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Manager Joined: 12 Jan 2016
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Given x,y,z are distinct positive real numbers [#permalink] 00:00

Question Stats: 53% (01:01) correct 46% (02:07) wrong based on 13 sessions
Given x,y,z are distinct positive real numbers

 Quantity A Quantity B (x²(y+z)+y²(x+z)+z²(x+y))/(xyz) 6

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

Last edited by Sonalika42 on 25 Sep 2016, 03:23, edited 2 times in total.
Intern Joined: 02 Jul 2016
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Re: Given x,y,z are distinct positive real numbers [#permalink]
D is the answer

Posted from my mobile device GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Re: Given x,y,z are distinct positive real numbers [#permalink]
Expert's post
The best way to approach this problem is by substitution.

Quantity A is $$\frac{(x^2(y+z)+y^2(x+z)+z^2(x+y))}{(xyz)}$$

or $$x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{x}+\frac{1}{z})+z(\frac{1}{x}+\frac{1}{y})$$

or $$\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}$$

rearrange

$$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}$$.

now Let $$\frac{x}{y}$$ = n. Then the first two terms become $$n + \frac{1}{n}$$. The minimum value of $$n + \frac{1}{n}$$ is 2 when n =1. Or and n can be one if x=y.

Hence a case x=y=z= 5( or any other real positive number) is not possible. Hence Quantity A is always greater

for x=1 y=1 z=2

$$\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{2}{1}+\frac{2}{1}+\frac{1}{2}$$

$$1+1+0.5+2+2+0.5$$ =$$7$$. Now Quantity A is 7.

Hence A is the correct option.
_________________

Sandy
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Re: Given x,y,z are distinct positive real numbers [#permalink]
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x,y and z are said to be distinct positive real number.So you can not make x,y or z values equal i think,sandy.I guess the answer to be A.
GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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Kudos [?]: 1865 , given: 397

Re: Given x,y,z are distinct positive real numbers [#permalink]
Expert's post
siam1993 wrote:
x,y and z are said to be distinct positive real number.So you can not make x,y or z values equal i think,sandy.I guess the answer to be A.

Hey,

Yeah you are correct. Missed the distinct part. Changed the answer.

Regards,
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Sandy
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Manager Joined: 27 Sep 2017
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Kudos [?]: 30 , given: 4

Re: Given x,y,z are distinct positive real numbers [#permalink]
Sonalika42 wrote:
Given x,y,z are distinct positive real numbers

 Quantity A Quantity B (x²(y+z)+y²(x+z)+z²(x+y))/(xyz) 6

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

My problem would be more like, since x, y, z are positive numbers which does NOT specify integers. Then it would be that x=0.2 y= 1 z= 2, though it seems that ans should be A. but can we guarantee there is no excpetion?
Intern Joined: 15 Mar 2018
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Re: Given x,y,z are distinct positive real numbers [#permalink]
Incorrect question, what if X Y Z are -1 0 1 respectively, the A will be undefined?

Zero is a real number .
Manager Joined: 27 Sep 2017
Posts: 112
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Kudos [?]: 30 , given: 4

Re: Given x,y,z are distinct positive real numbers [#permalink]
HEcom wrote:
Incorrect question, what if X Y Z are -1 0 1 respectively, the A will be undefined?

Zero is a real number .

pay attention to distinct "positive" real numbers

0 is not - or +
Intern Joined: 23 Nov 2017
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Kudos [?]: 46 , given: 0

Re: Given x,y,z are distinct positive real numbers [#permalink]
HEcom wrote:
Incorrect question, what if X Y Z are -1 0 1 respectively, the A will be undefined?

Zero is a real number .

Just for clarification. Zero is a real number but it is neither positive nor negative. The very definition of a positive number is a number greater than zero and the definition of a negative number is a number less than zero. Re: Given x,y,z are distinct positive real numbers   [#permalink] 29 Mar 2018, 13:04
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