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If x>y>0, which of the following must be true: [#permalink]
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11 Feb 2015, 06:40
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Re: If x>y>0, which of the following must be true: [#permalink]
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11 Feb 2015, 07:21
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]
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12 Feb 2015, 01:19
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]
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12 Feb 2015, 22:27
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]
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12 Feb 2015, 22:54
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]
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16 Feb 2015, 06:00
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 SOLUTIONSolution: 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]
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05 Dec 2015, 22:10
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 SOLUTIONSolution: 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]
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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]
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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. E
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Re: If x>y>0, which of the following must be true: [#permalink]
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Re: If x>y>0, which of the following must be true: [#permalink]
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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 SOLUTIONSolution: 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]
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