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

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Bunuel
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If x>y>0, which of the following must be true: [#permalink] New post   11 Feb 2015, 06:40
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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


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E
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Bunuel
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Re: If x>y>0, which of the following must be true: [#permalink] New post   06 Feb 2020, 03:38
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Anurag06 wrote:
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 ?

Please help!

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 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


I. x^2 > y^2 --> take the square root: |x| > |y|. Since given that both x and y are positive, then this translates to x > y. TRUE.

II. x^3 > y^3 --> take the cube root: x > y. TRUE.

III. |x| > y. Since x > 0, then x > y. TRUE.

Answer: E.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   11 Feb 2015, 07:21
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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


I would go wih E here

I. the issue is to check whether proper fractions can disprove this statement. Pick y=0,1 and pick x=0,101 \(x^2\) is going to be greater than \(y^2\) try with x= √0,6 and y= √0,5 \(x^2\) is greater than \(y^2\).

II. We don't have to worry about even/odd powers since our values are greater than zero. Thus our considerations on I. hold on II.

III. We don't have to worry about the modulus. |x| is x when x>0 this is our case. x>y thus |x| is greater than y.

Answer E.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   12 Feb 2015, 01:19
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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
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Re: If x>y>0, which of the following must be true: [#permalink] New post   12 Feb 2015, 22:27
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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
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Re: If x>y>0, which of the following must be true: [#permalink] New post   12 Feb 2015, 22:54
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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
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Re: If x>y>0, which of the following must be true: [#permalink] New post   16 Feb 2015, 06:00
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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.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   05 Dec 2015, 22:10
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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.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   05 Dec 2015, 22:16
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.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   28 Dec 2015, 13:28
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.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   10 Mar 2016, 11:04
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
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Re: If x>y>0, which of the following must be true: [#permalink] New post   17 Jul 2016, 14:52
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
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Re: If x>y>0, which of the following must be true: [#permalink] New post   06 Feb 2020, 03:33
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 ?

Please help!

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.
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Re: If x>y>0, which of the following must be true: [#permalink] New post   11 Aug 2021, 08:06
Hi Bunuel
If I consider 1>1/2>0
how answer can be E.
Please help
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Re: If x>y>0, which of the following must be true: [#permalink]
11 Aug 2021, 08:06
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