Author 
Message 
TAGS:


Manager
Joined: 12 Jan 2016
Posts: 144
Followers: 1
Kudos [?]:
73
[0], given: 17

Given x,y,z are distinct positive real numbers [#permalink]
23 Sep 2016, 09:10
Question Stats:
58% (01:01) correct
41% (02:02) wrong based on 12 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.
Last edited by Sonalika42 on 25 Sep 2016, 03:23, edited 2 times in total.




Intern
Joined: 02 Jul 2016
Posts: 25
Followers: 0
Kudos [?]:
15
[0], given: 2

Re: Given x,y,z are distinct positive real numbers [#permalink]
23 Sep 2016, 20:40
D is the answer Posted from my mobile device



GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4856
WE: Business Development (Energy and Utilities)
Followers: 102
Kudos [?]:
1731
[0], given: 397

Re: Given x,y,z are distinct positive real numbers [#permalink]
24 Sep 2016, 06:22
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 If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Intern
Joined: 21 Jun 2016
Posts: 1
Followers: 0
Kudos [?]:
1
[1]
, given: 0

Re: Given x,y,z are distinct positive real numbers [#permalink]
24 Sep 2016, 22:09
1
This post received KUDOS
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: 4856
WE: Business Development (Energy and Utilities)
Followers: 102
Kudos [?]:
1731
[0], given: 397

Re: Given x,y,z are distinct positive real numbers [#permalink]
25 Sep 2016, 02:20
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,
_________________
Sandy If you found this post useful, please let me know by pressing the Kudos Button
Try our free Online GRE Test



Manager
Joined: 27 Sep 2017
Posts: 112
Followers: 1
Kudos [?]:
29
[0], given: 4

Re: Given x,y,z are distinct positive real numbers [#permalink]
25 Mar 2018, 23:44
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
Posts: 32
Followers: 0
Kudos [?]:
7
[0], given: 1

Re: Given x,y,z are distinct positive real numbers [#permalink]
29 Mar 2018, 07:47
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
Followers: 1
Kudos [?]:
29
[0], given: 4

Re: Given x,y,z are distinct positive real numbers [#permalink]
29 Mar 2018, 07:53
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
Posts: 45
Followers: 0
Kudos [?]:
45
[0], given: 0

Re: Given x,y,z are distinct positive real numbers [#permalink]
29 Mar 2018, 13:04
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





