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Given x,y,z are distinct positive real numbers

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Given x,y,z are distinct positive real numbers [#permalink] New post 23 Sep 2016, 09:10
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63% (01:01) correct 36% (02:25) wrong based on 11 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.
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 23 Sep 2016, 20:40
D is the answer

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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 24 Sep 2016, 06:22
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.
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 24 Sep 2016, 22:09
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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.
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 25 Sep 2016, 02:20
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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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 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?
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 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 .
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 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 +
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Re: Given x,y,z are distinct positive real numbers [#permalink] New post 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
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