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Given x,y,z are distinct positive real numbers [#permalink]
23 Sep 2016, 09:10
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.
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]
23 Sep 2016, 20:40
D is the answer Posted from my mobile device



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



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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 .



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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 +



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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
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29 Mar 2018, 13:04





