]
1
Kudos
When two positive integers x & y are compared by the given condition x>y, then the only case when y>x will be when the power is -ve. So it satisfies all the options i, ii & iii.

Therefore option e
Math Expert
Joined: 02 Sep 2009
Posts: 93417
Own Kudos [?]: 626019 [3]
Given Kudos: 81940
Re: If x > y > 0, which of the following must be true: [#permalink]
2
Kudos
1
Bookmarks
Bunuel wrote:
If x>y>0, which of the following must be true:

I. x^2>y^2
II. x^3 > y^3
III. |x|>y

A. I only
B. II only
C. III only
D. II and III
E. I, II and III

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Solution: E.

The "trick" here is that you are apt to expect a trick. Clearly all three statements hold for integers (if x = 2 and y = 1, then x^2 = 4 and y^2 = 1, and x^3 = 8 and y^3 = 1). But you may expect for fractions to be different - if you square, say, 1/2, then it gets smaller (1/4). But, still, the larger fraction will remain larger when squared or cubed. Take for example 1/2 and 1/3 each squared. 1/2 --> 1/4, and 1/3 --> 1/9. The smaller fraction becomes even smaller. Statement III is true simply by definition - the absolute value of a positive number is just that number.
Manager
Joined: 03 Dec 2014
Posts: 74
Own Kudos [?]: 157 [1]
Given Kudos: 391
Location: India
Concentration: General Management, Leadership
GMAT 1: 620 Q48 V27
GPA: 1.9
WE:Engineering (Energy and Utilities)
Re: If x > y > 0, which of the following must be true: [#permalink]
1
Kudos
Bunuel wrote:
Bunuel wrote:
If x>y>0, which of the following must be true:

I. x^2>y^2
II. x^3 > y^3
III. |x|>y

A. I only
B. II only
C. III only
D. II and III
E. I, II and III

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Solution: E.

The "trick" here is that you are apt to expect a trick. Clearly all three statements hold for integers (if x = 2 and y = 1, then x^2 = 4 and y^2 = 1, and x^3 = 8 and y^3 = 1). But you may expect for fractions to be different - if you square, say, 1/2, then it gets smaller (1/4). But, still, the larger fraction will remain larger when squared or cubed. Take for example 1/2 and 1/3 each squared. 1/2 --> 1/4, and 1/3 --> 1/9. The smaller fraction becomes even smaller. Statement III is true simply by definition - the absolute value of a positive number is just that number.

if I consider X= 1/3 and y=1/2 , then, X square is not greater that y Square. and same is true for cube power. In my view only o'ption three holds must be true. please clarify. thanks in advance.
SVP
Joined: 20 Mar 2014
Posts: 2362
Own Kudos [?]: 3630 [0]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Re: If x > y > 0, which of the following must be true: [#permalink]
robu wrote:
if I consider X= 1/3 and y=1/2 , then, X square is not greater that y Square. and same is true for cube power. In my view only o'ption three holds must be true. please clarify. thanks in advance.

You can not consider x=1/3 and y=1/2 as this will make y>x but per the given condition x>y

Hope this helps.
Manager
Joined: 17 Jun 2015
Posts: 166
Own Kudos [?]: 200 [0]
Given Kudos: 176
GMAT 1: 540 Q39 V26
GMAT 2: 680 Q50 V31
Re: If x > y > 0, which of the following must be true: [#permalink]
Plug values

For any two numbers x and y such that x>y>0 the squares and cubes are also in the same inequality. However, x^2 is smaller than x in case of 0<x<1; so is the cube.

Also |x| = x for x>0 or -x for x <0. Given that x>0, this too is positive and so is the result.

Hence all options are true.
E
Alum
Joined: 12 Aug 2015
Posts: 2279
Own Kudos [?]: 3166 [0]
Given Kudos: 893
GRE 1: Q169 V154
Re: If x > y > 0, which of the following must be true: [#permalink]
We just need to remember here that if x and y are both greater than zero => and x>y => x^n>y^n is true
Retired Moderator
Joined: 02 Apr 2014
Status:I Declare War!!!
Posts: 218
Own Kudos [?]: 122 [0]
Given Kudos: 546
Location: United States
Concentration: Finance, Economics
GMAT Date: 03-18-2015
WE:Asset Management (Investment Banking)
Re: If x > y > 0, which of the following must be true: [#permalink]
Bunuel wrote:
Bunuel wrote:
If x>y>0, which of the following must be true:

I. x^2>y^2
II. x^3 > y^3
III. |x|>y

A. I only
B. II only
C. III only
D. II and III
E. I, II and III

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION

Solution: E.

The "trick" here is that you are apt to expect a trick. Clearly all three statements hold for integers (if x = 2 and y = 1, then x^2 = 4 and y^2 = 1, and x^3 = 8 and y^3 = 1). But you may expect for fractions to be different - if you square, say, 1/2, then it gets smaller (1/4). But, still, the larger fraction will remain larger when squared or cubed. Take for example 1/2 and 1/3 each squared. 1/2 --> 1/4, and 1/3 --> 1/9. The smaller fraction becomes even smaller. Statement III is true simply by definition - the absolute value of a positive number is just that number.

plz correct me if i am wrong, it means this is good for integers as well as for fractions too.. (provided positive)
thanks
Intern
Joined: 10 Dec 2019
Posts: 42
Own Kudos [?]: 5 [0]
Given Kudos: 15
Re: If x > y > 0, which of the following must be true: [#permalink]
Bunuel wrote:
Bunuel wrote:
If x>y>0, which of the following must be true:

I. x^2>y^2
II. x^3 > y^3
III. |x|>y

A. I only
B. II only
C. III only
D. II and III
E. I, II and III

Kudos for a correct solution.

Hi Bunuel

But no conditions of integers is mentioned here so if in terms of decimals the squares and cubes of values of x would rather go lower than y.

Also regarding distance |x| > = 0, but there is no relation between magnitude of y and distance of x mentioned either. So how are we assuming that |x|> y ?

VERITAS PREP OFFICIAL SOLUTION

Solution: E.

The "trick" here is that you are apt to expect a trick. Clearly all three statements hold for integers (if x = 2 and y = 1, then x^2 = 4 and y^2 = 1, and x^3 = 8 and y^3 = 1). But you may expect for fractions to be different - if you square, say, 1/2, then it gets smaller (1/4). But, still, the larger fraction will remain larger when squared or cubed. Take for example 1/2 and 1/3 each squared. 1/2 --> 1/4, and 1/3 --> 1/9. The smaller fraction becomes even smaller. Statement III is true simply by definition - the absolute value of a positive number is just that number.
Intern
Joined: 27 Jan 2021
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 95
Re: If x > y > 0, which of the following must be true: [#permalink]
Hi Bunuel
If I consider 1>1/2>0
how answer can be E.
Non-Human User
Joined: 09 Sep 2013
Posts: 33153
Own Kudos [?]: 829 [0]
Given Kudos: 0
Re: If x > y > 0, which of the following must be true: [#permalink]
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.
Re: If x > y > 0, which of the following must be true: [#permalink]
Moderator:
Math Expert
93417 posts