If x > y > 0, which of the following must be true: : Problem Solving (PS)

AverageGuy123

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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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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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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

Bunuel

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16 Feb 2015, 06:00

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Bunuel

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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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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05 Dec 2015, 22:10

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Bunuel

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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05 Dec 2015, 22:16

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robu

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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28 Dec 2015, 13:28

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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

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10 Mar 2016, 11:04

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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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17 Jul 2016, 14:52

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Bunuel

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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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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06 Feb 2020, 03:33

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Bunuel

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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11 Aug 2021, 08:06

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Hi Bunuel
If I consider 1>1/2>0
how answer can be E.
Please help

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